Física II – Sears, Zemansky, Young & Freedman. PHYSICS ACT. http//physicsact.wordpress.com Capítulo 12 Note: to obtain the numerical results given in this chapter, the following numerical values of certain physical quantities have been used; G = 6.673 × 10−11 N ⋅ m 2 kg 2 , g = 9.80 m s 2 and mE = 5.97 × 1024 kg. Use of other tabulated values for these quantities may result in an answer that differs in the third significant figure. The ratio will be the product of the ratio of the mass of the sun to the mass of the earth and the square of the ratio of the earth+moon radius to the sun+moon radius. Using the earth+sun radius as an average for the sun+moon radius, the ratio of the forces is  3.84 × 10 8 m    11  1.50 × 10 m  2  1.99 × 10 30 kg   = 2.18.  24  5.97 × 10 kg  Use of Eq. (12.1) gives m1 m2 (5.97 × 10 24 kg)(2150 kg) −11 2 2 = ( 6 . 673 × 10 N ⋅ m kg ) = 1.67 × 10 4 N. 2 5 6 2 r (7.8 × 10 m + 6.38 × 10 m) The ratio of this force to the satellite’s weight at the surface of the earth is (1.67 × 10 4 N) = 0.79 = 79%. (2150 kg)(9.80 m s 2 ) (This numerical result requires keeping one extra significant figure in the intermediate calculation.) The ratio, which is independent of the satellite mass, can be obtained directly as Fg = G 2 GmE m r 2 GmE  RE  = 2 =  , mg r g  r  yielding the same result. G (nm1 )(nm2 ) mm = G 1 2 2 = F12 . 2 (nr12 ) r12 The separation of the centers of the spheres is 2R, so the magnitude of the gravitational attraction is GM 2 (2 R ) 2 = GM 2 4 R 2 . a) Denoting the earth+sun separation as R and the distance from the earth as x, the distance for which the forces balance is obtained from GM S m GM E m = , x2 ( R − x) 2 which is solved for R = 2.59 × 108 m. x= MS 1+ ME b) The ship could not be at equilibrium for long, in that the point where the forces balance is moving in a circle, and to move in that circle requires some force. The spaceship could continue toward the sun with a good navigator on board. a) Taking force components to be positive to the right, use of Eq. (12.1) twice gives  (5.00 kg ) (10.0 kg ) , + Fg = (6.673 × 10−11 Ν ⋅ m 2 kg 2 ) (0.100 kg ) − 2 2  (0.400 m ) (0.600 m )  = −2.32 × 10−11 Ν with the minus sign indicating a net force to the left. b) No, the force found in part (a) is the net force due to the other two spheres. (6.673 × 10 −11 Ν. m 2 kg 2 ) (70kg ) (7.35 × 1022 kg ) = 2.4 × 10−3 Ν. (3.78 × 10 m ) 8 2 (333, 000) = 6.03 × 10− 4 (23, 500)2 Denote the earth+sun separation as r1 and the earth+moon separation as r2 . mS m  + 2E  = 6.30 × 1020 Ν, 2 r2   (r1 + r2 ) toward the sun. b)The earth+moon distance is sufficiently small compared to the earth+ sun distance (r2 << r2) that the vector from the earth to the moon can be taken to be perpendicular to the vector from the sun to the moon. The components of the gravitational force are then a)  (GmM )  GmM mS GmM mE = 4.34 × 1020 Ν, = 1.99 × 1020 Ν, 2 2 r1 r2 and so the force has magnitude 4.77 × 10 20 Ν and is directed 24.6° from the direction toward the sun. c)  mS m  − 2E  = 2.37 × 1020 Ν, 2 r2   (r1 − r2 ) (GmM )  toward the sun. FonA = 2 FB cos 45° + FD =2 = GmA mB cos 45° GmA mD + 2 2 rAB rAD 2(6.67 × 10−11 Nm 2 kg 2 )(800 kg) 2 cos 45° (0.10 m)2 + (6.67 × 10−11 Nm 2 kg 2 ) (800 kg) 2 (0.10 m)2 = 8.2 × 10−3 N, toward the center of the square m1 = m2 = m3 = 500 kg r12 = 0.10 m; r23 = 0.40 m F1 = G m1m2 = 1.668 × 10 −3 N r122 F3 = G m2 m3 = 1.043 × 10 −4 N 2 r23 F = F1 − F3 = 1.6 × 10 −3 N, to the left The direction of the force will be toward the larger mass, and the magnitude will be Gm2 m Gm1m 4Gm(m2 − m1 ) − = . (d 2) 2 (d 2) 2 d2 For convenience of calculation, recognize that the mass of the small sphere will cancel. The acceleration is then 2G (0.260 kg) 6.0 × = 2.1 × 10−9 m s 2 , −2 2 (10.0 × 10 m) 10.0 directed down. Equation (12.4) gives g= (6.763 × 10 )( (1.15 × 10 m ) −11 N ⋅ m 2 kg 2 1.5 × 1022 kg 6 2 ) = 0.757 m s 2 . To decrease the acceleration due to gravity by one+tenth, the distance from the earth must be increased by a factor of 10 , and so the distance above the surface of the earth is ( 10 − 1) R E = 1.38 × 107 m. a) Using g E = 9.80 m s 2 , Eq (12.4 ) gives m  R  Gm g v = 2 v = G v m E  E  Rv  mE   Rv  2 2  1   2   RE  Gm  m   R   1  = 2E  v   E  = g E (.815)  RE  mE   Rv   .949  2 = (9.80 m s 2 )(.905) = 8.87 m s 2 , where the subscripts v refer to the quantities pertinent to Venus. b) (8.87 m s 2 ) (5.00 kg) = 44.3N. a) See Exercise 12.16;  (8)2   = 0.369 m s 2 . g Titania = (9.80 m s 2 )    1700  b) ρTE = ρ mT mE r3 . rE3 , or rearranging and solving for density, ρT = ρE . T 512 ) = 1656 kg m3 , or about 0.39 ρE. (5500 kg m3 ) (1700 M= gR 2 G = 2.44 × 1021 kg and ρ = M ( 4 π 3 )R 3 (1 1700) m E mE = 1.30 × 103 Kg m 3 . r3 . (1 8rE E )3 = mmE r2 3 r = 600 × 10 m + RE so F = 610 N F =G w = mg = 735 N. At the surface of the earth, The gravity force is not zero in orbit. The satellite and the astronaut have the same acceleration so the astronaut’s apparent weight is zero. Get g on the neutron star GmM ns R2 GM ns g ns = R2 mg ns = Your weight would be wns = mg ns = mGM ns R2  675N =  2  9.8 m s  (6.67 × 1011 Nm 2 kg 2 )(1.99 × 10 30 kg)  (10 4 m) 2  = 9.1 × 1013 N From eq. (12.1), G = Fr 2 m1m 2 , and from Eq. (12.4), g = GmE RE2 ; combining and solving for RE , mE = gm1m2 RE2 = 5.98 × 1024 kg. 2 Fr a) From Example 12.4 the mass of the lander is 4000 kg. Assuming Phobos to be spherical, its mass in terms of its density ρ and radius R is (4π 3) ρR3 , and so the gravitational force is G (4π 3)(4000 kg) ρR 3 = G (4π 3)(4000 kg )(2000 kg m 3 )(12 × 10 3 m) = 27 N. 2 R b) The force calculated in part (a) is much less than the force exerted by Mars in Example 12.4. 2 GM R = 2(6.673 × 10−11 N ⋅ m 2 kg 2 ) (3.6 × 1012 kg) (700 m) = 0.83 m s. One could certainly walk that fast. a) F = GmE m r 2 and U = GmE m r , so the altitude above the surface of the earth is U F − RE = 9.36 × 105 m. b) Either of Eq. (12.1) or Eq. (12.9) can be used with the result of part (a ) to find m, or noting that U 2 = G 2 M E m 2 r 2 , m = U 2 FGM E = 2.55 × 103 kg. 2 The escape speed, from the results of Example 12.5, is 2GM R. a) 2(6.673 × 10 −11 N ⋅ m 2 kg 2 ) (6.42 × 10 23 kg ) (3.40 × 10 6 m) = 5.02 × 103 m s . b) 2(6.673 × 10 −11 N ⋅ m 2 kg 2 ) (1.90 × 10 27 kg) (6.91 × 107 m) = 6.06 × 10 4 m s. c) Both the kinetic energy and the gravitational potential energy are proportional to the mass. a) The kinetic energy is K = 12 mv 2 , or K = 12 (629 kg )(3.33 × 103 m s) 2 , or KE = 3.49 × 109 J. b) U = − GMm (6.673 × 10−11 N ⋅ m 2 kg 2 )(5.97 × 1024 kg)(629 kg) = , r 2.87 × 109 m or U = −8.73 × 10 7 J. a) Eliminating the orbit radius r between Equations (12.12) and (12.14) gives ( )( 2πGmE 2π 6.673 × 10−11 N ⋅ m 2 kg 2 5.97 × 1024 kg T= = v3 (6200 m s )3 ) = 1.05 × 104 s = 175 min. 2πv = 3.71 m s 2 . T b) Substitution into Eq. (12.14) gives T = 6.96 × 10 3 s, or 116 minutes. Using Eq. (12.12), (6.673 × 10 N ⋅ m kg )(5.97 × 10 (6.38 × 10 m + 7.80 × 10 m) −11 v= 2 6 2 5 24 kg ) = 7.46 × 10 3 m s. Applying Newton’s second law to the Earth ∑ F = ma; GmE ms v2 m = E r r2 2 rv 2πr ms = and v = TEarth G ms = = r ( ) 2 πr 2 TE G = 4π 2 r 3 GTE2 4π 2 (1.50 × 1011 m)3 4 (6.67 × 10 −11 Nm 2 kg 2 ) [(365.3d )( 8.64×d10 s )] 2 = 2.01 × 1030 kg ∑ F = ma c for the baseball. The net force is the gravity force exerted on the baseball by Deimos, so G v2 mmD = m RD RD2 v = GmD RD = (6.67 × 10−11 N ⋅ m 2 kg 2 ) (2.0 × 1015 kg) (6.0 × 103 m) = 4.7 m s A world+class sprinter runs 100 m in 10 s so have v = 10 m s; v = 4.7 m s for a thrown baseball is very achieveable. Apply Newton’s second law to Vulcan. v2 Gms mv m ∑ F = ma : = v r r2 2πr v= T Gms  2πr  =  r  T  T= 2 4π 2 r 3 Gms [ ] 4π 2 23 (5.79 × 1010 m) 3 = (6.67 × 10−11 Nm 2 kg 2 )(1.99 × 1030 kg)  1d  = 4.14 × 106 s  = 47.9 days  86,400 s  a) v = Gm r = (6.673 × 10 −11 N ⋅ m 2 kg 2 )(0.85 × 1.99 × 1030 kg) ((1.50 × 1011 m)(0.11)) = 8.27 × 10 4 m s. b) 2πr v = 1.25 × 106 s (about two weeks). From either Eq. (12.14) or Eq. (12.19), mS = 4π 2 r 3 4π 2 (1.08 × 1011 m)3 = GT 2 (6.673 × 10 −11 N ⋅ m 2 kg 2 ) ((224.7 d)(8.64 × 10 4 s d)) 2 = 1.98 × 1030 kg. a) The result follows directly from Fig. 12.18. b) (1 − 0.248)(5.92 × 1012 m) = 4.45 × 1012 m, (1 + 0.010)(4.50 × 1012 m) = 4.55 × 1012 m. c) T = 248 y. a) r= Gm1m2 = 7.07 × 1010 m. F b) From Eq. (12.19), using the result of part (a), 2π (7.07 × 1010 m)3 2 T= = 1.05 × 107 s = 121 days. 2 2 30 −11 (6.673 × 10 N ⋅ m kg )(1.90 × 10 kg) c) From Eq. (12.14) the radius is (8) 2 3 = four times that of the large planet’s orbit, or 2.83 × 1011 m. a) For a circular orbit, Eq. (12.12) predicts a speed of (6.673 × 10−11 N ⋅ m 2 kg 2 )(1.99 × 1030 kg) (43 × 109 m) = 56 km s. The speed doesn’t have this value, so the orbit is not circular. b) The escape speed for any object at this radius is 2 (56 km s) = 79 km s , so the spacecraft must be in a closed elliptical orbit. a) Divide the rod into differential masses dm at position l, measured from the right end of the rod. Then , dm = dl ( M L ), and dU = − Gm dm GmM dl =− . l+x L l+x Integrating, U= − GmM L dl GmM  L  =− ln1 + . Ol + x L x  ∫ L For x >> L, the natural logarithm is ~ (L x ) , and U → −Gm M x. b) The x+component of the gravitational force on the sphere is Fx = − δU GmM (− L x 2 ) GmM =− 2 , = δx L (1 + ( L x)) ( x + Lx) with the minus sign indicating an attractive force. As x >> L, the denominator in the above expression approaches x 2 , and Fx → Gm M x 2 , as expected. The derivative may also be taken by expressing  L ln1 +  = ln( x + L) − ln x x  at the cost of a little more algebra. a) Refer to the derivation of Eq. (12.26) and Fig. (12.22). In this case, the red ring in Fig. (12.22) has mass M and the common distance s is x 2 + a 2 . Then, U = − GMm x 2 + a 2 . b) When x >> a, the term in the square root approaches x 2 and U → − GMm x , as expected. GMmx δU =− 2 Fx = − c) , ( x + a 2 )3 2 , δx with the minus sign indicating an attractive force. d) when x >> a, the term inside the parentheses in the above expression approaches x 2 and Fx → − GMmx ( x 2 )3 2 − GMm = − GMm x 2 , as expected. e) The result of part (a) indicates that U = when a x = 0. This makes sense because the mass at the center is a constant distance a from the mass in the ring. The result of part (c) indicates that Fx = 0 when x = 0. At the center of the ring, all mass elements that comprise the ring attract the particle toward the respective parts of the ring, and the net force is zero. At the equator, the gravitational field and the radial acceleration are parallel, and taking the magnitude of the weight as given in Eq. (12.30) gives w = mg 0 − marad . The difference between the measured weight and the force of gravitational attraction is the term marad . The mass m is found by solving the first relation for m, m = g 0 −ωarad . Then, marad = w arad w = . g 0 − arad ( g 0 arad ) − 1 Using either g 0 = 9.80 m s 2 or calculating g 0 from Eq. (12.4) gives marad = 2.40 N. ( ) a) GmN m R 2 = 10.7 m s 2 (5.00 kg ) = 53.5 N, or 54 N to two figures. ( )  4π 2 2.5 × 10 7 m   = 52.0 N. b) m( g 0 − arad ) = (5.00 kg ) 10.7 m s 2 − 2  [ ( )( ) ] 16 h 3600 s h   GMm (RSc 2 2) mc 2 RS = = . r2 r2 2r 2 a) (5.00 kg )(3.00 × 108 m s )2 (1.4 × 10 −2 m ) = 350 N. b) ( 2 3.00 × 10 6 m ) 2 c) Solving Eq. (12.32) for M , RSc 2 (14.00 × 10− 3 m )(3.00 × 108 m s ) = = 9.44 × 1024 kg. −11 2 2 2G 2(6.673 × 10 N ⋅ m kg ) 2 M= a) From Eq. (12.12), Rv 2 (7.5 ly )(9.461 × 1015 m ly )(200 × 103 m s ) M= = (6.673 × 10−11 N ⋅ m2 kg 2 ) G 2 = 4.3 × 1037 kg = 2.1 × 107 M S . b) It would seem not. RS = c) 2GM 2v 2 R = 2 = 6.32 × 1010 m, 2 c c which does fit. Using the mass of the sun for M in Eq. (12.32) gives RS = ( )( m s) 2 6.673 × 10 −11 N ⋅ m 2 kg 2 1.99 × 1030 kg (3.00 × 10 8 2 ) = 2.95 km. That is, Eq. (12.32) may be rewritten  M  2Gmsun  M    = 2.95 km ×  . RS = 2 c  msun   msun  Using 3.0 km instead of 2.95 km is accurate to 1.7%. ( )( ) RS 2 6.67 × 10 −11 N m 2 kg 2 5.97 × 10 24 kg = = 1.4 × 10 − 9. 2 8 6 RE 3 × 10 m s 6.38 × 10 m ( )( ) a) From symmetry, the net gravitational force will be in the direction 45 ° from the + x +axis (bisecting the x and y axes), with magnitude   (2.0 kg) (1.0 kg) (6.673 × 10−11 N ⋅ m 2 kg 2 )(0.0150 kg)  sin 45° +2 2 2 (0.50 m)   (2(0.50 m) ) −12 = 9.67 × 10 N. b) The initial displacement is so large that the initial potential may be taken to be zero. From the work+energy theorem,  (2.0 kg ) (1.0 kg)  1 2 mv = Gm  +2 . (0.50 m)  2  2 (0.50 m) Canceling the factor of m and solving for v, and using the numerical values gives υ = 3.02 × 10−5 m s. The geometry of the 3+4+5 triangle is available to simplify some of the algebra, The components of the gravitational force are Fy = (6.673 × 10 −11 N ⋅ m 2 kg 2 )(0.500 kg)(80.0 kg ) 3 (5.000 m) 2 5 = 6.406 × 10 −11 N  (60.0 kg) (80.0 kg ) Fx = −(6.673 × 10−11 N ⋅ m 2 kg 2 )(0.500 kg)  + 2 (5.000 m) 2  (4.000 m) = −2.105 × 10 −10 N, so the magnitude is 2.20 × 10 −10 N and the direction of the net gravitational force is 163 ° counterclockwise from the + x + axis. b) A at x = 0, y = 1.39 m. 4 5  a) The direction from the origin to the point midway between the two large .100 m masses is arctan ( 00.200 m ) = 26.6°, which is not the angle(14.6°) found in the example. b) The common lever arm is 0.100 m, and the force on the upper mass is at an angle of 45° from the lever arm. The net torque is  (0.100 m)sin 45° (0.100 m)  (6.673 × 10−11 N ⋅ m 2 kg 2 )(0.0100 kg)(0.500 kg)  − 2 (0.200 m) 2   2(0.200 m) = −5.39 × 10−13 N ⋅ m, with the minus sign indicating a clockwise torque. c) There can be no net torque due to gravitational fields with respect to the center of gravity, and so the center of gravity in this case is not at the center of mass. a) The simplest way to approach this problem is to find the force between the spacecraft and the center of mass of the earth+moon system, which is 4.67 × 10 6 m from the center of the earth. The distance from the spacecraft to the center of mass of the earth+moon system is 3.82 × 10 8 m. Using the Law of Gravitation, the force on the spacecraft is 3.4 N, an angle of 0.61° from the earth+spacecraft line. This equilateral triangle arrangement of the earth, moon and spacecraft is a solution of the Lagrange Circular Restricted Three+Body Problem. The spacecraft is at one of the earth+moon system Lagrange points. The Trojan asteriods are found at the corresponding Jovian Lagrange points. b) The work is W = − GMm = − 6.673×10 r W = −1.31× 109 J. −11 N⋅m 2 / kg 2 )( 5.97×1024 kg + 7.35×1022 kg)(1250 kg) 3.84×108 m , or Denote the 25+kg sphere by a subscript 1 and the 100+kg sphere by a subscript 2. a) Linear momentum is conserved because we are ignoring all other forces, that is, the net external force on the system is zero. Hence, m1v1 = m2 v2. This relationship is useful in solving part (b) of this problem. b)From the work+ energy theorem, 1 1 1 Gm1m2  −  = (m1m12 + m2v22 )  rf ri  2 and from conservation of momentum the speeds are related by m1v1 = m2 v 2 . Using the conservation of momentum relation to eliminate v 2 in favor of v1 and simplifying yields 2Gm22 v = m1 + m2 2 1 1 1  r − r , i f with a similar expression for v 2 . Substitution of numerical values gives v1 = 1.63 × 10 −5 m s, v 2 = 4.08 × 10 −6 m s. The magnitude of the relative velocity is the sum of the speeds, 2.04 × 10 −5 m s. c) The distance the centers of the spheres travel (x1 and x 2 ) is proportional to their acceleration, and xx = aa = mm , or x1 = 4 x 2 . When the spheres finally make 1 1 2 2 2 1 contact, their centers will be a distance of 2 R apart, or x1 + x2 + 2 R = 40 m, or x2 + 4 x2 + 2 R = 40 m. Thus, x 2 = 8 m − 0.4 R, and x1 = 32 m − 1.6 R. Solving Eq. (12.14) for r, T  R 3 = GmE    2π  2  (27.3 d)(86,400 s d)  = (6.673 × 10−11 N ⋅ m 2 kg 2 )(5.97 × 1024 kg)   2π   25 3 = 5.614 × 10 m , from which r = 3.83 × 108 m. g= = ( 6.673×10 −11 N ⋅ m 2 kg 2 )( 20.0 kg) center of the sphere. (1.50 m) 2 2 = 5.93 × 10−10 N kg, directed toward the a) From Eq. (12.14), 2  86,164 s  T  r 3 = GmE   = (6.673 × 10−11 N ⋅ m 2 kg 2 ) (5.97 × 1024 kg)    2π   2π  = 7.492 × 1022 m3 , 2 and so h = r − RE = 3.58 × 107 m. Note that the period to use for the earth’s rotation is the siderial day, not the solar day (see Section 12.7). b) For these observers, the satellite is below the horizon. Equation 12.14 in the text will give us the planet’s mass: 2πr 3 2 T= GM P T2 = 4π 2 r 3 GM P MP = 4π 2 r 3 4π 2 (5.75 × 105 m + 4.80 × 106 m)3 = GT 2 (6.673 × 10−11 N ⋅ m 2 kg 2 )(5.8 × 103 s) 2 = 2.731 × 1024 kg , or about half earth’s mass. Now we can find the astronaut’s weight on the surface (The landing on the north pole removes any need to account for centripetal acceleration): w= GM p ma 2 p r = (6.673 × 10 −11 )( ) N ⋅ m 2 kg 2 2.731 × 1024 kg (85.6 kg ) (4.80 × 10 m ) 6 2 = 677 N In terms of the density ρ , the ratio M R is (4π 3) ρR 2 , and so the escape speed is v= (8π 3) (6.673 × 10−11 N ⋅ m 2 )( )( kg 2 2500 kg m3 150 × 103 m ) 2 = 177 m s. a) Following the hint, use as the escape velocity v = 2gh, where h is the height one can jump from the surface of the earth. Equating this to the expression for the escape speed found in Problem 12.55, 2 gh = 8π 3 gh , ρGR 2 , or R 2 = 3 4π ρG where g = 9.80 m s 2 is for the surface of the earth, not the asteroid. Using h = 1 m (variable for different people, of course), R = 3.7 km. As an alternative, if one’s jump speed is known, the analysis of Problem 12.55 shows that for the same density, the escape speed is proportional to the radius, and one’s jump speed as a fraction of 60 m s gives the largest radius as a fraction of 50 km. b) With a a = v 2 R, ρ = 4π3GR = 3.03 × 10 3 kg m 3 . a) The satellite is revolving west to east, in the same direction the earth is rotating. If the angular speed of the satellite is ωs and the angular speed of the earth is ωE , the angular speed ωrel of the satellite relative to you is ωrel = ωs − ωE . ωrel = (1 rev ) (12 h ) = (121 )rev h ωE = (121 )rev h ωs = ωrel + ωE = ( 18 )rev h = 2.18 × 10+4 rad s ∑ = m says G mmE v2 = m r2 r GmE Gm and with v = rω this gives r 3 = 2E ; r = 2.03 × 107 m r ω This is the radius of the satellite’s orbit. Its height h above the surface of the earth is h = r − RE = 1.39 × 107 m. v2 = b) Now the satellite is revolving opposite to the rotation of the earth. If west to east is positive, then ωrel = (− 121 )rev h ωs = ωrel + ωE = (− 241 ) rev h = −7.27 × 10 +5 rad s Gm r 3 = 2E gives r = 4.22 × 107 m and h = 3.59 × 107 m ω (a) Get radius of X : 14 (2πR ) = 18,850 km R = 1.20 × 107 m Astronant mass: m = ωg = 9.80943mNs = 96.2 kg Use astronant at north pole to get mass of 2 X: GmM x = mg x R2 mg x R 2 (915 N)(1.20 × 107 m) 2 Mx = = = 2.05 × 1025 kg −11 2 2 Gm (6.67 × 10 Nm kg )(96.2 kg) ∑ F = ma : Apply Newton’s second law to astronant on a scale at the equator of X. ∑ F = ma : Fgrav − Fscale = mv 2 R m( 2πRR ) 2πR 4π 2 mR → Fgrav − Fscale = = T R T2 4π 2 (96.2 kg)(1.20 × 10 7 m) 915.0 N − 850.0 N = T2  1 hr  T = 2.65 × 10 4 s  = 7.36 hr, which is one day  3600 s  2 v= (b) For satellite: ∑ F = ma → T= Gms mx r2 = ms v 2 r where v = 2πr T . Gmr x = ( 2Tπr ) 2 4π 2 r 3 4π 2 (1.20 × 107 m + 2 × 106 m)3 = Gmx (6.67 × 10−11 Nm 2 kg 2 )(2.05 × 1025 kg) T = 8.90 × 103 s = 2.47 hours The fractional error is 1− mgh g = 1− ( RE + h)( RE ). 1 1 GmmE RE − RE +h GmE ( ) At this point, it is advantageous to use the algebraic expression for g as given in Eq. (12.4) instead of numerical values to obtain the fractional difference as 1 − ( RE + h) RE = − h RE , so if the fractional difference is − 1%, h = (0.01) RE = 6.4 × 104 m. If the algebraic form for g in terms of the other parameters is not used, and the numerical values from Appendix F are used along with g = 9.80 m s 2 , h RE = 8.7 × 10−3 , which is qualitatively the same. (a) Get g on Mongo: It takes 4.00 s to reach the maximum height, where v = 0 then v0 − gt → 0 = 12.0 m s − g (4.00 s) g = 3.00 m s 2 Apply Newton’s second law to a falling object: GmM → M = gR 2 G 2 R 2πR = C → R = C 2π ∑ F = ma : mg = ( 8 ) 2 m (3.00 m s 2 ) 2.002×10 π M = gR G = = 4.56 × 10 25 kg 2 2 −11 6.67 × 10 N m kg 2 b) Apply Newton’s second law to the orbiting starship. ΣF = ma : GmM mv 2 = r2 r 2πr 4π 2 r 3 →T = GM T C r = R + 30,000 km = + 3.0 × 107 m 2π v= m 4π 2 ( 2.002×10 + 3.0 × 10 7 m) 3 π T= (6.67 × 10 −11 Nm 2 kg 2 )(4.56 × 10 25 kg) 8  1h  = 5.54 × 10 4 s   = 15.4 h  3600 s  At Sacramento, the gravity force on you is F1 = G mmE ⋅ RE2 At the top of Mount Everest, a height of h = 8800 m above sea level, the gravity force on you is F2 = G = mmE mmE =G 2 2 2 ( RE + h) R E (1 + h RE ) (1 + h RE ) − 2 ≈ 1 −  1 − 2h  2h  , F2 = F1  RE  RE  F1 − F2 2h = = 0.28 % F1 rE a) The total gravitational potential energy in this model is m mM  U = −Gm E + ⋅ REM − r   r b) See Exercise 12.5. The point where the net gravitational field vanishes is r= REM = 3.46 × 108 m. 1 + mM mE Using this value for r in the expression in part (a) and the work+energy theorem, including the initial potential energy of − Gm(mE RE + mM ( REM − RE )) gives 11.1 km s. c) The final distance from the earth is not RM , but the Earth+moon distance minus the radius of the moon, or 3.823 × 10 8 m. From the work+energy theorem, the rocket impacts the moon with a speed of 2.9 km s. One can solve this problem using energy conservation, units of J/kg for energy, 1 2 GM and basic concepts of orbits. E = K + U , or − GM 2 a = 2 v − r , where E , K and U are the energies per unit mass, v is the circular orbital velocity of 1655 m/s at the lunicentric distance of 1.79 × 10 6 m. The total energy at this distance is − 1.37 × 10 6 J Kg. When the velocity of the spacecraft is reduced by 20 m/s, the total energy becomes E= (6.673 × 10 −11 N ⋅ m 2 / kg 2 ) (7.35 × 10 22 kg) 1 , (1655 m / s − 20 m / s) 2 − (1.79 × 10 6 m) 2 , we can solve for a, a = 1.748 × 10 6 m, the semi or E = −1.40 × 10 6 J kg. Since E = − GM 2a –major axis of the new elliptical orbit. The old distance of 1.79 × 10 6 m is now the apolune distance, and the perilune can be found from r +r a = a 2 p , rp = 1.706 × 106 m. Obviously this is less than the radius of the moon, so the or, U = −2.818 × 106 J kg. spacecraft crashes! At the surface, U = − GM Rm Since the total energy at the surface is − 1.40 × 10 6 J kg, the kinetic energy at the surface is 1.415 × 10 6 J kg. So, 12 v 2 = 1.415 × 10 6 J kg, or v = 1.682 × 10 3 m s = 6057 km h. Combining Equations (12.13) and (3.28) and setting arad = 9.80 m s 2 (so that ω = 0 in Eq. (12.30)), T = 2π R = 5.07 × 10 3 s, a rad which is 84.5 min, or about an hour and a half. The change in gravitational potential energy is GmE m GmE m h − = −GmE m , (RE + h ) RE RE (RR + h ) so the speed of the hammer is, from the work+energy theorem, !U = 2GmE h . (RE + h )RE a) The energy the satellite has as it sits on the surface of the Earth is Ei = The energy it has when it is in orbit at a radius R ≈ RE is Ef = put it in orbit is the difference between these: W = Ef + Ei = − GmM E 2 RE GmM E 2 RE − GmM E RE . . The work needed to . b) The total energy of the satellite far away from the Earth is zero, so the additional GmM − GmM work needed is 0 − 2 RE E = 2 RE E . ( ) c) The work needed to put the satellite into orbit was the same as the work needed to put the satellite from orbit to the edge of the universe. The escape speed will be m m  v = 2G  E + s  = 4.35 × 104 m s.  RE RES  a) Making the simplifying assumption that the direction of launch is the direction of the earth’s motion in its orbit, the speed relative the earth is v− 2πRES 2π(1.50 × 1011 m) = 4.35 × 104 m s − = 1.37 × 104 m s. T (3.156 × 107 s) b) The rotational at Cape Canaveral is 2 π ( 6.38×10 6 m) cos 28.5° 86164 s 4 = 4.09 × 102 m s, so the speed relative to the surface of the earth is 1.33 × 10 m s. c) In French Guiana, the rotational speed is 4.63 × 10 2 m s, so the speed relative to the surface of the earth is 1.32 × 10 4 m s. a) The SI units of energy are kg ⋅ m 2 s 2 , so the SI units for φ are m 2 s 2 . Also, it is known from kinetic energy considerations that the dimensions of energy, kinetic or potential, are mass × speed 2 , so the dimensions of gravitational potential must be the same as speed 2 . b) φ = − Um = c) d) m!φ = 5.53 × 1010 GmE r . 1 1 !φ = GmE  −  = 3.68 × 10 6 J kg.  RE rf  J. (An extra figure was kept in the intermediate calculations.) a) The period of the asteroid is T = 2 πa 3 2 GM . Inserting 3 × 1011 m for a gives 2.84 y and 5 × 1011 m gives a period of 6.11 y. b) If the period is 5.93 y, then a = 4.90 × 1011 m. c) This happens because 0.4 = 2 5, another ratio of integers. So once every 5 orbits of the asteroid and 2 orbits of Jupiter, the asteroid is at its perijove distance. Solving when T = 4.74 y, a = 4.22 × 1011 m. a) In moving to a lower orbit by whatever means, gravity does positive work, v 2 Gm 1/ 2 and so the speed does increase. b) From = 2 E , v = (GmE ) r −1 / 2 , so r r − !r  −3 / 2  !r  GmE 1/ 2  ⋅ =  !v = (GmE )  − r 3 2   2  r  Note that a positive !r is given as a decrease in radius. Similarly, the kinetic energy is K = (1 2 )mv 2 = (1 2 )GmE m / r , and so !K = (1 2) GmE m / r 2 !r , !U = − GmE m / r 2 ( ( ) ) ( ) !r and W = !U + !K = − GmE m / 2r 2 !r , is agreement with part (a). c)v = GmE r = 7.72 × 103 m/s, !v = (!r 2 ) GmE r = 28.9 m/s, E = −GmE m / 2r = 3 ( ) − 8.95 × 1010 J (from Eq. (12.15)), !K = GmE m / 2r 2 (!r ) = 6.70 × 108 J, !U = −2!K = − 1.34 × 10 J and W = −!K = −6.70 × 10 J. d)As the term “burns up” suggests, the energy is converted to heat or is dissipated in the collisions of the debris with the grounds. 9 8 a) The stars are separated by the diameter of the circle d =2R, so the 2 . gravitational force is GM 4 R2 b) The gravitational force found in part (b) is related to the radial acceleration by Fg = Marad = Mv 2 R for each star, and substituting the expression for the force from part (a) and solving for v gives v = GM 4 R. The period is Τ= 2πR υ − GM 2 = 16π 2 R 3 GM = 4πR 3 2 GM . c) The initial gravitational potential energy is 2 R and the initial kinetic energy is 2(1 2) Mv 2 = .GM 2 4 R, so the total mechanical energy is − GM 2 2 R. If the stars have zero speed when they are very far apart, the energy needed to separate them is GM 2 4 R. a) The radii R1 and R2 are measured with respect to the center of mass, and so M 1 R1 = M 2 R2 , and R1 R2 = M 2 M 1 . b) If the periods were different, the stars would move around the circle with respect to one another, and their separations would not be constant; the orbits would not remain circular. Employing qualitative physical principles, the forces on each star are equal in magnitude, and in terms of the periods, the product of the mass and the radial accelerations are 4π2 M 1R1 4π2 M 2 R2 . = T12 T22 From the result of part (a), the numerators of these expressions are equal, and so the denominators are equal, and the periods are the same. To find the period in the symmetric from desired, there are many possible routes. An elegant method, using a bit of hindsight, GM M is to use the above expressions to relate the periods to the force Fg = ( R +1R )22 , so that 1 2 equivalent expressions for the period are M 2T 2 = 4π 2 R1(R1 + R2 )2 G M 1T 2 = 4π2 R2 ( R1 + R2 ) 2 . G Adding the expressions gives ( M 1 + M 2 )T 2 = 4π2 ( R1 + R2 )3 2π( R1 + R2 )3 2 . or T = G G(M1 + M 2 ) c) First we must find the radii of each orbit given the speed and period data. In a circular orbit, v = 2TπR , or R = vT 2π Thus , R α = ( 36×103 m s )(137 d ×86,400 s d ) 2π = 6.78 × 1010 m, and R β = (12×103 m s )(137 d×86, 400 s d) 2π = 2.26 × 1010 m. Now find the sum of the masses and use M α Rα = M β Rβ , and the fact that Rα = 3Rβ .( M α + M β ) = (M α + M β ) = 4 π 2. ( Rα + Rβ )3 T 2G , inserting the values of T, and the radii, 4π 2 ( 6.78×1010 m + 2.26×1010 m)3 (137 d ×86,400 s d ) 2 ( 6.673×10 −11 N⋅m 2 kg 2 ) .M α + M β = 3.12 × 10 30 kg. Since M β = M α Rα R β = 3M α ,4 M α = 3.12 × 10 30 kg, or M α = 7.80 × 10 29 kg, and M β = 2.34 × 10 30 d) Let α refer to the star and β refer to the black hole. Use the relationships derived in parts (a) and (b): Rβ = ( M α M β ) Rα = (0.67 3.8) Rα = (0.176) Rα , ( Rα + Rβ) = 3 ( M α + M β ) T 2G 4π 2 , and v = 2 πR T . For Rα , inserting the values for M and T and From conservation of energy, the speed at the closer distance is  1 1 v = v02 + 2Gms  −  = 6.8 × 104 m s .  rf ri  Using conservation of energy, GM S mM 1 GM S mM 1 = mM vp2 − , or mM va2 − 2 2 ra rp 1 1 vp = va2 − 2GM S  −  = 2.650 × 10 4 m s. r r  p   a The subscripts a and p denote aphelion and perihelion. To use conservation of angular momentum, note that at the extremes of distance (periheleion and aphelion), Mars’ velocity vector must be perpendicular to its radius vector, and so the magnitude of the angular momentum is L = mrv . Since L is constant, the product rv must be a constant, and so v p = va ra (2.492 × 1011 m) = 2.650 × 10 4 m s, = (2.198 × 10 4 m s) (2.067 × 1011 m) rp a confirmation of Kepler’s Laws. a) The semimajor axis is the average of the perigee and apogee distances, a = 12 (( RE + hp ) + ( RE + ha )) = 8.58 × 106 m. From Eq. (12.19) with the mass of the earth, the period of the orbit is T= 2πa 3 2 = 7.91 × 103 s, GM E a little more than two hours. b) See Problem 12.74; υp υa = ra rp = 1.53. c) The equation that represents conservation of energy (apart from a common factor of the mass of the spacecraft) is 2 1 2 GmE 1 2 GmE 1  rp  2 GmE = va − =   vp − vp − , 2 rp 2 ra 2  ra  ra where conservation of angular momentum has been used to eliminate va is favor of vp . Solving for vp2 and simplifying, vp2 = 2GmE ra = 7.71 × 107 m 2 s 2 , rp (rp + ra ) from which vp = 8.43 × 103 m s and va = 5.51 × 103 m s. d) The escape speed for a given distance is ve = 2GM r , and so the difference between escape speed and v p is, after some algebra, ve − v p = [ ] 2GmE 1 − 1 / 1 + (rp ra ) ⋅ rp Using the given values for the radii gives ve − vp = 2.41 × 103 m s. The similar calculation at apogee give ve − va = 3.26 × 103 m s, so it is more efficient to fire the rockets at perigee. Note that in the above, the escape speed ve is different at the two points, vpe = 1.09 × 10 4 m s and vae = 8.77 × 103 m s. a) From the value of g at the poles, ( )( g R2 11.1m s 2 2.556 × 107 m mU = U U = G 6.673 × 10 −11 N ⋅ m 2 kg 2 ( ) ) 2 = 1.09 × 10 26 kg. b) GmU r 2 = g U (RU r ) = 0.432 m s . c) GmM RM2 = 0.080 m s 2 . d) No; Miranda’s gravity is sufficient to retain objects released near its surface. 2 2 Using Eq. (12.15), with the mass M m instead of the mass of the earth, the energy needed is Gmm m  1 1  − 2  ri rf  6.673 × 10−11 N ⋅ m 2 kg 2 6.42 × 1023 kg (3000 kg ) = 2   1 1 × − 6 6 6 6 4.00 × 10 m + 3.40 × 10 m   2.00 × 10 m + 3.40 × 10 m = 3.22 × 109 J. !E = ( )( ) ( ) ( ) a) The semimajor axis is 4 × 1015 m and so the period is (6.673 × 10 ( ) kg )(1.99 × 10 2π 4 × 1015 m −11 N⋅m 2 2 32 30 kg ) = 1.38 × 1014 s, which is about 4 million years. b) Using the earth+sun distance as an estimate for the distance of closest approach, v = 2GmS RES = 4 × 104 m s. c) (1 2 ) mv 2 = GmS m R = 10 24 J. This is far larger than the energy of a volcanic eruption and is comparable to the energy of burning the fossil fuel. a) From Eq. (12.14) with the mass of the sun, ( )( )  6.673 × 10−11 N ⋅ m 2 kg 2 1.99 × 1030 kg  r=  2  × 3 × 104 y 3.156 × 107 s y 4π2  (( )( )) 13 = 1.4 × 1014 m. This is about 24 times the orbit radius of Pluto and about 1 250 of the way to Alpha Centauri. Outside the planet it behaves like a point mass, so at the surface: ∑ F = ma : GmM = mg → g = GM R 2 2 R Get M : M = ∫ dm = ∫ ρdV = ∫ ρ 4πr 2 dr. The density is ρ = ρ0 − br , where ρ 0 = 15.0 × 10 3 kg m 3 at the center at the surface, ρS = 2.0 × 103 kg m 3 , so b = 4π M = ∫ 0R ( ρ0 − br ) 4πr 2 dr = ρ0 R 3 − πbR 4 3 4  ρ − ρs   3 1 = πR 3 ρ0 − πR 4  0  = πR  ρ0 + ρs  3  R  3  g= ρ0 − ρ s R GM Gπ R 3 ( 13 ρ0 + ρs )  1 = = πRG ρ0 + ρs  2 2 R R  3   15.0 × 103 kg m 3 π (6.38 × 10 6 m )(6.67 × 10 −11 Nm 2 kg 2 ) + 2.0 × 103 kg m 3  3   2 = 9.36 m s The radius of the semicircle is R = L π Divide the semicircle up into small segments of length R dθ dM = (M L )R dθ = (M π )dθ d is the gravity force on m exerted by dM ∫ dFy = 0; the y+components from the upper half of the semicircle cancel the y+ components from the lower half. The x+components are all in the +x+direction and all add. mdM R2 Gmπm mdM cos θ = cosθ dθ dFx = G 2 L2 R π 2 Gmπm GmπM π 2 cos θdθ = (2) Fx = ∫ dFx = ∫ −π 2 − 2 π L2 L2 2πGmM F= L2 dF = G The direct calculation of the force that the sphere exerts on the ring is slightly more involved than the calculation of the force that the ring exerts on the ball. These forces are equal in magnitude but opposite in direction, so it will suffice to do the latter calculation. By symmetry, the force on the sphere will be along the axis of the ring in Fig. (12.34), toward the ring. Each mass element dM of the ring exerts a force of magnitude GmdM on the sphere, and the x+component of this force is a2 + x2 GmdM a2 + x2 x a +x 2 2 = GmdMx (a 2 + x2 ) 3/ 2 . As x >> a the denominator approaches x 3 and F → GMm , as expected, and so the force on x2 ( ) 3/ 2 , in the − x + direction. The sphere attracts the ring with the sphere is GmMx / a 2 + x 2 a force of the same magnitude. (This is an alternative but equivalent way of obtaining the result of parts (c) and (d) of Exercise 12.39.)2 Divide the rod into differential masses dm at position l, measured from the right end of the rod. Then, dm = dl (M L ) , and the contribution GmMdl . Integrating from l = 0 to l = L gives dFx from each piece is dFx = − (l + x )2 L F =− GmM L L dl ∫ (l + x ) 0 2 = GmM  1 1 GmM , − =−  L x + L x x( x + L ) with the negative sign indicating a force to the left. The magnitude is F = GmM x( x + L ) . As x >> , as expected. (This is an alternative but L, the denominator approaches x 2 and F → GmM x2 equivalent way of obtaining the result of part (b) Exercise 12.39.) a) From the result shown in Example 12.10, the force is attractive and its magnitude is proportional to the distance the object is from the center of the earth. Comparison with equations (6.8) and (7.9) show that the gravitational potential energy is given by GmE m 2 r . 2 RE3 This is also given by the integral of Fg from 0 to r with respect to distance. b) From part U (r ) = GmE m . Equating initial potential energy 1RE and final kinetic energy (initial kinetic energy and final potential energy are both zero) GmE gives v 2 = , so v = 7.90 × 103 m/s. RE (a), the initial gravitational potential energy is a) T = T + !T = , therefore (r + !r )3 2 = 2π GM E Since v = 2π r 3 2 GM E GM E r , !T = 3π ! r v 2π r 3 2 GM E r !r (1 + !rr )3 2 ≈ 2πGMr (1 + 32!rr ) = T + 3π GM . . v = GM E r 32 E −1 2 2π r 3 2 GM E , !v = π!r T E , therefore v − ! v = GM E (r + ! r ) −1 2 = GM E r −1 2 (1 + Since T = 12 ) ! r −1 2 r ≈ GM E r −1 2 (1 − !2 rr ) = v − GM E 2r 3 2 ! r. . b) Note: Because of the small change in r, several significant figures are needed to r3 2 (Eq.(12.14)), T = 2π r v , and v = GM see the results. Starting with T = 2 πGM r (Eq.(12.12)) find the velocity and period of the initial orbit: (6.673 × 10−11 N ⋅ m 2 kg 2 )(5.97 × 1024 kg) = 7.672 × 103 m s, and 6.776 × 106 m T = 2π r v = 5549 s = 92.5 min. We then can use the two derived equations to approximate the 3 π (100 m) ! T and ! v, ! T = 3 πv! r and !v = πT!r .!T = 7.672 = 0.1228 s, and !v ×10 3 m s v= = π !r T = π (100 m ) ( 5549 s ) = .05662 m s. Before the cable breaks, the shuttle will have traveled a distance d, d = (125 m 2 ) − (100 m 2 ) = 75 m. So, (75 m) (.05662 m s) = 1324.7 s = 22 min. It will take 22 minutes for the cable to break. c) The ISS is moving faster than the space shuttle, so the total angle it covers in an orbit must be 2π radians more than the angle that the space shuttle covers before they are ( v − !v ) t once again in line. Mathematically, vtr − ( r + !r ) = 2π . Using the binomial theorem and −1 ( v!T 3π ,t = ( v − !v ) t neglecting terms of order !v!r , vtr − r (1 + !rr ) ≈ t t= 2π r v!r    !v +  r   = shown. t = vT π !r v!r + T r T2 !T = . Since 2π r = vT and !r = ( 5549 s ) 2 ( 0.1228 s ) !v r ) + vr!2r = 2π . Therefore, vT π  v!T  2 π  v!T  +    t  3π  T  3π  = T2 !T , as was to be = 2.5 × 108 s = 2900 d = 7.9 y. It is highly doubtful the shuttle crew would survive the congressional hearings if they miss! a) To get from the circular orbit of the earth to the transfer orbit, the spacecraft’s energy must increase, and the rockets are fired in the direction opposite that of the motion, that is, in the direction that increases the speed. Once at the orbit of Mars, the energy needs to be increased again, and so the rockets need to be fired in the direction opposite that of the motion. From Fig. (12.37), the semimajor axis of the transfer orbit is the arithmetic average of the orbit radii of the earth and Mars, and so from Eq. (12.19), the energy of spacecraft while in the transfer orbit is intermediate between the energies of the circular orbits. Returning from Mars to the earth, the procedure is reversed, and the rockets are fired against the direction of motion. b) The time will be half the period as given in Eq. (12.19), with the semimajor axis a being the average of the orbit radii, a = 1.89 × 1011 m, so t= T π (1.89 × 1011 m)3 2 = = 2.24 × 107 s, −11 2 2 30 2 (6.673 × 10 N ⋅ m kg )(1.99 × 10 kg) which is more than 8 12 months. c) During this time, Mars will pass through an angle of 7 ×10 s ) (360°) ( 687( 2d.24 = 135.9° , and the spacecraft passes through an angle of 180° , so the )(86 , 400 s d ) angle between the earth+sun line and the Mars+sun line must be 44.1° . a) There are many ways of approaching this problem; two will be given here. I) Denote the orbit radius as r and the distance from this radius to either ear as δ . Each ear, of mass m , can be modeled as subject to two forces, the gravitational force from the black hole and the tension force (actually the force from the body tissues), denoted by F . Then, the force equations for the two ears are GMm − F = mω 2 (r − δ ) 2 (r − δ ) GMm + F = mω 2 (r + δ ), (r + δ ) 2 where ω is the common angular frequency. The first equation reflects the fact that one ear is closer to the black hole, is subject to a larger gravitational force, has a smaller acceleration, and needs the force F to keep it in the circle of radius r − δ . The second equation reflects the fact that the outer ear is further from the black hole and is moving in a circle of larger radius and needs the force F to keep in in the circle of radius r + δ . Dividing the first equation by r − δ and the second by r + δ and equating the resulting expressions eliminates ω , and after a good deal of algebra, (r + δ ) . F = (3GMmδ ) 2 (r − δ 2 ) 2 At this point it is prudent to neglect δ in the sum and difference, but recognize that F is δ = 2.1 kN. (Using the result of Exercise proportional to δ , and numerically F = 3GMm r3 12.39 to express the gravitational force in terms of the Schwartzschild radius gives the same result to two figures.) II) Using the same notation, GMm − F = mω 2 (r + δ ), 2 (r + δ ) where δ can be of either sign. Replace the product mω 2 with the value for δ = 0, mω 2 = GMm r 3 and solve for [ ] r + δ 1  GMm −2 = r + δ − r (1 + (δ r)) ⋅ F = (GMm)  3 − 2 3 (r + δ)  r  r Using the binomial theorem to expand the term in square brackets in powers of δ r , GMm GMm F = 3 [r + δ − r (1 − 2(δ r ) )] = 3 (3δ), r r the same result as above. Method (I) avoids using the binomial theorem or Taylor series expansions; the approximations are made only when numerical values are inserted and higher powers of δ are found to be numerically insignificant. As suggested in the problem, divide the disk into rings of radius r and thickness dr. Each ring has an area dA = 2πr dr and mass dM = πMa 2 dA = 2aM2 r dr. The magnitude of the force that this small ring exerts on the mass m is then (G m dM )( x (r 2 + x 2 )3 2 ), the expression found in Problem 12.82, with dM instead of M and the variable r instead of a. rdr 2GMmx . Thus, the contribution dF to the force is dF = 2 2 a ( x + r 2 )3 2 The total force F is then the integral over the range of r; 2GMmx a r F = ∫ dF = dr. 2 2 ∫ 0 (x + r 2 )3 2 a The integral (either by looking in a table or making the substitution u = r 2 + a 2 ) is  1 1  a x r 1 ∫0 ( x 2 + r 2 ) 3 2 dr =  x − a 2 + x 2  = x 1 − a 2 + x 2 . Substitution yields the result  x 2GMm  − F= 1  . a2  a2 + x2  The second term in brackets can be written as 1 2 −1 2 1a ≈1−   2 x 2 = (1 + (a x) ) 1 + ( a x) 2 if x >> a, where the binomial approximation (or first+order Taylor series expansion) has been used. Substitution of this into the above form gives F≈ as it should. GMm , x2 From symmetry, the component of the gravitational force parallel to the rod is zero. To find the perpendicular component, divide the rod into segments of length dx and mass dm = dx 2ML , positioned at a distance x from the center of the rod. The magnitude of the gravitational force from each segment is Gm dM GmM dx dF = 2 = . 2L x 2 + a 2 x + a2 The component of dF perpendicular to the rod is dF 2a 2 , and so the net gravitational x +a force is L L GmMa dx . 2 ∫ 2 L − L ( x + a 2 )3 2 −L The integral can be found in a table, or found by making the substitution x = a tanθ. Then, dx = a sec 2θ dθ , ( x 2 + a 2 ) = a 2 sec 2θ , and so F = ∫ dF = dx a sec 2θ dθ x 1 1 ∫ ( x 2 + a 2 ) 3 2 = ∫ a 3 sec 3θ = a 2 ∫ cosθ dθ = a 2 sinθ = a 2 x 2 + a 2 , and the definite integral is GmM F= . a a 2 + L2 , as expected. When a >> L, the term in the square root approaches a 2 and F → GmM a2 Capítulo 13 : a) T= 1 4 ( 220 Hz) b) 1 f = 4.55 × 10−3 s, ω = 2π T = 2πf = 1.38 × 103 rad s. = 1.14 × 10−3 s, ω = 2πf = 5.53 × 103 rad s. a) Since the glider is released form rest, its initial displacement (0.120 m) is the amplitude. b) The glider will return to its original position after another 0.80 s, so the period is 1.60 s. c) The frequency is the reciprocal of the period (Eq. (13.2)), f = 1.601 s = 0.625 Hz. The period is ω= 2π T 0.50 s 440 = 1.14 × 10 −3 s and the angular frequency is = 5.53 × 103 rad s. (a) From the graph of its motion, the object completes one full cycle in 2.0 s; its period is thus 2.0 s and its frequency = 1 period = 0.5 s −1 . (b) The displacement varies from − 0.20 m to + 0.20 m, so the amplitude is 0.20 m. (c) 2.0 s (see part a) This displacement is 14 of a period. T = 1 f = 0.200 s, so t = 0.0500 s. The period will be twice the time given as being between the times at which the glider is at the equilibrium position (see Fig. (13.8)); 2  2π  2π  k = ω m =   m =  T   2(2.60 s) 2 a) T = 1 f 2   (0.200 kg) = 0.292 N m.  = 0.167 s. b) ω = 2πf = 37.7 rad s. c) m = k ω2 = 0.084 kg. Solving Eq. (13.12) for k, 2 2  2π   2π  3 k = m  = (0.600 kg)   = 1.05 × 10 N m. T   0.150 s  From Eq. (13.12) and Eq. (13.10), T = 2π 0.500 kg 140 N m = 0.375 s, f = T1 = 2.66 Hz, ω = 2πf = 16.7 rad s. a) ax = d 2x dt 2 = −ω2 A sin(ωt + β ) = −ω2 x, so x(t ) is a solution to Eq. (13.4) if ω2 = mk . b) a = 2 Aω a constant, so Eq. (13.4) is not satisfied. c) vx = ax = dv x dt dx dt = iωi ( ωt + β ) , = (iω) 2 Aei ( ωt + β ) = −ω2 x, so x (t ) is a solution to Eq. (13.4) if ω2 = k m ⋅ a) x = (3.0 mm) cos ((2π )(440 Hz)t ) b) (3.0 × 10−3 m)(2π )(440 Hz) = 8.29 m s, (3.0 mm)(2π ) 2 (440 Hz) 2 = 2.29 × 104 m s 2 . c) j (t ) = (6.34 × 107 m s 3 ) sin((2π )(440 Hz)t ), jmax = 6.34 × 107 m s 3 . a) From Eq. (13.19), A = v0 ω = v0 k m = 0.98 m. b) Equation (13.18) is indeterminant, but from Eq. (13.14), φ = ± π2 , and from Eq. (13.17), sin φ > 0, so φ = + π2 . c) cos (ωt + (π 2)) = − sin ωt , so x = (−0.98 m) sin((12.2 rad s)t )). With the same value for ω , Eq. (13.19) gives A = 0.383 m and Eq. (13.18) gives and x = (0.383 m) cos ((12.2 rad/s)t + 1.02 rad ).   (−4.00 m/s)  = 1.02 rad = 58.5°,  (0.200 m) 300 N/m/2.00 kg   φ = arctan − and x = (0.383 m) cos ((12.2 rad/s)t + 1.02 rad). 2 For SHM, ax = −ω2 x = −(2πf ) 2 x = −(2π (2.5 Hz) ) (1.1 × 10−2 m) = −2.71 m/s 2 . b) From Eq. (13.19) the amplitude is 1.46 cm, and from Eq. (13.18) the phase angle is 0.715 rad. The angular frequency is 2πf = 15.7 rad/s, so x = (1.46 cm) cos ((15.7 rad/s)t + 0.715 rad) vx = (−22.9 cm s) sin ((15.7 rad/s)t + 0.715 rad) ax = (−359 cm/s2 ) cos ((15.7 rad/s)t + 0.715 rad) . The equation describing the motion is x = A sin ωt; this is best found from either inspection or from Eq. (13.14) (Eq. (13.18) involves an infinite argument of the arctangent). Even so, x is determined only up to the sign, but that does not affect the result of this exercise. The distance from the equilibrium position is A sin (2π (t T )) = (0.600 m )sin (4π 5) = 0.353 m. Empty chair: T = 2π m k k= 4π 2 m 4π 2 (42.5 kg) = = 993 N/m T2 (1.30 s) 2 With person in chair: T = 2π m k T 2 k (2.54 s) 2 (993 N/m) = = 162 kg 4π 2 4π 2 mperson = 162 kg − 42.5 kg = 120 kg m= T = 2π m k , m = 0.400 kg Use a x = −2.70 m/s 2 to calculate k : ma x (0.400 kg)(−2.70 m/s 2 ) =− = +3.60 N/m x 0.300 m T = 2π m k = 2.09 s − kx = max gives k = − We have vx (t ) = (3.60 cm/s)sin((4.71 s −1 ) t − π 2). Comparing this to the general form of the velocity for SHM: − ωA = 3.60 cm/s (a) (b) (c ) amax ω = 4.71 s −1 φ = −π 2 T = 2π ω = 2π 4.71 s −1 = 1.33 s 3.60 cm s 3.60 cm s A= = = 0.764 cm ω 4.71 s −1 = ω 2 A = (4.71 s −1 ) 2 (0.764 cm) = 16.9 cm s 2 a) x(t ) = (7.40 cm) cos((4.16 rad s)t − 2.42 rad) When t = T , (4.16 rad s)T = 2π so T = 1.51 s b) T = 2π m k so k = m(2π T ) 2 = 26.0 N m c) A = 7.40 cm = 0.0740 m 1 2 mv 2 + 12 kx 2 = 12 kA2 gives vmax = A k m = 0.308 m s d) F = − kx so Fmax = kA = 1.92 N e) x(t ) evaluated at t = 1.00 s gives x = −0.0125 m v = ± k m A2 − x 2 = ± 26.0 1.50 (0.0740) 2 − (0.0125) 2 m s = ±0.303 m s Speed is 0.303 m s . a = − kx m = −(26.0 1.50)(−0.0125) m s 2 = +0.216 m s 2 See Exercise 13.15; t = (arccos(− 1.5 6))(0.3 (2π )) = 0.0871 s. a) Dividing Eq. (13.17) by ω , x0 = A cos φ, v0 = − A sin φ. ω Squaring and adding, v02 = A2 , 2 ω which is the same as Eq. (13.19). b) At time t = 0, Eq. (13.21) becomes 1 2 1 2 1 2 1 k 2 1 2 kA = mv0 + kx0 = v0 + kx0 , 2 2 2 2 ω2 2 2 where m = kω (Eq. (13.10)) has been used. Dividing by k 2 gives Eq. (13.19). x02 + a) vmax = (2πf ) A = (2π (392 Hz))(0.60 × 10−3 m) = 1.48 m s. 1 1 b) K max = m(Vmax ) 2 = (2.7 × 10 −5 kg)(1.48 m s) 2 = 2.96 × 10 −5 J. 2 2 a) Setting 12 mv 2 = 12 kx 2 in Eq. (13.21) and solving for x gives x = ± A 2 . Eliminating x in favor of v with the same relation gives vx = ± kA2 2m = ± ωA2 . b) This happens four times each cycle, corresponding the four possible combinations of + and – in the results of part (a). The time between the occurrences is one=fourth of a period or 2 2 T / 4 = 42ωπ = 2πω . c) U = 14 E , K = 34 E U = kA8 , K = 3kA8 ( ) a) From Eq. (13.23), vmax = k A= m 450 N m (0.040 m) = 1.20 m/s. 0.500 kg b) From Eq. (13.22), 450 N v= (0.040 m) 2 − (−0.015 m) 2 = 1.11 m/s. 0.500 kg c) The extremes of acceleration occur at the extremes of motion, when x = ± A, and kA (450 N/m)(0.040 m) a max = = = 36 m/s 2 m (0.500 kg) ( 450 N/m)( −0.015 m) = 13.5 m/s2 . d) From Eq. (13.4), ax = − (0.500 kg) e) From Eq. (13.31), E = 12 (450 N/m)(0.040 m) 2 = 0.36 J. a) amax = ω2 A = (2πf ) 2 A = (2π(0.85 Hz) ) (18.0 × 10 −2 m) = 5.13 m/s 2 . vmax = 2 ωA = 2πfA = 0.961 m/s . b) ax = −(2πf ) 2 x = −2.57 m/s2 , v = (2πf ) A2 − x 2 = (2π (0.85 Hz) ) (18.0 × 10− 2 m)2 − (9.0 × 10− 2 m)2 = 0.833 m/s. c) The fraction of one period is (1 2π ) arcsin (12.0 18.0), and so the time is (T 2π ) × arcsin (12.0 18.0) = 1.37 × 10 −1 s. Note that this is also arcsin ( x A) ω . d) The conservation of energy equation can be written 12 kA 2 = 12 mv 2 + 12 kx 2 . We are given amplitude, frequency in Hz, and various values of x . We could calculate velocity from this information if we use the relationship k m = ω2 = 4π 2 f 2 and rewrite the conservation equation as 1 2 A2 = 1 v2 2 4π 2 f 2 + 12 x 2 . Using energy principles is generally a good approach when we are dealing with velocities and positions as opposed to accelerations and time when using dynamics is often easier. In the example, A2 = A1 M M +m and now we want A2 = 12 A1 . So 12 = M M +m , or m = 3M . For the energy, E 2 = 12 kA22 , but since A2 = 12 A1 , E 2 = 14 E1 , or 34 E1 is lost to heat. a) b) x02 + 1 2 mv 2 + 12 kx 2 = 0.0284 J . v02 (0.300 m/s)2 2 = + = 0.014 m. ( 0 . 012 m) ω2 (300 N/m) (0.150 kg) c) ωA = k mA = 0.615 m s ⋅ At the time in question we have x = A cos (ωt + φ) = 0.600 m v = −ωA sin(ωt + φ) = 2.20 m s a = −ω2 A cos (ωt + φ) = −8.40 m s 2 Using the displacement and acceleration equations: − ω2 A cos (ωt + φ) = −ω2 (0.600 m) = −8.40 m s 2 ω2 = 14.0 and ω = 3.742 s −1 To find A, multiply the velocity equation by ω : − ω2 A sin (ωt + φ ) = (3.742 s −1 ) (2.20 m s) = 8.232 m s 2 Next square both this new equation and the acceleration equation and add them: ω4 A2 sin 2 (ωt + φ) + ω4 A2 cos 2 (ωt + φ) = (8.232 m s 2 ) 2 + (−8.40 m s 2 ) 2 = ω 4 A2 sin 2 (ωt + φ) + cos 2 (ωt + φ) ω4 A2 = 67.77 m 2 s 4 + 70.56 m 2 s 4 = 138.3 m 2 s 4 138.3 m 2 s 4 138.3 m 2 s 4 = = 0.7054 m 2 ω4 (3.742 s −1 ) 4 A = 0.840 m A2 = The object will therefore travel 0.840 m − 0.600 m = 0.240 m to the right before stopping at its maximum amplitude. vmax = A k m Use T to find k m : T = 2π m k so k m = (2π T ) 2 = 158 s − 2 Use amax to find A : amax = kA m so A = amax (k m) = 0.0405 m. Then vmax = A k m = 0.509 m s Using k = F0 L0 m= a) k = from the calibration data, ( F0 L0 ) (200 N) (1.25 × 10−1 m) = = 6.00 kg. (2πf ) 2 (2π (2.60 Hz))2 mg (650 kg) (9.80 m s 2 ) = = 531 × 103 N m. ?l (0.120 m) b) T = 2π = m = 2π k l 0.120 m = 2π = 0.695 s. g 9.80 m s 2 a) At the top of the motion, the spring is unstretched and so has no potential energy, the cat is not moving and so has no kinetic energy, and the gravitational potential energy relative to the bottom is 2mgA = 2(4.00 kg)(9.80 m/s 2 ) × (0.050 m) = 3.92 J . This is the total energy, and is the same total for each part. b) U grav = 0, K = 0, so U spring = 3.92 J . c) At equilibrium the spring is stretched half as much as it was for part (a), and so U spring = 14 (3.92 J) = 0.98 J, U grav = 12 (3.92 J) = 1.96 J, and so K = 0.98 J . The elongation is the weight divided by the spring constant, w mg gT 2 l = = 2 = 2 = 3.97 cm . k ω m 4π See Exercise 9.40. a) The mass would decrease by a factor of (1 3) 3 = 1 27 and so 2 the moment of inertia would decrease by a factor of (1 27)(1 3) = (1 243) , and for the same spring constant, the frequency and angular frequency would increase by a factor of 243 = 15.6 . b) The torsion constant would need to be decreased by a factor of 243, or changed by a factor of 0.00412 (approximately). a) With the approximations given, I = mR 2 = 2.72 × 10−8 kg ⋅ m 2 , or 2.7 × 10−8 kg ⋅ m 2 to two figures. b) κ = (2πf ) 2 I = (2π 2 Hz) 2 (2.72 × 10−8 kg ⋅ m 2 ) = 4.3 × 10−6 N ⋅ m rad . Solving Eq. (13.24) for κ in terms of the period, 2  2π  κ=  I T  2  2π  −3 −2 2 =  ((1 2)(2.00 × 10 kg)(2.20 × 10 m) )  1.00 s  = 1.91 × 10− 5 N ⋅ m/rad. I= κ 0.450 N ⋅ m/rad = = 0.0152 kg ⋅ m 2 . 2 (2πf ) (2π (125) (265 s) )2 The equation θ = Θcos (ωt + φ) describes angular SHM. In this problem, φ = 0. a) dθ dt = −ω Θ sin(ω t ) and d 2θ dt 2 = −ω2 Θ cos(ω t ). b) When the angular displacement is Θ, Θ = Θ cos(ω t ) , and this occurs at t = 0, so dθ d 2θ = 0 since sin(0) = 0, and 2 = −ω2Θ, since cos(0) = 1. dt dt Θ Θ 2 , = Θ cos(ω t ), or 12 = cos(ω t ). When the angular displacement is 2 dθ − ωΘ 3 3 d 2θ − ω 2 Θ = , and 2 = , since cos(ω t ) = 1 2. since sin(ω t ) = dt 2 2 dt 2 This corresponds to a displacement of 60° . Using the same procedure used to obtain Eq. (13.29), the potential may be expressed as U = U 0 [(1 + x R0 ) −12 − 2(1 + x R0 ) −6 ]. Note that at r = R0 , U = −U 0 . Using the appropriate forms of the binomial theorem for | x R0 | << 1,  (− 12)(− 13) (x R )2   0  1 − 12( x R0 ) + 2    U ≈ U0   (− 6)(− 7 ) (x R )2   0 − 21 − 6( x R0 ) + 2     36  = U 0 − 1 + 2 x 2  R0   1 = kx 2 − U 0 . 2 where k = 72U 0 / R 2 has been used. Note that terms in u 2 from Eq. (13.28) must be kept ; the fact that the first=order terms vanish is another indication that R0 is an extreme (in this case a minimum) of U. f = 1 2π k 1 = (m 2) 2π 2(580 N/m) = 1.33 × 1014 Hz. (1.008)(1.66 × 10− 27 kg) T = 2π L g , so for a different acceleration due to gravity g ′, T ′ = T g g ′ = (1.60 s ) 9.80 m s 2 3.71 m s 2 = 2.60 s. a) To the given precision, the small=angle approximation is valid. The highest speed is at the bottom of the arc, which occurs after a quarter period, T4 = π2 Lg = 0.25 s. b) The same as calculated in (a), 0.25 s. The period is independent of amplitude. Besides approximating the pendulum motion as SHM, assume that the angle is sufficiently small that the length of the spring does not change while swinging in the arc. Denote the angular frequency of the vertical motion as ω0 = = 12 ω0 = kg 4w k m = kg ω and ω′ = g L , which is solved for L = 4 w k . But L is the length of the stretched spring; the unstretched length is L0 = L − w k = 3 w k = 3(1.00 N ) (1.50 N/m ) = 2.00 m. The period of the pendulum is T = (136 s ) 100 = 1.36 s. Then, g= 4π 2 L 4π 2 (.5 m ) = = 10.67 m s 2 . 2 T2 (1.36 s ) From the parallel axis theorem, the moment of inertia of the hoop about the nail is I = MR 2 + MR 2 = 2MR 2 , so T = 2π 2 R g , with d = R in Eq.(13.39 ). Solving for R, 2 R = gT 2 8π = 0.496 m. For the situation described, I = mL2 and d = L in Eq. (13.39); canceling the factor of m and one factor of L in the square root gives Eq. (13.34). a) Solving Eq. (13.39) for I, 2 2 ( ) T   0.940 s  2 2 I =   mgd =   (1.80 kg ) 9.80 m s (0.250 m ) = 0.0987 kg ⋅ m .  2π   2π  b) The small=angle approximation will not give three=figure accuracy for Θ = 0.400 rad. From energy considerations, 1 mgd (1 − cos Θ ) = IF 2max . 2 Expressing max in terms of the period of small=angle oscillations, this becomes 2 max 2  2π   2π  = 2  (1 − cos Θ ) = 2  (1 − cos(0.40 rad )) = 2.66 rad s .  0.940 s  T  Using the given expression for I in Eq. (13.39), with d=R (and of course m=M), T = 2π 5 R 3 g = 0.58 s. From Eq. (13.39), 2 ( ) 2 T   120 s 100  2 I = mgd   = (1.80 kg ) 9.80 m s 2 (0.200 m )   = 0.129 kg.m . 2π  2π    a) From Eq. (13.43), (2.50 N m ) − (0.90 kg s )2 = 2.47 rad s , so f ′ = ω′ = 0.393 Hz. (0.300 kg ) 4(0.300 kg )2 2π km = 2 (2.50 N m )(0.300 kg ) = 1.73 kg s . ω′ = b) b = 2 From Eq. (13.42) A2 = A1 exp (− 2bm t ). Solving for b, 2m  A1  2(0.050 kg )  0.300 m   = 0.0220 kg s. ln  = ln (5.00 s) t  0.100 m   A2  As a check, note that the oscillation frequency is the same as the undamped frequency to 4.8 ×10 −3%, so Eq. (13.42) is valid. b= a) With φ = 0, x(0) = A. dx   b cos ω′t − ω′ sin ω′t , = Ae − ( b 2 m ) t − vx = b) dt   2m and at t = 0, v = − Ab 2m ; the graph of x versus t near t = 0 slopes down. ax = c)   b 2  ω′b dvx sin ω′t , = Ae − ( b 2 m ) t  2 − ω′2  cos ω ' t + 2m dt    4m and at t = 0,  b2   b2 k ax = A 2 − ω′2  = A 2 − . m  4m   2m (Note that this is (− bv0 − kx0 ) m.) This will be negative if b < 2km , zero if b = 2km and positive if b > 2km . The graph in the three cases will be curved down, not curved, or curved up, respectively. A At resonance, Eq. (13.46) reduces to A = Fmax bωd . a) 31 . b) 2 A1. Note that the resonance frequency is independent of the value of b (see Fig. (13.27)). a) The damping constant has the same units as force divided by speed, or kg ⋅ m s 2 [m s] = [kg s ]. ⋅ b)The units of km are the same as [[kg s 2 ][kg]]1 2 = [kg s ], [ ] the same as those for b. c) ωd2 = k m. (i) bωd = 0.2 k , so A = Fmax (0.2k ) = 5 Fmax k . (ii) bωd = 0.4k , so A = Fmax (0.4 k ) = 2.5 Fmax k , as shown in Fig.(13.27). The resonant frequency is k m = (2.1 × 106 N m) 108 kg) = 139 rad s = 22.2 Hz, and this package does not meet the criterion. a) 2  π rad s    0.100 m   2 3 a = Aω =    = 6.72 × 10 m s .   (3500 rev min )    2  30 rev min    2  π rad s  b) ma = 3.02 × 10 3 N. c) ωA = (3500 rev min ) (.05 m)  = 18.3 m s .  30 rev min  K = 12 mv 2 = ( 12 )(.45 kg)(18.3 m s) 2 = 75.6 J. d) At the midpoint of the stroke, cos( ω t)=0 rad s and so ωt = π 2, thus t = π 2ω. ω = (3500 rev min )( 30π rev min ) = t= 3 2 ( 350 ) s. Then P = K 350 π 3 rad s, so 3 t , or P = 75.6 J ( 2(350) s) = 1.76 × 10 4 W. e) If the frequency doubles, the acceleration and hence the needed force will quadruple (12.1 × 10 3 N). The maximum speed increases by a factor of 2 since v α ω, so the speed will be 36.7 m/s. Because the kinetic energy depends on the square of the velocity, the kinetic energy will increase by a factor of four (302 J). But, because the time to reach the midpoint is halved, due to the doubled velocity, the power increases by a factor of eight (141 kW). Denote the mass of the passengers by m and the (unknown) mass of the car by M. The spring cosntant is then k = mg l . The period of oscillation of the empty car is TE = 2π M k and the period of the loaded car is TL = 2π M +m l 2 = TE2 + (2π ) , so k g TE = TL2 − (2π ) 2 l = 1.003 s. g a) For SHM, the period, frequency and angular frequency are independent of amplitude, and are not changed. b) From Eq. (13.31), the energy is decreased by a factor of 14 . c) From Eq. (13.23), the maximum speed is decreased by a factor of 12 d) Initially, the speed at A1 4 was ω ( A1 2)2 − ( A1 4)2 = 15 4 3 4 ωA1; after the amplitude is reduced, the speed is ωA1 , so the speed is decreased by a factor of 1 5 (this result is valid at x = − A1 4 as well). e) The potential energy depends on position and is unchanged. From the result of part (d), the kinetic energy is decreased by a factor of 15 . This distance L is L = mg k ; the period of the oscillatory motion is m L = 2π , k g which is the period of oscillation of a simple pendulum of lentgh L. T = 2π a) Rewriting Eq. (13.22) in terms of the period and solving, T= 2π A 2 − x 2 = 1.68 s. v b) Using the result of part (a), 2  vT  A 2 −   = 0.0904 m.  2π  c) If the block is just on the verge of slipping, the friction force is its maximum, 2 2 f = ,s n = ,s mg. Setting this equal to ma = mA(2π T ) gives ,s = A(2π T ) g = 0.143. x= a) The normal force on the cowboy must always be upward if he is not holding on. He leaves the saddle when the normal force goes to zero (that is, when he is no longer in contact with the saddle, and the contact force vanishes). At this point the cowboy is in free fall, and so his acceleration is − g ; this must have been the acceleration just before he left contact with the saddle, and so this is also the saddle’s acceleration. b) x = + a (2π f ) 2 = +(9.80 m s 2 ) 2π (1.50 Hz)) 2 = 0.110 m. c) The cowboy’s speed will be the saddle’s speed, v = (2πf ) A2 − x 2 = 2.11 m s. d) Taking t = 0 at the time when the cowboy leaves, the position of the saddle as a function of time is given by Eq. g (13.13), with cos φ = − 2 ; this is checked by setting t = 0 and finding that ω A g a x = ω 2 = − ω 2 . The cowboy’s position is xc = x0 + v0t − ( g 2)t 2 . Finding the time at which the cowboy and the saddle are again in contact involves a transcendental equation which must be solved numerically; specifically, (0.110 m) + (2.11m s)t − (4.90 m s 2 )t 2 = (0.25 m) cos ((9.42 rad s)t − 1.11 rad), which has as its least non=zero solution t = 0.538 s. e) The speed of the saddle is (−2.36 m s) sin (ω t + φ) = 1.72 m s , and the cowboy’s speed is (2.11 m s) − (9.80 m s 2 ) × (0.538 s) = −3.16 m s, giving a relative speed of 4.87 m s (extra figures were kept in the intermediate calculations). The maximum acceleration of both blocks, assuming that the top block does not slip, is amax = kA (m + M ), and so the maximum force on the top block is ( m +mM ) kA = ,smg , and so the maximum amplitude is Amax = ,s (m + M ) g k. (a) Momentum conservation during the collision: mv0 = (2m)V 1 1 V = v0 = (2.00 m s) = 1.00 m s 2 2 Energy conservation after the collision: 1 1 MV 2 = kx 2 2 2 x= MV 2 (20.0 kg)(1.00 m s) 2 = = 0.500 m (amplitude) k 80.0 N m ω = 2πf = k M f = 1 1 80.0 N m k M = = 0.318 Hz 2π 2π 20.0 kg T= 1 1 = = 3.14 s f 0.318 Hz (b) It takes 1 2 period to first return: 12 (3.14 s) = 1.57 s a) m → m 2 Splits at x = 0 where energy is all kinetic energy, E = 12 mv 2 , so E → E 2 k stays same E = 12 kA2 so A = 2 E k Then E → E 2 means A → A 2 T = 2π m k so m → m 2 means T → T 2 b) m → m 2 Splits at x = A where all the energy is potential energy in the spring, so E doesn’t change. E = 12 kA2 so A stays the same. T = 2π m k so T → T 2, as in part (a). c) In example 13.5, the mass increased. This means that T increases rather than decreases. When the mass is added at x = 0, the energy and amplitude change. When the mass is added at x = ± A, the energy and amplitude remain the same. This is the same as in this problem. a) For space considerations, this figure is not precisely to the scale suggested in the problem. The following answers are found algebraically, to be used as a check on the graphical method. 2E 2(0.200 J) A= = = 0.200 m. b) k (10.0 N/m) c) v0 = − E 4 = 0.050 J. d) If U = 12 E , x = 2 K0 m and x0 = 2U 0 k ( ) = 0.141 m. e) From Eq. (13.18), using , − and φ = arctan A 2 v0 = ω x0 0.429 = 0.580 rad . 2K0 m k m 2U 0 k = K0 = 0.429 U0 a) The quantity l is the amount that the origin of coordinates has been moved from the unstretched length of the spring, so the spring is stretched a distance l − x (see Fig. (13.16 ( c ))) and the elastic potential energy is U el = (1 2)k ( l − x) 2 . b) Since 1 2 1 2 kx + ( l ) − k lx + mgx − mgx0 . 2 2 l = mg k , the two terms proportional to x cancel, and U = U el + mg ( x − x0 ) = U= 1 2 1 2 kx + k ( l ) − mgx0 . 2 2 c) An additive constant to the mechanical energy does not change the dependence of the force on x, Fx = − dU dx , and so the relations expressing Newton’s laws and the resulting equations of motion are unchanged. The “spring constant” for this wire is k = f = 1 2π k 1 = m 2π g 1 = l 2π mg l , so 9.80 m s 2 = 11.1 Hz. 2.00 × 10−3 m a) 2TπA = 0.150 m s. b) a = −(2π T ) x = −0.112 m s 2 . The time to go from equilibrium to half the amplitude is sin ωt = (1 2 ), or ωt = π 6 rad, or one=twelfth of a period. The needed time is twice this, or one=sixth of a period, 0.70 s. d) l = mgk = ωg2 = (2 πgr )2 = 4.38 m. 2 Expressing Eq. (13.13) in terms of the frequency, and with φ = 0, and taking two derivatives,  2πt   x = (0.240 m ) cos  1 . 50 s    2π (0.240 m )   2πt   2πt   sin   = −(1.00530 m s ) sin   vx = −  (1.50 s )   1.50 s   1.50 s  2  2π   2πt   2πt   (0.240 m ) cos  = − 4.2110 m s 2 cos . ax = −  1.50 s   1.50 s   1.50 s  ( ) a) Substitution gives x = −0.120 m, or using t = T3 gives x = A cos 120° = b) Substitution gives max = +(0.0200 kg ) 2.106 m s 2 = 4.21 × 10−2 N, in the + x = direction. ( ) ( −A 2 . ) c) t = 2Tπ arccos −3AA 4 = 0.577 s. d) Using the time found in part (c) , v = 0.665 m s (Eq.(13.22) of course gives the same result). a) For the totally inelastic collision, the final speed v in terms of the initial speed V = 2 gh is v =V M m+M ( ) = 2 9.80 m s 2 (0.40 m )( 22..24 ) = 2.57 m s, or 2.6 m s to two figures. b) When the steak hits, the pan is v 02 2 ω Mg k above the new equilibrium position. The ratio is v 2 (k (m + M )) = 2 ghM 2 k (m + M ), and so the amplitude of oscillation is 2 2 ghM 2  Mg  A=   +  k  k (m + M ) (  (2.2 kg ) 9.80 m/s 2 =  (400 N m )  = 0.206 m. )  2  +  2(9.80 m s 2 )(0.40 m)(2.2 kg) 2 (400 N m)(2.4 kg) (This avoids the intermediate calculation of the speed.) c) Using the total mass, T = 2π (m + M ) k = 0.487 s. f = 0.600 Hz, m = 400 kg; f = 12 mk gives k = 5685 N/m. This is the effective force constant of the two springs. a) After the gravel sack falls off, the remaining mass attached to the springs is 225 kg. The force constant of the springs is unaffected, so f = 0.800 Hz. To find the new amplitude use energy considerations to find the distance downward that the beam travels after the gravel falls off. Before the sack falls off, the amount x0 that the spring is stretched at equilibrium is given by mg − kx0 , so x 0 = mg k = (400 kg ) (9.80 m/s 2 ) (5685 N/m ) = 0.6895 m. The maximum upward displacement of the beam is A = 0.400 m. above this point, so at this point the spring is stretched 0.2895 m. With the new mass, the mass 225 kg of the beam alone, at equilibrium the spring is stretched mg k = (225 kg) (9.80 m s 2 ) (5685 N m) = 0.6895 m. The new amplitude is therefore 0.3879 m − 0.2895 m = 0.098 m. The beam moves 0.098 m above and below the new equilibrium position. Energy calculations show that v = 0 when the beam is 0.098 m above and below the equilibrium point. b) The remaining mass and the spring constant is the same in part (a), so the new frequency is again 0.800 Hz. The sack falls off when the spring is stretched 0.6895 m. And the speed of the beam at this point is v = A k m = (0.400 m ) = (5685 N/m ) (400 kg ) = 1.508 m/s. . Take y = 0 at this point. The total energy of the beam at this point, just after the sack falls off, is 2 E = K + U el + U g = 12 (225 kg ) 1.508 m/s2 + 12 (5695 N/m)(0.6895 m ) + 0 = 1608 J. Let this be point 1. Let point 2 be where the beam has moved upward a distance d and where 2 v = 0 . E2 = 12 k (0.6985 m − d ) + mgd . E1 = E2 gives d = 0.7275 m . At this end point of motion the spring is compressed 0.7275 m – 0.6895 m =0.0380 m. At the new equilibrium position the spring is stretched 0.3879 m, so the new amplitude is 0.3789 m + 0.0380 m = 0.426 m. Energy calculations show that v is also zero when the beam is 0.426 m below the equilibrium position. ( ) The pendulum swings through 12 cycle in 1.42 s, so T = 2.84 s. L = 1.85 m. Use T to find g: 2 T = 2π L g so g = L(2π T ) = 9.055 m/s 2 Use g to find the mass M p of Newtonia: g = GM p / Rp2 2πRp = 5.14 × 107 m, so Rp = 8.18 × 106 m mp = gRp2 = 9.08 × 1024 kg G a) Solving Eq. (13.12) for m , and using k = 2 F l 2  T  F  1  40.0 N = 4.05 kg. = m=    2π  l  2π  0.250 m b) t = (0.35)T , and so x = − Asin2π (0.35) = −0.0405 m. Since t > T4 , the mass has already passed the lowest point of its motion, and is on the way up. c) Taking upward forces to be positive, Fspring − mg = − kx, where x is the displacement from equilibrium , so Fspring = −(160 N m)(−0.030 m) + (4.05 kg)(9.80 m/s 2 ) = 44.5 N. Of the many ways to find the time interval, a convenient method is to take φ = 0 in Eq. (13.13) and find that for x = A 2, cos ω t = cos(2πt / T ) = 21 and so t = T / 6 . The time interval available is from − t to t , and T / 3 = 1.17 s. See Problem 12.84; using x as the variable instead of r , GM E g dU GM E m . x, so ω2 = F ( x) = − =− = 3 3 RE RE dx RE The period is then 2π RE 6.38 × 106 m T= = 2π = 2π = 5070 s, ω g 9.80 m/s 2 or 84.5 min. Take only the positive root (to get the least time), so that dx = dt dx A −x 2 ∫ A 0 2 dx A −x 2 2 k A2 − x 2 , or m = k dt m = k t1 dt = m ∫0 arcsin(1) = k (t1 ) m k t1 m π k = t1 , 2 m where the integral was taken from Appendix C. The above may be rearranged to show that t1 = π k = T4 , which is expected. 2 m a) x x 0 0 U = − ∫ F dx = c ∫ x 3 dx = c 4 x . 4 a) From conservation of energy, 12 mv 2 = 4c ( A4 − x 4 ) , and using the technique of Problem 13.77, the separated equation is dx c dt. = 2m A4 − x 4 Integrating from 0 to A with respect to x and from 0 to T 4 with respect to t , A ∫ 0 dx A4 − x 4 c T . 2m 4 = To use the hint, let u = Ax , so that dx = a du and the upper limit of the u − integral is u = 1. Factoring A 2 out of the square root, 1 du 1 1.31 c = = T, ∫ A 0 1 − u4 A 32m which may be expressed as T = motion is not simple harmonic. 7.41 A m c . c) The period does depend on amplitude, and the As shown in Fig. (13.5(b )), v = −vtan sinθ. With vtan = Aω and θ = ωt + φ, this is Eq. (13.15). a) Taking positive displacements and forces to be upwad, 2 n − mg = ma, a = −(2πf ) x, so ( ) n = m g − (2πf ) A cos((2πf ) t + φ ) . 2 a) The fact that the ball bounces means that the ball is no longer in contact with the lens, and that the normal force goes to zero periodically. This occurs when the amplitude of the acceleration is equal to g , or when g = (2πf b ) A. 2 a) For the center of mass to be at rest, the total momentum must be zero, so the momentum vectors must be of equal magnitude but opposite directions, and the momenta can be represented as and − . K tot = 2 × b) p2 p2 = . 2m 2(m 2 ) c) The argument of part (a) is valid for any masses. The kinetic energy is K tot = p2 p2 p 2  m1 + m 2  + = 2m1 2m 2 2  m1 m 2  p2  = .  2(m1 m 2 (m1 + m2 ))  R07  1  dU Fr = − = Α 9  − 2 . dr  r  r  a) b) Setting the above expression for Fr equal to zero, the term in square brackets 1 R7 vanishes, so that r 90 = 2 , or R07 = r 7 , and r = R0 . r U (R0 ) = − c) 7Α = −7.57 × 10−19 J. 8 R0 d) The above expression for Fr can be expressed as A Fr = 2 R0  r  −9  r  −2    −     R0   R0   [ = A (1 + (x R0 ))−9 − (1 + (x R0 ))− 2 2 R0 ≈ A [(1 − 9(x R0 )) − (1 − 2(x R0 ))] R02 = A (− 7 x R0 ) R02 ]  7A = − 3  x.  R0  e) f = 1 1 k m= 2π 2π 7A = 8.39 × 1012 Η z. 3 R0 m 1  1 dU = A 2 − . dx (r − 2 R0 )2   r b) Setting the term in square brackets equal to zero, and ignoring solutions with r < 0 or r > 2 R0 , r = 2 R0 − r , or r = R0 . a) Fr = − c) The above expression for Fr may be written as A Fr = 2 R0 −2  r  −2  r     −  − 2    R0   R0   [ ] = A (1 + (x R0 ))− 2 − (1 − (x R0 ))− 2 2 R0 ≈ A [(1 − 2(x R0 )) − (1 − (− 2)(x R0 ))] R02  4A  = −  3  x,  R0  corresponding to a force constant of k = 4 A R0 . d) The frequency of small oscillations 3 3 would be f = (1 2π ) k m = (1 π ) A mR0 . a) As the mass approaches the origin, the motion is that of a mass attached to a spring of spring constant k, and the time to reach the origin is π2 m k . After passing through the origin, the motion is that of a mass attached to a spring of spring constant 2k and the time it takes to reach the other extreme of the motions is π2 m 2k . The period is twice the sum of these times, or T = π m k (1 + ). The period does not depend on the 1 2 amplitude, but the motion is not simple harmonic. B) From conservation of energy, if the negative extreme is A′, 12 kA2 = 12 (2k ) A′2 , so A′ = − A2 ; the motion is not symmetric about the origin. There are many equivalent ways to find the period of this oscillation. Energy considerations give an elegant result. Using the force and torque equations, taking torques about the contact point, saves a few intermediate steps. Following the hint, take torques about the cylinder axis, with positive torques counterclockwise; the direction of positive rotation is then such that α = Ra , and the friction force f that causes this torque acts in the –x=direction. The equations to solve are then Max = − f − kx, fR = I cmα, a = Rα, Which are solved for ax = kx k =− x, 2 M +I R (3 2) M where I = I cm = (1 2) MR 2 has been used for the combination of cylinders. Comparison with Eq. (13.8) gives T = 2π ω = 2π 3M 2k . Energy conservation during downward swing: m2 gh0 = 12 m2v 2 2 v = 2 gh0 = 2(9.8 m s )(0.100 m) = 1.40 m s Momentum conservation during collision: m2v = (m2 + m3 )V V= (2.00 kg)(1.40 m s) m2v = = 0.560 m s 5.00 kg m2 + m3 Energy conservation during upward swing: 1 MV 2 2 (0.560 m s) 2 hf = V 2 2 g = = 0.0160 m = 1.60 cm 2(9.80 m s 2 ) Mghf = 48.4 cm 50.0 cm θ = 14.5° cos θ = f = 1 g 1 9.80 m s 2 = = 0.705 Hz 2π l 2π 0.500 m T = 2π I mgd , m = 3M d = ycg = d= m1 y1 + m2 y2 m1 + m2 2 M ([1.55 m ] 2 ) + M (1.55 m + [1.55 m] 2) = 1.292 m 3M I + I1 + I 2 I1 = 1 3 (2M )(1.55 m )2 = (1.602 m 2 )M I 2, cm = 121 M (1.55 m ) 2 The parallel=axis theorem (Eq. 9.19) gives 2 I 2 = I 2,cm + M (1.55 m + [1.55 m] 2) = 5.06 m 2 M ( ) I = I1 + I 2 = 7.208 m 2 M ( ) (7.208 m ) M = 2.74 s. (3M )(9.80 m s )(1.292 m ) 2 Then T = 2π I mgd = 2π 2 This is smaller than T = 2.9 s found in Example 13.10. The torque on the rod about the pivot (with angles positive in the direction indicated in the figure) is τ = −(k L2 θ ) L2 . Setting this equal to the rate of change of angular momentum, Iα = I d 2θ dt 2 , d 2θ L2 4 3k = − k θ = − θ, 2 dt I M where the moment of inertia for a slender rod about its center, I = 121 ML2 has been used. It follows that ω2 = 3K M , and T = 2π ω = 2π M 3k . The period of the simple pendulum (the clapper) must be the same as that of the bell; equating the expression in Eq. (13.34) to that in Eq. (13.39) and solving for L gives L = Ι md = (18.0 kg ⋅ m 2 ) ((34.0 kg)(0.60 m)) = 0.882 m. Note that the mass of the bell, not the clapper, is used. As with any simple pendulum, the period of small oscillations of the clapper is independent of its mass. The moment of inertia about the pivot is 2(1 3) ML2 = (2 3) ML2 , and the center of gravity when balanced is a distance d = L (2 2 ) below the pivot (see Problem 8.95). From Eq. (13.39), the frequency is f = 1 1 = T 2π 3g 1 = 4 2 L 4π 3g ⋅ 2L a) L = g (T 2π ) 2 = 3.97 m. b)There are many possibilities. One is to have a uniform thin rod pivoted about an axis perpendicular to the rod a distance d from its center. Using the desired period in Eq. (13.39) gives a quadratic in d, and using the maximum size for the length of the rod gives a pivot point a distance of 5.25 mm, which is on the edge of practicality. Using a “dumbbell,” two spheres separated by a light rod of length L gives a slight improvement to d=1.6 cm (neglecting the radii of the spheres in comparison to the length of the rod; see Problem 13.94). Using the notation 2bm = γ, mk = ω2 and taking derivatives of Eq. (13.42) (setting the phase angle φ = 0 does not affect the result), x = Αe − γt cosω′ t vx = − Αe − γt (ω′ sin ω′ t + γ cos ω′ t ) ax = − Αe − γt ((ω′ 2 − γ 2 ) cos ω′ t − 2ω′ γ sin ω′ t ). Using these expression in the left side of Eq. (13.41), − kx − bvx = Αe − γt ( − k cos ω′ t + (2γ m)ω′ sin ω′ t + 2mγ 2 cos ω′ t ) = mΑ e − γt ((2γ 2 − ω 2 ) cos ω′ t + 2γω′ sin ω′ t ). The factor (2γ 2 − ω 2 ) is γ 2 − ω′2 (this is Eq. (13.43)), and so − kx − bvx = mΑ e − γt ((γ 2 − ω′ 2 ) cos ω′ t + 2γω′ sin ω′ t ) = max . a) In Eq. (13.38), d=x and from the parallel axis theorem, I = m( L 12 + x 2 ) , so ω2 = ( L2 12gx)+ x 2 . b) Differentiating the ratio ω2 g = 2 x ( L2 12 ) + x 2 with respect to x and setting the result equal to zero gives 1 2 x2 = , or 2 x 2 = x 2 + L2 12, 2 2 2 2 2 ( L 12) + x (( L 12) + x ) Which is solved for x = L 12. ω2 L 12 6 3 = = = c) When x is the value that maximizes ω the ratio 2 g 2 L 12 L 12 L, ( so the length is L = 3g = 0.430 m. ω2 ) is a) From the parellel axis theorem, the moment of inertia about the pivot point M L2 + (2 5)R 2 . ( ) Using this in Eq. (13.39), With d = L gives. T = 2π L2 + (2 5)R 2 L = 2π 1 + 2 R 2 5L2 = Tsp 1 + 2 R 2 5 L2 . gL g b) Letting 1 + 2 R 2 5L2 = 1.001 and solving for the ratio L R (or approximating the square root as 1 + R 2 5L2 ) gives c) (14.1)(1.270 cm ) = 18.0 cm. L R = 14.1. a) The net force on the block at equilibrium is zero, and so one spring (the one with k1 = 2.00 Ν m ) must be stretched three times as much as the one with k2 = 6.00 Ν m . The sum of the elongations is 0.200 m, and so one spring stretches 0.150 m and the other stretches 0.050 m, and so the equilibrium lengths are 0.350 m and 0.250 m. b) There are many ways to approach this problem, all of which of course lead to the result of Problem 13.96(b). The most direct way is to let x1 = 0.150 m and x2 = 0.050 m, the results of part (a). When the block in Fig.(13.35) is displaced a distance x to the right, the net force on the block is − k1 ( x1 + x ) + k 2 ( x 2 − x ) = [k1 x1 − k 2 x 2 ] − (k1 + k 2 )x. From the result of part (a), the term in square brackets is zero, and so the net force is − (k1 + k 2 )x, the effective spring constant is k eff = k1 + k 2 and the period of vibration is T = 2π 0.100 kg 8.00 Ν m = 0.702 s. In each situation, imagine the mass moves a distance x, the springs move distances x1 and x2 , with forces F1 = −k1 x1 , F2 = − k2 x2 . a) x = x1 = x2 , F = F1 + F2 = −(k1 + k2 ) x, so keff = k1 + k2 . b) Despite the orientation of the springs, and the fact that one will be compressed when the other is extended, x = x1 + x2 , and the above result is still valid; k eff = k1 + k 2 . c) For massless springs, the force on the block must be equal to the tension in any point F F of the spring combination, and F = F1 = F2 , and so x1 = − , x2 = − , and k2 k1 and κeff 1 1 k + k2 x = − +  F = − 1 F k1k2  k1 k2  κ κ = κ11+ κ22 . d) The result of part (c) shows that when a spring is cut in half, the effective spring constant doubles, and so the frequency increases by a factor of 2. a) Using the hint, g 1   T + T ≈ 2π L  g −1 2 − g −3 2 g  = T − T ,   2 2g so T = −(1 2)(T g ) g. This result can also be obtained from T 2 g = 4π 2 L, from which (2T T )g + T 2 g = 0. Therefore, ( T T = − 12  2 T 2 and g + g = g 1 −  = 9.80 m s T   ) g g . b) The clock runs slow; T > 0,  2(4.00 s )  1 −  = 9.7991 m s 2 .  (86,400 s )  g <0 Denote the position of a piece of the spring by l ; l = 0 is the fixed point and l = L is the moving end of the spring. Then the velocity of the point corresponding to l , l denoted u, is u (l ) = v (when the spring is moving, l will be a function of time, and so L u is an implicit function of time). a) dm = ML dl , and so dK = 1 1 Mv 2 2 l dl , dm u 2 = 2 2 L3 and K = ∫ dK = L Mv 2 2 Mv 2 l dl = . 2 L3 ∫0 6 + kx dx = 0, or ma + kx = 0, which is Eq. (13.4). c) m is replaced by b) mv dv dt dt ω= 3k M and M ′ = M 3 M 3 , so . a) With I = (1 3)ML2 and d = L 2 in Eq. (13.39 ), T0 = 2π 2 L 3 g. With the (( ) ) addedmass, I = M L2 3 + y 2 , m = 2M and d = (L 4) + y 2, T = 2π × (L 2 3+ y 2 ) (g (L 2 + y )) and r= T = T0 L2 + 3 y 2 . L2 + 2 yL b) From the expression found in part a), T = T0 when y = 23 L. At this point, a simple pendulum with length y would have the same period as the meter stick without the added mass; the two bodies oscillate with the same period and do not affect the other’s motion. Let the two distances from the center of mass be d1 and d 2 . There are then two relations of the form of Eq. (13.39); with I1 = I cm + md12 and I 2 = I cm + md 22 , these relations may be rewritten as ( = 4π (I ) + md ). mgd1T 2 = 4π 2 I cm + md12 mgd 2T 2 2 cm 2 2 Subtracting the expressions gives ( ) mg (d1 − d 2 )T 2 = 4π 2 m d12 − d 22 = 4π 2 m(d1 − d 2 )(d1 + d 2 ), and dividing by the common factor of m(d1 − d 2 ) and letting d1 + d 2 = L gives the desired result. a) The spring, when stretched, provides an inward force; using ω′2 l for the magnitude of the inward radial acceleration, kl0 mω′l = k (l − l0 ), or l = . k − mω′2 b) The spring will tend to become unboundedly long. Let r = R0 + x, so that r − R0 = x and F = A[e −2bx − e − bx ]. When x is small compared to b −1 , expanding the exponential function gives F ≈ A [(1 − 2bx ) − (1 − bx )] = − Abx, corresponding to a force constant of Ab = 579.2 N m or 579 N m to three figures. This is close to the value given in Exercise 13.40. Capítulo 14 w = mg = ρVg ( ) ( )( ) = 7.8 × 103 kg m3 (0.858 m )π 1.43 × 10−2 m 9.80 m s 2 = 41.8 N or 42 N to two places. A cart is not necessary. ρ= ρ = Vm = ( 2 ) ) m m 7.35 × 1022 kg = 4 3= = 3.33 × 103 kg m 3 . 3 6 4 V 3 πr π 1.74 × 10 m 3 (0.0158 kg ) (5.0×15.0×30.0 ) mm 3 ( 3 = 7.02 × 103 kg m . You were cheated. The length L of a side of the cube is 1 1 3  m 3  40.0 kg  = 12.3 cm. L = V =   =  3 3  ρ  21.4 × 10 kg m  1 3 m = ρV = 43 πr 3 ρ Same mass means ra3 ρa = r13 ρ1 (a = aluminum, l = lead ) 13 ra  ρ1  =  r1  ρa  13  11.3 × 103   =  3   2.7 × 10  D= a) = 1.6 1.99 × 1030 kg 1.99 × 1030 kg M sun = = 3 4 1.412 × 1027 m3 Vsun π 6.96 × 108 m 3 ( ) = 1.409 × 10 3 kg m 3 D= b) 1.99 × 1030 kg 4 3 ( π 2.00 × 104 m ) 3 = 1.99 × 1030 kg = 0.594 × 1017 kg m 3 3.351 × 1013 m 3 = 5.94 × 1016 kg m 3 p − p0 = ρgh h= p − p0 1.00 × 105 Pa = = 9.91m ρg (1030 kg m 3 ) (9.80 m s 2 ) The pressure difference between the top and bottom of the tube must be at least 5980 Pa in order to force fluid into the vein: ρgh = 5980 Pa h= 5980 Pa 5980 N m 2 = = 0.581m gh (1050 kg m 3 ) (9.80 m s 2 ) ( )( ) ( )( ) a) ρgh = 600 kg m 3 9.80 m s 2 (0.12 m ) = 706 Pa. b) 706 Pa + 1000 kg m3 9.80 m s 2 (0.250 m ) = 3.16 × 103 Pa. a) The pressure used to find the area is the gauge pressure, and so the total area is (16.5 × 10 3 N) = 805 cm 2 ⋅ 3 (205 × 10 Pa ) b) With the extra weight, repeating the above calculation gives 1250 cm 2 . 2 a) ρgh = (1.03 × 103 kg m 3 )(9.80 m s )(250 m) = 2.52 × 106 Pa. b) The pressure difference is the gauge pressure, and the net force due to the water and the air is (2.52 × 106 Pa )(π (0.15 m) 2 ) = 1.78 × 105 N. p = ρgh = (1.00 × 103 kg m3 )(9.80 m s 2 )(640 m) = 6.27 × 106 Pa = 61.9 atm. a) pa + ρgy2 = 980 × 102 Pa + (13.6 × 103 kg m3 )(9.80 m s 2 )(7.00 × 10−2 m) = 1.07 × 10 5 Pa. b) Repeating the calcultion with y = y 2 − y1 = 4.00 cm instead of y 2 gives 1.03 × 10 5 Pa. c) The absolute pressure is that found in part (b), 1.03 × 10 5 Pa. d) ( y2 − y1 ) ρg = 5.33 × 103 Pa (this is not the same as the difference between the results of parts (a) and (b) due to roundoff error). ρgh = (1.00 × 103kg m3 )(9.80 m s 2 )(6.1 m) = 6.0 × 10 4 Pa. With just the mercury, the gauge pressure at the bottom of the cylinder is p = p 0 + p m ghm⋅ With the water to a depth hw , the gauge pressure at the bottom of the cylinder is p = p0 + ρm ghm + pw ghw . If this is to be double the first value, then ρw ghw = ρm ghm. hw = hm ( ρm ρw ) = (0.0500 m)(13.6 ×103 1.00 × 103 ) = 0.680 m The volume of water is V = hA = (0.680 m)(12.0 × 10−4 m 2 ) = 8.16 × 10−4 m3 = 816 cm3 a) Gauge pressure is the excess pressure above atmospheric pressure. The pressure difference between the surface of the water and the bottom is due to the weight of the water and is still 2500 Pa after the pressure increase above the surface. But the surface pressure increase is also transmitted to the fluid, making the total difference from atmospheric 2500 Pa+1500 Pa = 4000 Pa. b) The pressure due to the water alone is 2500 Pa = ρgh. Thus h= 2500 N m 2 = 0.255 m (1000 kg m3 ) (9.80 m s 2 ) To keep the bottom gauge pressure at 2500 Pa after the 1500 Pa increase at the surface, the pressure due to the water’s weight must be reduced to 1000 Pa: 1000 N m 2 h= = 0.102 m (1000 kg m 3 )(9.80 m s 2 ) Thus the water must be lowered by 0.255 m − 0.102 m = 0.153 m The force is the difference between the upward force of the water and the downward forces of the air and the weight. The difference between the pressure inside and out is the gauge pressure, so F = ( ρgh) A − w = (1.03 × 103 ) (9.80 m s 2 ) (30 m) (0.75 m 2 ) − 300 N = 2.27 × 105 N. [130 × 10 = 1.79 × 10 5 N. 3 ] Pa + (1.00 × 103 kg m3 )(3.71 m s 2 )(14.2 m) − 93 × 103 Pa (2.00 m 2 ) The depth of the kerosene is the difference in pressure, divided by the product ρg = mg V , (16.4 × 103 N) (0.0700 m 2 ) − 2.01 × 105 Pa h= = 4.14 m. (205 kg)(9.80 m s 2 ) (0.250 m3 ) p= F mg (1200 kg)(9.80 m s 2 ) = = = 1.66 × 105 Pa = 1.64 atm. A π ( d 2) 2 π (0.15 m)2 The buoyant force must be equal to the total weight; ρwaterVg = ρiceVg + mg , so V= m 45.0 kg = = 0.563 m 3 , ρwater − ρice 1000 kg m 3 − 920 kg m3 or 0.56 m 3 to two figures. The buoyant force is B = 17.50 N − 11.20 N = 6.30 N, and V= B ρwater g = (6.30 N) = 6.43 × 10− 4 m3 . 3 2 (1.00 × 10 kg m )(9.80 m s ) 3 The density is ρ= m w g w  17.50  3 3 = = ρwater = (1.00 × 103 kg m 3 )   = 2.78 × 10 kg m . 6 . 30 V B ρwater g B   a) The displaced fluid must weigh more than the object, so ρ < ρ fluid . b) If the ship does not leak, much of the water will be displaced by air or cargo, and the average density of the floating ship is less than that of water. c) Let the portion submerged have ρ volume V, and the total volume be V0 . Then, ρVo = ρfluid V , so VV0 = ρfluid ⋅ The fraction above the fluid is then 1 − P Pfluid . If p → 0, the entire object floats, and if ρ → ρ fluid , none of the object is above the surface. d) Using the result of part (c), 1− ρ ρ fluid = 1− (0.042 kg) (5.0 × 4.0 × 3.0 × 10 <6 m 3 ) = 0.32 = 32%. 1030 kg m 3 a) B = ρwater gV = (1.00 × 103 kg m3 )(9.80 m s 2 )(0.650 m3 ) = 6370 N. N < 900 N = 558 kg. b) m = wg = B −g T = 6370 9.80 m s 2 c) (See Exercise 14.23.) If the submerged volume is V ′, V′ = w ρwater g and V′ w 5470 N = = = 0.859 = 85.9%. V ρwater gV 6370 N a) ρoil ghoil = 116 Pa. (( ) ( ) )( ) b) 790 kg m3 (0.100 m ) + 1000 kg m3 (0.0150 m ) 9.80 m s 2 = 921 Pa. c) 2 w ( pbottom − ptop )A (805 Pa )(0.100 m ) m= = = = 0.822 kg. g g 9.80 m s 2 The density of the block is p = ( 0.822 kg (0.10 m )3 ( ) = 822 m 3 . Note that is the same as the average kg ) density of the fluid displaced, (0.85) 790 kg m 3 + (0.15) (1000 kg m3 ) . a) Neglecting the density of the air, V= (89 N ) m wg w = = = = 3.36 × 10− 3 m 3 , 2 3 3 ρ ρ gρ 9.80 m s 2.7 × 10 kg m ( )( ) or 3.4 × 10−3 m3 to two figures.   ρ  1.00  b) T = w −B = w − gρwaterV = ω1 − water  = (89 N )1 −  = 56.0 N. ρaluminum  2.7    a) The pressure at the top of the block is p = p 0 + ρgh, where h is the depth of the top of the block below the surface. h is greater for block Β , so the pressure is greater at the top of block Β . b) B = ρ flVobj g . The blocks have the same volume Vobj so experience the same buoyant force. c) T − w + B = 0 so T = w −B. w = ρVg. The object have the same V but ρ is larger for brass than for aluminum so w is larger for the brass block. B is the same for both, so T is larger for the brass block, block B. The rock displaces a volume of water whose weight is 39.2 N < 28.4 N = 10.8 N. The mass of this much water is thus 10.8 N 9.80 m s 2 = 1.102 kg and its volume, equal to the rock’s volume, is 1.102 kg = 1.102 × 10−3 m 3 1.00 × 103 kg m3 The weight of unknown liquid displaced is 39.2 N − 18.6 N = 20.6 N, and its mass is 20.6 N 9.80 m s 2 = 2.102 kg. The liquid’s density is thus 2.102 kg 1.102 × 10−3 m3 = 1.91 × 103 kg m3 , or roughly twice the density of water. v1 Α1 = v2 Α2 , v2 = v1 ( Α1 Α2 ) Α1 = π (0.80 cm)2 , Α2 = 20π (0.10 cm)2 v2 = (3.0 m s) π (0.80)2 = 9.6 m s 20π (0.10) 2 v2 = v1 Α1 (3.50 m s)(0.0700 m 2 ) 0.245 m3 s = = ⋅ Α2 Α2 Α2 a) (i) Α2 = 0.1050 m 2 , v2 = 2.33 m s. (ii) Α2 = 0.047 m 2 , v2 = 5.21 m s. b) v1 Α1t = υ2 Α2t = (0.245 m 3 s) (3600 s) = 882 m3 . a) v = dV dt (1.20 m3 s) = = 16.98. A π (0.150 m) 2 b) r2 = r1 v1 v2 = (dV dt ) πv2 = 0.317 m. a) From the equation preceding Eq. (14.10), dividing by the time interval dt gives Eq. (14.12). b) The volume flow rate decreases by 1.50% (to two figures). The hole is given as being “small,”and this may be taken to mean that the velocity of the seawater at the top of the tank is zero, and Eq. (14.18) gives v = 2( gy + ( p ρ)) 2((9.80 m s 2 )(11.0 m) + (3.00)(1.013 × 10 5 Pa) (1.03 × 10 3 kg m )) = 28.4 m s. 3 Note that y = 0 and p = pa were used at the bottom of the tank, so that p was the given gauge pressure at the top of the tank. a) From Eq. (14.18), v = 2 gh = 2(9.80 m s 2 )(14.0 m) = 16.6 m s. b) vΑ = (16.57 m s)(π (0.30 × 10−2 m)2 ) = 4.69 × 10−4 m3 s. Note that an extra figure was kept in the intermediate calculation. The assumption may be taken to mean that v1 = 0 in Eq. (14.17). At the maximum height, v2 = 0, and using gauge pressure for p1 and p2 , p2 = 0 (the water is open to the atmosphere), p1 = ρgy2 = 1.47 × 105 Pa. Using v2 = 14 v1 in Eq. (14.17), p2 = p1 +   15  1 ρ(v12 − v22 ) + ρg ( y1 − y2 ) = p1 + ρ  υ12 + g ( y1 − y2 ) 2   32   15  = 5.00 × 104 Pa + (1.00 × 103 kg m3 )  (3.00 m s) 2 + (9.80 m s 2 )(11.0 m)   32  5 = 1.62 × 10 Pa. Neglecting the thickness of the wing (so that y1 = y2 in Eq. (14.17)), the pressure difference is p = (1 2) ρ(v22 − v12 ) = 780 Pa. The net upward force is then (780 Pa) × (16.2 m 2 ) − (1340 kg)(9.80 m s 2 ) = −496 N. a) 0.355 kg 0.355×10 −3 m 3 1.30 kg s 1000 kg m 3 from ( 220 )(0.355 kg ) 60.0 s = 1.30 kg s. b) The density of the liquid is = 1000 kg m3 , and so the volume flow rate is = 1.30 × 10−3 m3 s = 1.30 L s. ( 220 )(0.355 L ) 60.0 s This result may also be obtained −3 3 1.30×10 m s = 1.30 L s. c) v1 = 2.00×10 −4 m 2 = 6.50 m s, v2 = v1 4 = 1.63 m s. 1 d) p1 = p2 + ρ v22 − v12 + ρg ( y2 − y1 ) 2 2 2 = 152 kPa + (1 2) 1000 kg m 3 (1.63 m s ) − (6.50 m s ) ( ( + 1000 kg = 119 kPa ) ( )( m )(9.80 m s ) (− 1.35 m ) 3 ) 2 The water is discharged at a rate of v1 = 4.65×10 −4 m 3 s 1.32×10 −3 m 2 = 0.352 m s. The pipe is given as horizonatal, so the speed at the constriction is v2 = v12 + 2 p ρ = 8.95 m s, keeping an extra figure, so the cross<section are at the constriction is 4.65×10 −4 m 3 s = 5.19 × 10− 5 m 2 , and the radius is r = A π = 0.41 cm. 8.95 m s From Eq. (14.17), with y1 = y 2 , p2 = p1 + 1 1  v2  3 ρ v12 − v22 = p1 + ρ v12 − 1  = p1 + ρv12 2 2  4 8 ( ) ( ) 3 2 1.00 × 10 3 kg m 3 (2.50 m s ) = 2.03 × 10 4 Pa, 8 v where the continutity relation v 2 = 1 has been used. 2 = 1.80 × 10 4 Pa + Let point 1 be where r1 = 4.00 cm and point 2 be where r2 = 2.00 cm. The volume flow rate has the value 7200 cm 3 s at all points in the pipe. v1 A1 = v1πr12 = 7200 cm3 , so v1 = 1.43 m s v2 A2 = v2πr22 = 7200 cm3 , so v2 = 5.73 m s 1 1 p1 + ρgy1 + ρv12 = p2 + ρgy2 + ρv22 2 2 1 y1 = y2 and p2 = 2.40 × 105 Pa, so p2 = p1 + ρ(v12 − v22 ) = 2.25 × 105 Pa 2 2 a) The cross<sectional area presented by a sphere is π D4 , therefore F = ( p0 − p )π D4 . b) The force on each hemisphere due to the atmosphere is 2 ( π 5.00 × 10− 2 m ) (1.013 × 10 2 5 ) Pa (0.975) = 776Ν. ( )( )( ) )( ( )( a) ρgh = 1.03 × 103 kg m 3 9.80 × m s 2 10.92 × 103 m = 1.10 × 108 Pa. b) The fractional change in volume is the negative of the fractional change in density. The density at that depth is then ( ρ = ρ0 (1 + k p ) = 1.03 × 103 kg m3 1 + 1.16 × 108 Pa 45.8 × 10−11 Pa −1 )) = 1.08 × 10 kg m , 3 3 A fractional increase of 5.0%. Note that to three figures, the gauge pressure and absolute pressure are the same. a) The weight of the water is ( )( ρgV = 1.00 × 103 kg m3 9.80 m s 2 ) ((5.00 m)(4.0 m)(3.0 m )) = 5.88 × 10 5 N, or 5.9 × 10 5 N to two figures. b) Integration gives the expected result the force is what it would be if the pressure were uniform and equal to the pressure at the midpoint; d 2 = 1.00 × 103 kg m3 9.80 m s 2 F = ρgA ( )( ) ((4.0 m )(3.0 m ))(1.50 m) = 1.76 × 10 5 N, or1.8 × 10 5 N to two figures. Let the width be w and the depth at the bottom of the gate be H . The force on a strip of vertical thickness dh at a depth h is then dF = ρgh(wdh ) and the torque about the hinge is dτ = ρgwh(h − H 2 )dh; integrating from h = 0 to h = H gives τ = ρgωH 3 12 = 2.61 × 10 4 N ⋅ m. a) See problem 14.45; the net force is ∫ dF from h = 0 to h = H , F = ρgωH 2 2 = ρgAH 2, where A = ωH . b) The torque on a strip of vertical thickness dh about the bottom is dτ = dF (H − h ) = ρgwh(H − h )dh, and integrating from h = 0 to h = H gives τ = ρgwH 3 6 = ρgAH 2 6. c) The force depends on the width and the square of the depth, and the torque about the bottom depends on the width and the cube of the depth; the surface area of the lake does not affect either result (for a given width). The acceleration due to gravity on the planet is p p g= = m ρd V d and so the planet’s mass is gR 2 pVR 2 M = = G mGd The cylindrical rod has mass M , radius R, and length L with a density that is proportional to the square of the distance from one end, ρ = Cx 2 . a) M = ∫ ρdV = ∫ Cx 2 dV . The volume element dV = πR 2 dx. Then the integral becomes M = ∫ 0L Cx 2πR 2 dx. Integrating gives M = CπR 2 ∫ 0L x 2 dx = CπR 2 for C , C = 3M πR 2 L3 . b) The density at the x = L end is ρ = Cx 2 = L3 . Solving 3 ( )(L ) = ( ). The denominator is 3M πR 2 L3 2 3M πR 2 L just the total volume V , so ρ = 3M V , or three times the average density, M V . So the average density is one<third the density at the x = L end of the rod. a) At r = 0, the model predicts ρ = A = 12,700 kg m 3 and at r = R, the model predicts ρ = A − BR = 12,700 kg m 3 − (1.50 × 10 −3 kg m 4 )(6.37 × 10 6 m) = 3.15 × 10 3 kg m 3 . b), c) R  AR 3 BR 4   4πR 3   3BR    A − − =  M = ∫ dm = 4π ∫ [ A − Br ]r 2 dr = 4π   4   3  4   3 0  4π (6.37 × 106 m)3   3(1.50 × 10−3 kg m 4 )(6.37 × 106 m)   12,700 kg m3 − =   3 4    24 = 5.99 × 10 kg, which is within 0.36% of the earth’s mass. d) If m (r ) is used to denote the mass contained in a sphere of radius r , then g = Gm ( r ) r 2 . Using the same integration as that in part (b), with an upper limit of r instead of R gives the result. e) g = 0 at r = 0, and g at r = R, g = Gm( R) R 2 = (6.673 × 10−11 N ⋅ m 2 kg 2 ) (5.99 × 1024 kg) (6.37 × 106 m)2 = 9.85 m s 2 . f) 3Br 2   4πG   3Br  dg  4πG  d  ; = =  A −   Ar −  4   3  2  dr  3  dr  setting ths equal to zero gives r = 2 A 3B = 5.64 × 10 6 m , and at this radius  3   2 A   4πG  2 A   g =     A −   B  4   3B    3  3B   4πGA2 = 9B 4π (6.673 × 10−11 N ⋅ m 2 kg 2 ) (12,700 kg m 3 ) 2 = 10.02 m s 2 . = −3 4 9(1.50 × 10 kg m ) a) Equation (14.4), with the radius r instead of height y, becomes dp = − ρg dr = − ρgs (r R)dr. This form shows that the pressure decreases with increasing radius. Integrating, with p = 0 at r = R, p=− ρg s R r ∫ r dr = R ρg s R ∫ R r r dr = b) Using the above expression with r = 0 and ρ = p(0) = ρg s 2 ( R − r 2 ). 2R M V = 3M 4πR 3 , 3(5.97 × 1024 kg)(9.80 m s 2 ) = 1.71 × 1011 Pa. 8π (6.38 × 106 m)2 c) While the same order of magnitude, this is not in very good agreement with the estimated value. In more realistic density models (see Problem 14.49 or Problem 9.99), the concentration of mass at lower radii leads to a higher pressure. a) ρwater ghwater = (1.00 × 103 kg m3 )(9.80 m s 2 )(15.0 × 10 −2 m) = 1.47 × 103 Pa. b) The gauge pressure at a depth of 15.0 cm − h below the top of the mercury column must be that found in part (a); ρHg g (15.0 cm − h) = ρwater g (15.0 cm), which is solved for h = 13.9 cm. Following the hint, h F = ∫ ( ρgy )(2πR)dy = ρgπRh 2 o where R and h are the radius and height of the tank (the fact that 2 R = h is more or less coincidental). Using the given numerical values gives F = 5.07 × 10 8 N. For the barge to be completely submerged, the mass of water displaced would need to be ρwaterV = (1.00 × 103 kg m 3 )(22 × 40 × 12 m3 ) = 1.056 × 107 kg. The mass of the barge itself is (7.8 × 103 kg m3 ) ((2(22 + 40) × 12 + 22 × 40) × 4.0 × 10−2 m3 ) = 7.39 × 105 kg, so the barge can hold 9.82 × 10 6 kg of coal. This mass of coal occupies a solid volume of 6.55 × 10 3 m 3 , which is less than the volume of the interior of the barge (1.06 × 10 4 m 3 ), but the coal must not be too loosely packed. The difference between the densities must provide the “lift” of 5800 N (see Problem 14.59). The average density of the gases in the balloon is then ρave = 1.23 kg m 3 − (5800 N) = 0.96 kg m3 . 2 3 (9.80 m s )(2200 m ) a) The submerged volume V ′ is w ρ water g , so V ′ w ρwater g m (900 kg) = = = = 0.30 = 30% ⋅ 3 V V ρwaterV (1.00 × 10 kg m3 ) (3.0 m3 ) b) As the car is about to sink, the weight of the water displaced is equal to the weight of the car plus the weight of the water inside the car. If the volume of water inside the car is V ′′ , Vρwater g = w + V ′′pwater g , or V ′′ w = 1− = 1 − 0.30 = 0.70 = 70% ⋅ V Vpwater g a) The volume displaced must be that which has the same weight and mass as the ice, 1.009.70gmgmcm 3 = 9.70 cm3 (note that the choice of the form for the density of water avoids conversion of units). b) No; when melted, it is as if the volume displaced by the 9.70 gm of melted ice displaces the same volume, and the water level does not change. c) 9.70 gm 1.05 gm cm 3 = 9.24 cm3 ⋅ d) The melted water takes up more volume than the salt water displaced, and so 0.46 cm3 flows over. A way of considering this situation (as a thought experiment only) is that the less dense water “floats” on the salt water, and as there is insufficient volume to contain the melted ice, some spills over. The total mass of the lead and wood must be the mass of the water displaced, or VPb ρ Pb + Vwood ρ wood = (VPb + Vwood ) ρ water ; solving for the volume VPb , VPb = Vwood ρ water − ρ wood ρ Pb − ρ water = (1.2 × 10− 2 m3 ) 1.00 × 103 kg m3 − 600 kg m3 11.3 × 103 kg m3 − 1.00 × 103 kg m3 = 4.66 × 10−4 m3 , which has a mass of 5.27 kg. The fraction f of the volume that floats above the fluid is f = 1 − ρ ρ , fluid where ρ is the average density of the hydrometer (see Problem 14.23 or Problem 14.55), 1 . Thus, if two fluids are observed to have which can be expressed as ρ fluid = ρ 1− f 1 − f1 . In this form, it’s clear that a larger f 2 floating fraction f1 and f 2 , ρ2 = ρ1 1 − f2 corresponds to a larger density; more of the stem is above the fluid. Using cm 2 ) ( 3.20 cm)(0.400 cm 2 ) f1 = (8.00 (cm)(0.400 = 0 . 242 , f = = 0.097 gives 3 2 13.2 cm ) (13.2 cm 3 ) ρalcohol = (0.839) ρwater = 839 kg m 3 . a) The “lift” is V ( ρair − ρH 2 ) g , from which V= 120,000 N = 11.0 × 10 3 m 3 . (1.20 kg m − 0.0899 kg m 3 )(9.80 m s 2 ) 3 b) For the same volume, the “lift” would be different by the ratio of the density differences,  ρ − ρHe (120,000 N) air  ρair − ρH 2    = 11.2 × 104 N.   This increase in lift is not worth the hazards associated with use of hydrogen. M . ρA b) The buoyant force is ρgA( L + x) = Mg + F , and using the result of part (a) and F . solving for x gives x = ρgA a) Archimedes’ principle states ρgLA = Mg , so L = c) The “spring constant,” that is, the proportionality between the displacement x and the applied force F, is k = ρgA, and the period of oscillation is M M = 2π . k ρgA T = 2π a) x = (70.0 kg ) w mg m = = = = 0.107 m. 3 ρgA ρgA ρA 1.03 × 10 kg m3 π (0.450 m )2 ( ) b) Note that in part (c) of Problem 14.60, M is the mass of the buoy, not the mass of the man, and A is the cross<section area of the buoy, not the amplitude. The period is then T = 2π (1.03 × 10 (950 kg ) 3 kg m 3 )(9.80 m s )π (0.450 m ) 2 2 = 2.42 s. To save some intermediate calculation, let the density, mass and volume of the life preserver be ρ 0 , m and v, and the same quantities for the person be ρ1 , M and V . Then, equating the buoyant force and the weight, and dividing out the common factor of g, ρwater ((0.80 )V + v ) = ρ0v + ρ1V , Eliminating V in favor of ρ1 and M , and eliminating m in favor of ρ0 and v,   M ρ0v + M = ρwater  (0.80) + v . ρ1   Solving for ρ 0 ,    1 M ρ0 =  ρwater  (0.80) + v  − M  ρ1 v     M ρ = ρwater − 1 − (0.80 ) water  v  ρ1  = 1.03 × 103 kg m 3 − 75.0 kg 0.400 m3  1.03 × 103 kg m3  1 − (0.80)  980 kg m3   = 732 kg m3 ⋅ To the given precision, the density of air is negligible compared to that of brass, but not compared to that of the wood. The fact that the density of brass may not be known the three<figure precision does not matter; the mass of the brass is given to three figures. The weight of the brass is the difference between the weight of the wood and the buoyant force of the air on the wood, and canceling a common factor of g , Vwood ( ρwood − ρair ) = M brass, and M wood = ρwoodVwood = M brass  ρwood ρ  = M brass 1 − air  ρwood − ρair ρwood   −1  1.20 kg m 3   = 0.0958 kg. = (0.0950 kg)1 − 3   150 kg m  −1 The buoyant force on the mass A, divided by g , must be 7.50 kg − 1.00 kg − 1.80 kg = 4.70 kg (see Example 14.6), so the mass block is 4.70 kg + 3.50 kg = 8.20 kg. a) The mass of the liquid displaced by the block is 4.70 kg, so the density of the liquid is 4.70 kg 3.80×10 < 3 m 3 = 1.24 × 10 3 kg m 3 . b) Scale D will read the mass of the block, 8.20 kg, as found above. Scale E will read the sum of the masses of the beaker and liquid, 2.80 kg. Neglecting the buoyancy of the air, the weight in air is g ( ρAuVAu + ρA1VA1 ) = 45.0 N. and the buoyant force when suspended in water is ρwater (VAu + VA1 ) g = 45.0 N − 39.0 N = 6.0 N. These are two equations in the two unknowns VAu and VA1. Multiplying the second by ρ A1 and the first by ρ water and subtracting to eliminate the VA1 term gives ρwaterVAu g ( ρAu − ρA1 ) = ρwater (45.0 N) − ρA1 (6.0 N) ρAu wAu = ρAu gVAu = ( ρwater (45.0 N) − ρAu (6.0)) ρwater ( ρAu − ρA1 ) (19.3) = ((1.00)(45.0 N) − (2.7)(6.0 N)) (1.00)(19.3 − 2.7) = 33.5 N. Note that in the numerical determination of wAu , specific gravities were used instead of densities. The ball’s volume is 4 4 V = πr 3 = π (12.0 cm)3 = 7238 cm3 3 3 As it floats, it displaces a weight of water equal to its weight. a) By pushing the ball under water, you displace an additional amount of water equal to 84% of the ball’s volume or (0.84)(7238 cm 3 ) = 6080 cm 3 . This much water has a mass of 6080 g = 6.080 kg and weighs (6.080 kg)(9.80 m s 2 ) = 59.6 N, which is how hard you’ll have to push to submerge the ball. b) The upward force on the ball in excess of its own weight was found in part (a): 59.6 N. The ball’s mass is equal to the mass of water displaced when the ball is floating: (0.16)(7238 cm3 )(1.00 g cm3 ) = 1158 g = 1.158 kg, and its acceleration upon release is thus a= Fnet 59.6 N = = 51.5 m s 2 m 1.158 kg a) The weight of the crown of its volume V is w = ρcrown gV , and when suspended the apparent weight is the difference between the weight and the buoyant force, fw = fρcrown gV = ( ρcrown − ρwater ) gV . Dividing by the common factors leads to − ρwater + ρcrown = fρcrown or ρcrown 1 = . ρwater 1 − f As f → 0, the apparent weight approaches zero, which means the crown tends to float; from the above result, the specific gravity of the crown tends to 1. As f → 1, the apparent weight is the same as the weight, which means that the buoyant force is negligble compared to the weight, and the specific gravity of the crown is very large, as reflected in the above expression. b) Solving the above equations for f in terms of the ρ water , and so the weight of the crown would be specific gravity, f = 1 − ρcrown (1 − (1 19.3))(12.9 N ) = 12.2 N. c) Approximating the average density by that of lead for a “thin” gold plate, the apparent weight would be (1 − (1 11.3))(12.9 N ) = 11.8 N. a) See problem 14.67. Replacing f with, respectively, wwater w and wfluid w gives ρsteel w ρ w = , steel = , ρfluid w < wfluid ρfluid w < wwater and dividing the second of these by the first gives ρfluid w < wfluid = . ρwater w < wwater b) When wfluid is greater than wwater, the term on the right in the above expression is less than one, indicating that the fluids is less dense than water, and this is consistent with the buoyant force when suspended in liquid being less than that when suspended in water. If the density of the fluid is the same as that of water wfluid = wwater , as expected. Similarly, if wfluid is less than wwater , the term on the right in the above expression is greater than one, indicating the the fluid is denser than water. c) Writing the result of part (a) as ρfluid 1 − ffluid = . ρwater 1 − f water and solving for f fluid , f fluid = 1 − ρfluid (1 − f water ) = 1 − (1.220)(0.128) = 0.844 = 84.4%. ρwater a) Let the total volume be V; neglecting the density of the air, the buoyant force in terms of the weight is   (w g ) + V0 , B = ρwater gV = ρwater g    ρm or V0 = b) B ρ water g − w ρ Cu g −4 B ρwater g − w ⋅ ρw g = 2.52 × 10 m . Since the total volume of the casting is 3 cavities are 12.4% of the total volume. B ρ water g , the a) Let d be the depth of the oil layer, h the depth that the cube is submerged in the water, and L be the length of a side of the cube. Then, setting the buoyant force equal to the weight, canceling the common factors of g and the cross<section area and supressing units, (1000)h + (750)d = (550) L.d , h and L are related by d + h + (0.35) L = L, so h = (0.65) L − (0.65)(1000) − (550) d. Substitution into the first relation gives d = L (1000) − (750) = 2L 5.00 = 0.040 m. b) The gauge pressure at the lower face must be sufficient to support the block (the oil exerts only sideways forces directly on the block), and p = ρ wood gL = (550 kg m3 )(9.80 m s 2 )(0.100 m) = 539 Pa. As a check, the gauge pressure, found from the depths and densities of the fluids, is ((0.040 m)(750 kg m 3 ) + (0.025 m)(1000 kg m 3 ))(9.80 m s 2 ) = 539 Pa. The ship will rise; the total mass of water displaced by the barge<anchor combination must be the same, and when the anchor is dropped overboard, it displaces some water and so the barge itself displaces less water, and so rises. To find the amount the barge rises, let the original depth of the barge in the water be h0 = (mb + ma ) ( ρ water A), where mb and ma are the masses of the barge and the anchor, and A is the area of the bottom of the barge. When the anchor is dropped, the buoyant force on the barge is less than what it was by an amount equal to the buoyant force on the anchor; symbolically, h′ρ water Ag = h0 ρ water Ag − (ma ρsteel )ρ water g , which is solved for h = h0 − h′ = or about 0.56 mm. ma ρ steel A = (35.0 kg ) (7860 kg m 3 )(8.00 m ) 2 = 5.57 × 10 −4 m, a) The average density of a filled barrel is 15.0 kg ρ oil + Vm = 750 kg m3 + 0.120 = 875 kg m3 , which is less than the density of seawater, m 3 so the barrel floats. b) The fraction that floats (see Problem 14.23) is 1− ρ ave 875 kg m 3 =1− = 0.150 = 15.0%. ρ water 1030 kg m 3 32.0 kg = 1172 mkg3 which means the barrel sinks. c) The average density is 910 mkg3 + 0.120 m3 In order to lift it, a tension T = (1177 mkg3 )(0.120 m3 )(9.80 sm2 ) − (1030 mkg3 )(0.120 m3 )(9.80 sm2 ) = 173 N is required. 1− ρB ρL a) See Exercise 14.23; the fraction of the volume that remains unsubmerged is . b) Let the depth of the liquid be x and the depth of the water be y. Then ρLgx + ρwgy = ρ B gL and x + y = L. Therefore x = L − y and y = ( ρL − ρB ) L ρ L − ρω . c) .6 − 7.8 y = 13 (0.10 m) = 0.046 m. 13.6 −1.0 V , A where A is the surface area of the water in the lock. V is the volume of water that has the same weight as the metal, so a) The change is height y is related to the displaced volume y= = V by y = V w ρwater g w = = A A ρwater gA (2.50 × 106 N) = 0.213 m. (1.00x103 kg m3 )(9.80 m s 2 )((60.0 m)(20.0 m)) b) In this case, V is the volume of the metal; in the above expression, ρ water is replaced by ρ metal = 9.00 ρ water , which gives y′ = water sinks by this amount. y 9 , and y − y′ = 8 9 y = 0.189 m; the a) Consider the fluid in the horizontal part of the tube. This fluid, with mass ρAl , is subject to a net force due to the pressure difference between the ends of the tube, which is the difference between the gauge pressures at the bottoms of the ends of the tubes. This difference is ρg ( yL − yR ), and the net force on the horizontal part of the fluid is ρ g ( y L − y R ) A = ρ Ala , or ( yL − yR ) = a l. g b) Again consider the fluid in the horizontal part of the tube. As in part (a), the fluid is accelerating; the center of mass has a radial acceleration of magnitude a rad = ω 2 l 2, and so the difference in heights between the columns is (ω 2 l 2)(l g ) = ω 2 l 2 2 g . Anticipating Problem, 14.77, an equivalent way to do part (b) is to break the fluid in the horizontal part of the tube into elements of thickness dr; the pressure difference between the sides of this piece is dp = ρ (ω 2 r )dr (see Problem 14.78), and integrating from r = 0 to r = l gives p = ρω 2 l 2 2, giving the same result. c) At any point, Newton’s second law gives dpA = pAdla from which the area A cancels out. Therefore the cross<sectional area does not affect the result, even if it varies. p = pal between the ends. This is Integrating the above result from 0 to l gives related to the height of the columns through p = pg y from which p cancels out. a) The change in pressure with respect to the vertical distance supplies the force necessary to keep a fluid element in vertical equilibrium (opposing the weight). For the rotating fluid, the change in pressure with respect to radius supplies the force necessary to ∂ keep a fluid element accelerating toward the axis; specifically, dp = ∂ pp dr = ρa dr , and using a = ω2 r gives ∂p ∂p = ρω 2 r. b) Let the pressure at y = 0, r = 0 be p a (atmospheric pressure); integrating the expression for ∂p ∂p from part (a) gives p(r , y = 0 ) = pa + ρω2 2. r 2 c) In Eq. (14.5), p2 = pa , p1 = p (r , y = 0) as found in part (b), y1 = 0 and y 2 = h(r ), the height of the liquid above the y = 0 plane. Using the result of part (b) gives h( r ) = ω 2 r 2 2 g . a) The net inward force is ( p + dp ) A − pA = Adp, and the mass of the fluid element is ρAdr ′. Using Newton’s second law, with the inward radial acceleration of ω 2 r ' , gives dp = ρω2 r ′dr ′. b) Integrating the above expression, ∫ p p0 r dp = ∫ ρω2 r ′dr ′ r0  ρω 2  2  r − r 20 , p − p0 =   2  which is the desired result. c) Using the same reasoning as in Section 14.3 (and Problem 14.78), the net force on the object must be the same as that on a fluid element of the same shape. Such a fluid element is accelerating inward with an acceleration of magnitude ω 2 Rcm, and so the force on the object is ρVω2 Rcm . d) If ρR cm > ρob Rcmob, the inward ( ) force is greater than that needed to keep the object moving in a circle with radius Rcmob at angular frequency ω , and the object moves inward. If ρRcm < ρob Rcmob , , the net force is insufficient to keep the object in the circular motion at that radius, and the object moves outward. e) Objects with lower densities will tend to move toward the center, and objects with higher densities will tend to move away from the center. (Note that increasing x corresponds to moving toward the back of the car.) a) The mass of air in the volume element is ρdV = ρAdx , and the net force on the element in the forward direction is ( p + dp )A − pA = Adp. From Newton’s second law, Adp = ( ρA dx)a, from which dp = ρadx. b) With ρ given to be constant, and with p = p0 at x = 0, p = p0 + ρax. c) Using ρ = 1.2 kg/m 3 in the result of part (b) gives (1.2 kg m3 )(5.0 m s 2 ) (2.5 m ) = 15.0 Pa ~ 15 × 10<5 patm , so the fractional pressure difference is negligble. d) Following the argument in Section 14<4, the force on the balloon must be the same as the force on the same volume of air; this force is the product of the mass ρV and the acceleration, or ρVa. e) The acceleration of the balloon is the force found in part (d) divided by the mass ρbalV , or ( ρ ρbal )a. The acceleration relative to the car is the difference between this acceleration and the car’s acceleration, arel = [( ρ ρbal ) − 1]a. f) For a balloon filled with air, ( ρ ρbal ) < 1 (air balloons tend to sink in still air), and so the quantity in square brackets in the result of part (e) is negative; the balloon moves to the back of the car. For a helium balloon, the quantity in square brackets is positive, and the balloon moves to the front of the car. If the block were uniform, the buoyant force would be along a line directed through its geometric center, and the fact that the center of gravity is not at the geometric center does not affect the buoyant force. This means that the torque about the geometric center is due to the offset of the center of gravity, and is equal to the product of the block’s weight and the horizontal displacement of the center of gravity from the geometric center, (0.075 m) 2. The block’s mass is half of its volume times the density of water, so the net torque is (0.30 m)3 (1000 kg m3 ) 0.075 m = 7.02 N ⋅ m, (9.80 m s 2 ) 2 2 or 7.0 N ⋅ m to two figures. Note that the buoyant force and the block’s weight form a couple, and the torque is the same about any axis. a) As in Example 14.8, the speed of efflux is 2gh. After leaving the tank, the water is in free fall, and the time it takes any portion of the water to reach the ground is t = 2( H − h) g , in which time the water travels a horizontal distance R = vt = 2 h( H − h). b) Note that if h′ = H − h, h′( H − h′ ) = ( H − h)h, and so h′ = H − h gives the same range. A hole H − h below the water surface is a distance h above the bottom of the tank. The water will rise until the rate at which the water flows out of the hole is the rate at which water is added; A 2 gh = dV , dt which is solved for 2 2 −4 3 1  dV dt  1  2.40 × 10 m s    = = = 13.1 cm. h   −4 2  2   A  2 g  1.50 × 10 m  2(9.80 m s ) Note that the result is independent of the diameter of the bucket. a) v3 A3 = 2 g ( y1 − y3 ) A3 = 2(9.80 m s 2 )(8.00 m) (0.0160 m 2 ) = 0.200 m3 s . b) Since p 3 is atmospheric, the gauge pressure at point 2 is 2 1 1 2   A3   8 2 2 p2 = ρ (v 3 − v2 ) = ρv3 1 −   = ρg ( y1 − y3 ),   A2   9 2 2   using the expression for υ3 found above. Subsititution of numerical values gives p2 = 6.97 × 10 4 Pa. The pressure difference, neglecting the thickness of the wing, is 2 2 p = (1 2) ρ(vtop − vbottom ), and solving for the speed on the top of the wing gives vtop = (120 m s) 2 + 2(2000 Pa) (1.20 kg m 3 ) = 133 m s . The pressure difference is comparable to that due to an altitude change of about 200 m, so ignoring the thickness of the wing is valid. a) Using the constancy of angular momentum, the product of the radius and  30  speed is constant, so the speed at the rim is about (200 km h)   = 17 km h. b) The  350  pressure is lower at the eye, by an amount 2 c) v2 2g  1m s  1  = 1.8 × 103 Pa. p = (1.2 kg m 3 ) ((200 km h ) 2 − (17 km h) 2 ) 2  3.6 km h  = 160 m to two figures. d) The pressure at higher altitudes is even lower. The speed of efflux at point D is 2 gh1 , and so is 8gh1 at C. The gauge pressure at C is then ρgh1 − 4 ρgh1 = −3 ρgh1, and this is the gauge pressure at E. The height of the fluid in the column is 3h1 . a) v = dV dt A , so the speeds are 6.00 × 10−3 m 3 s 6.00 × 10−3 m3 s = 1.50 m s . = 6 . 00 m s and 40.0 × 10− 4 m 2 10.0 × 10− 4 m 2 b) p = 12 ρ (v12 − v 22 ) = 1.688 × 10 4 Pa, or 1.69 × 10 4 Pa to three figures. c) h = p ρH g g = (1.688×10 4 Pa) (13.6×10 3 kg m 3 )( 9.80 m s 2 ) = 12.7 cm. a) The speed of the liquid as a function of the distance y that it has fallen is v = v + 2 gy , and the cross<section area of the flow is inversely proportional to this speed. The radius is then inversely proportional to the square root of the speed, and if the radius of the pipe is r0 , the radius r of the stream a distance y below the pipe is 2 0 r= r0 v0 (v02 + 2 gy )1 4  2 gy  = r0 1 + 2  v0   −1 4 . b) From the result of part (a), the height is found from (1 + 2 gy v02 )1 4 = 2, or y= 15v02 15(1.2 m s) 2 = = 1.10 m. 2g 2(9.80 m s 2 ) a) The volume V of the rock is V= B ρ water g = w − T ((3.00 kg)(9.80 m s 2 ) − 21.0 N) = = 8.57 × 10− 4 m3 . 3 3 2 ρ water g (1.00 × 10 kg m )(9.80 m s ) In the accelerated frames, all of the quantities that depend on g (weights, buoyant forces, gauge pressures and hence tensions) may be replaced by g ′ = g + a, with the ′ g positive direction taken upward. Thus, the tension is T = mg ′ − B′ = (m − ρV ) g ′ = T0 g , where T0 = 21.0 N. + 2.50 b) g ′ = g + a; for a = 2.50 m s 2 , T = (21.0 N) 9.809.80 = 26.4 N. c) For a = −2.50 m s 2 , T = (21.0 N) 9.809.−802.50 = 15.6 N. d) If a = − g , g ′ = 0 and T = 0. a) The tension in the cord plus the weight must be equal to the buoyant force, so T = Vg ( ρ water − ρ foam ) = (1 2)(0.20 m) 2 (0.50 m)(9.80 m s 2 )(1000 kg m 3 − 180 kg m 3 ) = 80.4 N. b) The depth of the bottom of the styrofoam is not given; let this depth be h0 . Denote the length of the piece of foam by L and the length of the two sides by l. The pressure force on the bottom of the foam is then ( p0 + ρgh0 ) L 2l and is directed up. The pressure on each side is not constant; the force can be found by integrating, or using the result of Problem 14.44 or Problem 14.46. Although these problems found forces on vertical surfaces, the result that the force is the product of the average pressure and the area is valid. The average pressure is p0 + ρg (h0 − (l (2 2 ))), and the force on one side has magnitude ( p0 + ρg (h0 − l (2 2 ))) Ll ( ) and is directed perpendicular to the side, at an angle of 45.0° from the vertical. The force on the other side has the same magnitude, but has a horizontal component that is opposite that of the other side. The horizontal component of the net buoyant force is zero, and the vertical component is B = ( p 0 + ρgh0 ) Ll 2 − 2(cos 45.0°)( p 0 + ρg (h0 − l (2 2 ))) Ll = ρg the weight of the water displaced. Ll 2 , 2 When the level of the water is a height y above the opening, the efflux speed is 2gy , and dV = π (d 2) 2 2 gy . As the tank drains, the height decreases, and dt π (d 2) 2 gy dV dt dy d =− =− = −  2 gy . 2 dt A π ( D 2) D This is a separable differential equation, and the time T to drain the tank is found from 2 2 dy d = −  y  D 2 2 g dt , which integrates to [2 y ] 0 H d = −   D 2 2 gT , or 2 D 2 H D T =  =  2g  d  d 2 2H . g a) The fact that the water first moves upwards before leaving the siphon does not change the efflux speed, 2gh . b) Water will not flow if the absolute (not gauge) pressure would be negative. The hose is open to the atmosphere at the bottom, so the pressure at the top of the siphon is pa − ρg ( H + h), where the assumption that the cross< section area is constant has been used to equate the speed of the liquid at the top and bottom. Setting p = 0 and solving for H gives H = ( pa ρg ) − h. Any bubbles will cause inaccuracies. At the bubble, the pressure at the surfaces of the water will be the same, but the levels need not be the same. The use of a hose as a level assumes that pressure is the same at all point that are at the same level, an assumption that is invalidated by the bubble. Capítulo 15 a) The period is twice the time to go from one extreme to the other, and v = f λ = λ T = (6.00 m) (5.0 s) = 1.20 m s, or 1.2 m s to two figures. b) The amplitude is half the total vertical distance, 0.310 m. c) The amplitude does not affect the wave speed; the new amplitude is 0.150 m. d) For the waves to exist, the water level cannot be level (horizontal), and the boat would tend to move along a wave toward the lower level, alternately in the direction of and opposed to the direction of the wave motion. fλ=v f = v 1500 m s = = 1.5 × 106 Hz λ 0.001 m a) λ = v f = (344 m s) (784 Hz) = 0.439 m. b) f = v λ = (344 m s) (6.55 × 10−5 m) = 5.25 × 106 Hz. Denoting the speed of light by c, λ = cf , and a) 3.00×10 8 m s 540×10 3 Hz = 556 m. 8 3.00×10 m s b) 104 = 2.87 m. .5×10 6 Hz a) λ max = (344 m s) (20.0 Hz) = 17.2 m, λ min = (344 m s) (20,000 Hz) = 1.72 cm. b) λ max = (1480 m s) (20.0 Hz) = 74.0 m, λ min = (1480 m s) (20,000 Hz) = 74.0 mm. c) f = 1 T Comparison with Eq. (15.4) gives a) 6.50 mm, b) 28.0 cm, 1 = 0.0360 = 27.8 Hz and from Eq. (15.1), d) v = (0.280 m)(27.8 Hz) = 7.78 m s , s e) + x direction. a) f = v λ = (8.00 m s) (0.320 m) = 25.0 Hz, T = 1 f = 1 (25.0 Hz) = 4.00 × 10−2 s, k = 2π λ = (2π ) (0.320 m) = 19.6 rad m.   x . b) y ( x, t ) = (0.0700 m) cos 2π  t (25.0 Hz) + 0.320 m   c) (0.0700 m) cos [2π ((0.150 s)(25.0 Hz) + (0.360 m) (0.320 m))] = −4.95 cm. d) The argument in the square brackets in the expression used in part (c) is 2π (4.875), and the displacement will next be zero when the argument is 10π; the time is then T (5 − x λ) = (1 25.0 Hz)(5 − (0.360 m) (0.320 m)) = 0.1550 s and the elapsed time is 0.0050 s, e) T 2 = 0.02 s. a) b) a) and so b) and so ∂2 y ∂x 2 = ∂y = − Ak sin(kx + ωt ) ∂x ∂2 y = − Ak 2 cos(kx + ωt ) ∂x 2 ∂y = − Aω sin (kx + ωt ) ∂t ∂2 y = − Aω2 cos(kx + ωt ), 2 ∂t 2 k2 ∂ y 2 ω ∂t 2 , and y ( x, t ) is a solution of Eq. (15.12) with v = ω k . ∂y = + Ak cos(kx + ωt ) ∂x ∂2 y = − Ak 2 sin(kx + ωt ) 2 ∂x ∂y = + Aω cos(kx + ωt ) ∂t ∂2 y = − Aω2 sin (kx + ωt ), ∂t 2 ∂2 y ∂x 2 = 2 k2 ∂ y ω 2 ∂t 2 , and y ( x, t ) is a solution of Eq. (15.12) with v = ω k . c) Both waves are moving in the − x 1direction, as explained in the discussion preceding Eq. (15.8). d) Taking derivatives yields v y ( x, t ) = −ωA cos (kx + ωt ) and a y ( x, t ) = −ω 2 A sin (kx + ωt ). a) The relevant expressions are y ( x, t ) = A cos(kx − ωt ) ∂y = ωA sin (kx − ωt ) vy = ∂t ∂ 2 y ∂v a y = 2 = y = −ω2 A cos (ωt − kx). ∂t ∂t b) (Take A, k and ω to be positive. At x t = 0, the wave is represented by (19.7(a)); point (i) in the problem corresponds to the origin, and points (ii)1(vii) correspond to the points in the figure labeled 117.) (i) v y = ωA cos(0) = ωA, and the particle is moving upward (in the positive y1direction). a y = −ω 2 A sin(0) = 0, and the particle is instantaneously not accelerating. (ii) v y = ωA cos(− π 4) = ωA a y = −ω2 A sin(− π 4) = ω2 A 2 , and the particle is moving up. 2 , and the particle is speeding up. (iii) v y = ωA cos(− π 2) = 0, and the particle is instantaneously at rest. a y = −ω2 A sin( − π 2) = ω2 A, and the particle is speeding up. (iv) v y = ωA cos(− 3π 4) = − ωA a y = −ω A sin(− 3π 4) = ω A 2 2 2 , and the particle is moving down. 2 , and the particle is slowing down ( v y is becoming less negative). (v) v y = ωA cos(−π ) = −ωA and the particle is moving down. a y = −ω 2 A sin(−π ) = 0, and the particle is instantaneously not accelerating. (vi) v y = ωA cos(− 5π 4) = − ωA 2 and the particle is moving down. a y = −ω 2 A sin(− 5π 4) = −ω2 A 2 and the particle is speeding up ( v y and a y have the same sign). (vii) v y = ωA cos(− 3π 2) = 0, and the particle is instantaneously at rest. a y = −ω2 A sin(− 3π 2) = −ω2 A and the particle is speeding up. (viii) v y = ωA cos(− 7π 4) = ωA a y = −ω 2 A sin(− 7π 4) = − ω 2 A opposite signs). 2 , and the particle is moving upward. 2 and the particle is slowing down ( v y and a y have Reading from the graph, a) A = 4.0 mm, b) T = 0.040 s. c) A displacement of 0.090 m corresponds to a time interval of 0.025 s; that is, the part of the wave represented by the point where the red curve crosses the origin corresponds to the point where the blue curve crosses the t1axis ( y = 0) at t = 0.025 s, and in this time the wave has traveled 0.090 m, and so the wave speed is 3.6 m s and the wavelength is vT = (3.6 m s)(0.040 s) = 0.14 m . d) 0.090 m 0.015 s = 6.0 m s and the wavelength is 0.24 m. d) No; there could be many wavelengths between the places where y (t ) is measured. a) 2π  λ  x t  A cos 2π  −  = + A cos  x − t  λ  T  λ T  = + A cos where 2π (x − vt ), λ λ = λf = v has been used. T ∂y 2πv 2π = A sin ( x − vt ). ∂t λ λ c) The speed is the greatest when the cosine is 1, and that speed is 2πvA λ . This will be equal to v if A = λ 2π , less than v if A < λ 2π and greater than v if A > λ 2π . b) vy = a) t = 0 : 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 0.000 −0.212 −0.300 −0.212 0.000 0.212 0.300 0.212 0.000 b) i) t = 0.400 s: 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 0.285 0.136 −0.093 −0.267 −0.285 −0.136 0.093 0.267 0.285 ii ) t = 0.800 s : 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 0.176 0.296 0.243 0.047 −0.176 −0.296 −0.243 −0.047 0.176 Solving Eq. (15.13) for the force F , F = v2 = ( f λ )2 =  0.120 kg  ((40.0 Hz) (0.750 m))2 = 43.2 Ν.  2.50 m  a) Neglecting the mass of the string, the tension in the string is the weight of the pulley, and the speed of a transverse wave on the string is v= F = (1.50 kg)(9.80 m s 2 ) = 16.3 m s. (0.0550 kg m) b) λ = v f = (16.3 m s) (120 Hz) = 0.136 m. c) The speed is proportional to the square root of the tension, and hence to the square root of the suspended mass; the answers change by a factor of 2 , to 23.1 m s and 0.192 m. a) v = F = (140.0 Ν ) (10.0 m) (0.800 kg) = 41.8 m s. b) λ = v f = (41.8 m s) (1.20 Hz) = 34.9 m. c) The speed is larger by a factor of 2 , and so for the same wavelength, the frequency must be multiplied by 2 , or 1.70 Hz. Denoting the suspended mass by M and the string mass by m, the time for the pulse to reach the other end is t= L = v L = Mg (m L) mL (0.800 kg)(14.0 m) = = 0.390 s. Mg (7.50 kg)(9.80 m s 2 ) a) The tension at the bottom of the rope is due to the weight of the load, and the speed is the same 88.5 m s as found in Example 15.4 b) The tension at the middle of the rope is (21.0 kg ) (9.80 m s 2 ) = 205.8 N (keeping an extra figure) and the speed of the rope is 90.7 m s. c) The tension at the top of the rope is (22.0 kg)(9.80 m s 2 ) = 215.6 m s and the speed is 92.9 m s . (See Challenge Problem (15.80) for the effects of varying tension on the time it takes to send signals.) a) v = F = (5.00 N) (0.0500 kg m) = 10.0 m s b) λ = v f = (10.0 m s) (40.0 Hz) = 0.250 m c) y ( x, t ) = A cos(kx − ωt ) (Note : y (0.0) = A, as specified.) k = 2π λ = 8.00π rad m; ω = 2πf = 80.0π rad s 1 y ( x, t ) = (3.00 cm)cos[π (8.00 rad m) x − (80.0π rad s)t ] d) v y = + Aω sin(kx − ωt ) and a y = − Aω2cos(kx − ωt ) a y , max = Aω2 = A(2πf ) 2 = 1890 m s 2 e) a y , maxis much larger than g, so ok to ignore gravity. a) Using Eq.(15.25), 1 Pave = F ω 2 A2 2  3.00 × 10−3 kg   (25.0 N) (2π(120.0 Hz)) 2 (1.6 × 10− 3 m) 2 0.80 m   = 0.223 W, = 1 2 or 0.22 W to two figures. W. b) Halving the amplitude quarters the average power, to 0.056 Fig. 15.13 plots P( x, t ) = F ω2 A2 sin 2 (kx − ωt ) at x = 0. For x = 0, P ( x, t ) = F ω2 A2 sin 2 (ωt ) = Pmax sin 2 (ωt ) When x = λ 4, kx = (2π λ ) (λ 4) = π 2. sin (π 2 − ωt ) = cos ωt , so P(λ 4, t ) = Pmax cos 2 ωt The graph is shifted by T 4 but is otherwise the same. The instantaneous power is still never negative and Pav = 12 Pmax , the same as at x = 0. r2 = r1 Ι1 Ι2 = (7.5 m) 0.11 W m 2 1.0 W m 2 = 2.5 m, so it is possible to move r1 − r2 = 7.5 m − 2.5 m = 5.0 m closer to the source. a) Ι1r 12 = Ι 2 r 2 2 Ι 2 = Ι1 (r1 r2 ) 2 = (0.026 W m 2 )(4.3 m 3.1 m) 2 = 0.050 W m 2 b) P = 4πr 2 Ι = 4π (4.3m ) 2 (0.026 W m 2 ) = 6.04 W Energy = Pt = (6.04 W )(3600 s) = 2.2 × 10 4 J (a) A = 2.30 mm. (b) f = v= ω k = 742 rad s 6.98 rad m ω 2π = 742 rad s 2π 118 Hz. (c) λ = 2π k = 2π 6.98 rad m = 0.90 m. (d) = 106 m s. (e) The wave is traveling in the –x direction because the phase of y (x,t) has the form kx + ωt. (f) The linear mass density is = (3.38 × 10 −3 kg ) (1.35 m ) = 2.504 × 10 −3 kg m , so the tension is F = v 2 = (2.504 × 10 −3 kg m)(106.3 m s) 2 = 28.3 N (keeping an extra figure in v for accuracy). (g) Pav = 1 2 F ω 2 A2 = 1 2 (2.50 × 10−3 kg m)(28.3 N) (742 rad s) 2 (2.30 × 10−3 m) 2 = 0.39 W. I = 0.250 W m 2 at r = 15.0 m P = 4πr 2 I = 4π (15.0 m) 2 (0.250 W m 2 ) = 707 W a) The wave form for the given times, respectively, is shown. b) a) The wave form for the given times, respectively, is shown. b) Let the wave traveling in the + x direction be y1 ( x, t ) = A cos (kx − ωt). The wave traveling in the − x direction is inverted due to reflection from the fixed end of the string at x = 0, so it has the form y2 ( x, t ) = − A cos(kx + ωt ). The wave function of the resulting standing wave is then y ( x, t ) = y1 ( x, t ) + y2 ( x, t ) , where A = 2.46 mm, ω = 2π T = 2π (3.65 × 10−3 s) = 1.72 × 103 rad s, k = ω v = (1.72 × 103 rad s)(111 m s) = 15.5 rad m. a) The nodes correspond to the places where y = 0 for all t in Eq. (15.1); that is, sin kxnode = 0 or kxnode = nπ , n an integer . With k = 0.75π rad m, xnode = (1.333 m)n and for n = 0, 1, 2, ..., xnode = 0, 1.333 m, 2.67 m, 4.00 m, 5.33 m, 6.67 m,... b) The antinodes correspond to the points where cos kx = 0, which are halfway between any two adjacent nodes, at 0.667 m, 2.00 m, 3.33 m, 4.67 m, 6.00 m, ... ∂2 y ∂2 y 2 [ ] k A sin ωt sin kx , = − = −ω2 [ Asw sin ωt ]sin kx, sw ∂x 2 ∂t 2 − ω2 ω so for y ( x, t ) to be a solution of Eq. (15.12), − k 2 = 2 , and v = . v k b) A standing wave is built up by the superposition of traveling waves, to which the relationship v = ω k applies. a) a) The amplitude of the standing wave is Asw = 0.85 cm, the wavelength is twice the distance between adjacent antinodes, and so Eq. (15.28) is y ( x, t ) = (0.85 cm) sin((2π 0.075 s)t ) sin(2πx 30.0 cm). b) c) v = λ f = λ T = (30.0 cm) (0.0750 s) = 4.00 m/s. (0.850 cm) sin(2π (10.5 cm) (30.0 cm)) = 0.688 cm. y1 + y2 = A [− cos(kx + ωt ) + cos(kx − ωt )] = A [− cos kx cos ωt + sin kx sin ωt + cos kx cos ωt + sin kx sin ωt ] = 2 A sin kx sin ωt. The wave equation is a linear equation, as it is linear in the derivatives, and differentiation is a linear operation. Specifically, ∂y ∂( y1 + y2 ) ∂y1 ∂y2 = = + . ∂x ∂x ∂x ∂x Repeating the differentiation to second order in both x and t, ∂ 2 y1 ∂ 2 y2 ∂2 y = + 2 , ∂x 2 ∂x ∂x 2 ∂ 2 y1 ∂ 2 y 2 ∂2 y = + 2 . ∂t 2 ∂t 2 ∂t The functions y1 and y2 are given as being solutions to the wave equation; that is, ∂2 y ∂ 2 y1 ∂ 2 y2  1  ∂ 2 y1  1  ∂ 2 y 2 = + 2 + 2 = 2 2 2 ∂x ∂x 2 ∂x 2  v  ∂t  v  ∂t  1 = 2 v 2 2   ∂ y1 ∂ y 2   2 + 2  ∂t    ∂t 2  1 ∂ y = 2  2  v  ∂t and so y = y1 + y 2 is a solution of Eq. (15.12). a) From Eq. (15.35), f1 = b) 10 , 000 Hz 408 Hz 1 2L FL 1 = 2(0.400 m) m (800 N)(0.400 m) =408 Hz. (3.00 × 10−3 kg ) = 24.5, so the 24 th harmonic may be heard, but not the 25 th . a) In the fundamental mode, λ = 2 L = 1.60 m and so v = f λ = (60.0 Hz)(1.60 m) = 96.0 m s. b) F = v 2 = v 2 m L = (96.0 m s) 2 (0.0400 kg) (0.800 m) = 461 Ν. The ends of the stick are free, so they must be displacement antinodes. 1st harmonic: L= 1 λ1 → λ1 = 2 L = 4.0 m 2 2nd harmonic: L = 1λ 2 → λ 2 = L = 2.0 m rd 3 harmonic: L= 3 2L λ3 → λ3 = = 1.33 m 2 3 a) b) Eq. (15.28) gives the general equation for a standing wave on a string: y ( x, t ) = ( Asw sin kx ) sinωt Asw = 2 A, so A = ASW 2 = (5.60 cm) 2 = 2.80 cm c) The sketch in part (a) shows that L = 3(λ 2) k = 2π λ, λ = 2π k Comparison of y ( x, t ) given in the problem to Eq.(15.28) gives k = 0.0340 rad cm. So, λ = 2π (0.0340 rad cm) = 184.8 cm L = 3(λ 3) = 277 cm d) λ = 185 cm, from part (c) ω = 50.0 rad s so f = ω 2π = 7.96 Hz period T = 1 f = 0.126 s v = f λ = 1470 cm s e) v y = dy dt = ωAsw sin kx cos ωt v y , max = ωASW = (50.0 rad s)(5.60 cm) = 280 cm s f) f 3 = 7.96 Hz = 3 f1 , so f1 = 2.65 Hz is the fundamental f8 = 8 f1 = 21.2 Hz; ω8 = 2πf8 = 133 rad s λ = v f = (1470 cm s) (21.2 Hz) = 69.3 cm and k = 2π λ = 0.0906 rad cm. y ( x, t ) = (5.60 cm) sin ([0.0906 rad cm]x) sin ([133 rad s]t ) (a) A = 12 ASW = 12 (4.44 mm) = 2.22 mm. (b) λ = ( c) f = ω 2π = 754 rad m 2π = 120 Hz. (d) v = ω k = 754 rad m 32.5 rad m 2π k = 2π 32.5 rad m = 0.193 m. = 23.2 m s. (e) If the wave traveling in the + x direction is written as y1 ( x, t ) = A cos(kx − ωt ), then the wave traveling in the − x direction is y2 ( x, t ) = − A cos(kx + ωt ), where A = 2.22 mm from (a), and k = 32.5 rad m and ω = 754 rad s. (f) The harmonic cannot be determined because the length of the string is not specified. a) The traveling wave is y ( x, t ) = (2.30 m) cos ([6.98 rad m]x) + [742 rad s]t ) A = 2.30 mm so ASW = 4.60 mm; k = 6.98 rad m and ω = 742 rad s The general equation for a standing wave is y ( x, t ) = ( ASW sin kx) sin ωt , so y ( x, t ) = (4.60 mm) sin([6.98 rad m]x)sin([742 rad s]t ) b) L = 1.35 m (from Exercise 15.24) λ = 2π k = 0.900 m L = 3(λ 2), so this is the 3rd harmonic c) For this 3rd harmonic, f = ω 2π = 118 Hz f 3 = 3 f1 so f1 = (118 Hz) 3 = 39.3 Hz The condition that x = L is a node becomes kn L = nπ. The wave number and wavelength are related by knλ n = 2π , and so λ n = 2 L n. a) The product of the frequency and the string length is a constant for a given string, equal to half of the wave speed, so to play a note with frequency 587 Hz, x = (60.0 cm) (440 Hz) (587 Hz) = 45.0 cm. b) Lower frequency requires longer length of string free to vibrate. Full length of string gives 440 Hz, so this is the lowest note possible. a) (i) x = λ 2 is a node, and there is no motion. (ii) x = λ 4 and vmax = A(2πf ) = 2πfA, amax = (2πf )vmax = 4π 2 f 2 A. (iii) cos π4 = is an antinode, 1 2 , and this factor multiplies the results of (ii), so vmax = 2π fA, amax = 2 2π 2 f 2 A . b) The amplitude is A sin kx, or (i)0, (ii) A, (iii) A 2. c) The time between the extremes of the motion is the same for any point on the string (although the period of the zero motion at a node might be considered indeterminate) and is 21f . a) λ1 = 2 L = 3.00 m, f1 = v 2L = ( 48.0 m s ) 2 (1.50 m ) = 16.0 Hz. b) λ 3 = λ1 3 = 1.00 m, f 2 = 3 f1 = 48.0 Hz. c) λ 4 = λ1 4 = 0.75 m, f3 = 4 f1 = 64.0 Hz. a) For the fundamental mode, the wavelength is twice the length of the string, and v = f λ = 2 fL = 2(245 Hz)(0.635 m) = 311 m s. b) The frequency of the fundamental mode is proportional to the speed and hence to the square root of the tension; (245 Hz) 1.01 = 246 Hz. c) The frequency will be the same, 245 Hz. The wavelength will be λ air = vair f = (344 m s) (245 Ηz) = 1.40 m, which is larger than the wavelength of standing wave on the string by a factor of the ration of the speeds. a) f = v λ = (36.0 m s) (1.80 m) = 20.0 Hz, ω = 2πf = 126 rad s, k = ω v = 2π λ = 3.49 rad m. b) y ( x, t ) = A cos (kx − ωt ) = (2.50 mm)cos [(3.49 rad m) x − (126 rad s)t ]. c)At x = 0, y (0, t ) = A cos ωt = (2.50 mm) cos [(126 rad s)t ]. From this form it may be seen that at x = 0, t = 0, ∂∂yt > 0. d) At x = 1.35 m = 3λ 4, kx = 3π 2 and y (3λ 4, t ) = A cos [3π 2 − ωt ]. e) See Exercise 15.12; ωA = 0.315 m s. f) From the result of part (d ), y = 0 mm. v y = − 0.315 m s. a) From comparison with Eq. (15.4 ), A = 0.75 cm, λ = f = 125 Hz, T = 1 f 2 0.400 cm = 5.00 cm, = 0.00800 s and v − λf = 6.25 m s. b) c) To stay with a wavefront as t increases, x and so the wave is moving in the − x 1 direction. d) From Eq. (15.13), the tension is F = v 2 = (0.50 kg m) (6.25 m s) 2 = 19.5 N. e) Pav = 1 2 F ω2 A2 = 54.2 W. a) Speed in each segment is v = F t = L v. The travel times then, are t1 = L ttotal = L 1 F + 2L 1 F + 12 L 1 F = 72 L 1 F 1 F . The time to travel through a segment is , t2 = L 4 1 F , and t3 = L . b) No, because the tension is uniform throughout each piece. 1 4F . Adding gives The amplitude given is not needed, it just ensures that the wave disturbance is small. Both strings have the same tension F , and the same length L = 1.5 m. The wave takes different times t1 and t2 to travel along each string, so the design requirements is t1 + t 2 = 0.20 s. Using t = L v and v = F ( ) = FL m gives m1 + m2 L F = 0.20 s, with m1 = 90 × 10 −3 kg and m1 = 10 × 10 −3 kg . Solving for F gives F = 6.0 N. a) y ( x, t ) = A cos(kx − ωt ) v y = dy dt = + Aω sin(kx − ωt ) v y , max = Aω = 2πfA f = v and v = λ F  1  FL , so f =   (m L ) λ M  2πA  FL v y , max =    λ  M b) To double v y , max increase F by a factor of 4 The maximum vertical acceleration must be at least g . Because a = ω A, g = ω2 Amin and thus Amin = g ω2. Using ω = 2πf = 2πv λ and v = F 2 becomes Amin = gλ2 4π 2 F . , this a) See Exercise 15.10; a y = ∂2 y ∂t 2 = −ω2 y, and so k ′ = mω2 = x ω2 . 2 b) 4π 2 F  2πv  2 ω2 = (2πf ) =  =  λ2  λ  and so k ′ = (4π 2 F λ2 ) x. The effective force constant k ′ is independent of amplitude, as for a simple harmonic oscillator, and is proportional to the tension that provides the restoring force. The factor of 1 λ2 indicates that the curvature of the string creates the restoring force on a segment of the string. More specifically, one factor of 1 λ is due to the curvature, and a factor of 1 (λ ) represents the mass in one wavelength, which determines the frequency of the overall oscillation of the string. The mass m = x also contains a factor of , and so the effective spring constant per unit length is independent of . a), b) c) The displacement is a maximum when the term in parentheses in the denominator is zero; the denominator is the sum of two squares and is minimized when x = vt , and the maximum displacement is A. At x = 4.50 cm, the displacement is a maximum at t = (4.50 × 10−2 m) (20.0 m s) = 2.25 × 10−3 s. The displacement will be half of the maximum when ( x − vt ) 2 = A2 , or t = ( x ± A) v = 1.75 × 10−3 s and 2.75 × 10−3 s. d) Of the many ways to obtain the result, the method presented saves some algebra and minor calculus, relying on the chain rule for partial derivatives. Specifically, let ∂f dg ∂u dg ∂f dg ∂u dg u = u ( x, t ) = x − vt , so that if f ( x, t ) = g(u ), = = and = = − v. ∂x du ∂t du ∂t du ∂t du (In this form it may be seen that any function of this form satisfies the wave equation; see Problem 15.59.) In this case, y ( x, t ) = A3 ( A2 + u 2 )−1 , and so − 2 A3u ∂y , = 2 ∂x ( A + u 2 ) 2 ∂y 2 A3u =v 2 , ∂t ( A + u 2 )2 2 A3 ( A2 − 3u 2 ) ∂2 y = − ( A2 + u 2 ) 3 ∂x 2 3 2 2 ∂2 y 2 2 A ( A − 3u ) = − v , ∂t 2 ( A2 + u 2 ) 2 and so the given form for y ( x, t ) is a solution to the wave equation with speed v. a) and b) (1): The curve appears to be horizontal, and v y = 0. As the wave moves, the point will begin to move downward, and a y < 0. (2): As the wave moves in the + x 1direction (to the right in Fig. (15.34)), the particle will move upward so v y > 0. The portion of the curve to the left of the point is steeper, so a y > 0. (3) The point is moving down, and will increase its speed as the wave moves; v y < 0, a y < 0. (4) The curve appears to be horizontal, and v y = 0. As the wave moves, the point will move away from the x 1axis, and a y >0. (5) The point is moving downward, and will increase its speed as the wave moves; v y < 0, a y < 0. (6) The particle is moving upward, but the curve that represents the wave appears to have no curvature, so v y > 0 and a y = 0. c) The accelerations, which are related to the curvatures, will not change. The transverse velocities will all change sign. (a ) The wave travels a horizontal distance d in a time t= d d 8.00 m = = = 0.333 s. v λ f (0.600 m )(40.0 Hz ) (b) A point on the string will travel a vertical distance of 4 A each cycle. Although the transverse velocity v y ( x, t ) is not constant, a distance of h = 8.00 m corresponds to a whole number of cycles, n = h (4 A) = (8.00 m) ((4(5.00 × 10−3 m)) = 400, so the amount of time is t = nT = n f = (400) (40.0 Hz) = 10.0 s. (c ) The answer for (a ) is independent of amplitude. For (b), the time is halved if the amplitude is doubled. a) y 2 ( x, y ) + z 2 ( x, y ) = A2 The trajectory is a circle of radius A. At t = 0, y (0,0) = A, z (0,0) = 0. At t = π 2ω, y (0, π 2ω) = 0, z (0, π 2ω) = − A. At t = π ω, y (0, π ω) = − A, z (0, π 2ω) = 0. At t = 3π 2ω, y (0, 3π 2ω) = 0, z (0, 3π 2ω) = + A b) v y = dy dt = + Aω sin(kx − ωt ), vz = dz dt = − Aω cos(kx − ωt ) v = v y2 + v z2 = Aω, so the speed is constant. = yˆ + z ˆ ⋅ = yv y + zvz = A2ω sin (kx − ωt ) cos(kx − ωt ) − A2ω cos(kx − ωt ) sin(kx − ωt ) ⋅ = 0, so is tangent to the circular path. c) a y = dv y dt = − Aω2 cos(kx − ωt ), az = dvz dt = − Aω2 sin( kx − ωt ) ⋅ = ya y + zaz = − A2ω2 [cos 2 (kx − ωt ) + sin 2 (kx − ωt )] = − A2ω2 r = A, a = Aω2 , so ⋅ = −ra ⋅ = ra cos φ so φ = 180° and is opposite in direction to ; is radially inward. y 2 + z 2 = A 2 , so the path is again circular, but the particle rotates in the opposite sense compared to part (a ). The speed of light is so large compared to the speed of sound that the travel time of the light from the lightning or the radio signal may be neglected. Them, the distance from the storm to the dorm is (344 m s)(4.43 s) = 1523.92 m and the distance from the storm to the ballpark is (344 m s)(3.00 s) = 1032 m. The angle that the direction from the storm to the ballpark makes with the north direction is found from these distances using the law of cosines;  (1523.92 m) 2 − (1032 m) 2 − (1120 m) 2   = 90.07°, − 2(1032 m) (1120 m)   θ = arccos  so the storm can be considered to be due west of the park. a) As time goes on, someone moving with the wave would need to move in such a way that the wave appears to have the same shape. If this motion can be described by x = vt + c, with c a constant (not the speed light),then y ( x, t ) = f (c), and the waveform is the same to such observer. b) See Problem 15.54. The derivation is completed by taking the second partials, ∂2 y 1 d 2 f , = ∂x 2 v 2 du 2 ∂2 y d 2 f = , ∂t 2 du 2 so y ( x, t ) = f (t − x / v) is a solution to the wave equation with wave speed v . of the form y ( x, t ) = f (u ), with u = t − x v and f (u ) = De − C 2 (t −( B c ) x ) 2 c) This is , and the result of part (b) may be used to determine the speed v = C B immediately. a) b) ∂∂yt = ωΑ sin( kx − ωt + φ ). c) No; φ = π 4 or φ = 3π 4 would both give Α 2. If the particle is known to be moving downward, the result of part b) shows that cos φ < 0, and so φ = 3π 4. d) To identify φ uniquely, the quadrant in which φ is must be known. In physical terms, the signs of both the position and velocity, and the magnitude of either, are necessary to determine φ (within additive multiples of 2π ). a) F =F F = F v = F k ω and substituting this into Eq. (15.33) gives the result. by a b) Quadrupling the tension for F to F ′ = 4 F increases the speed v = F factor of 2, so the new frequency ω′ and new wave number k ′ are related to ω and k by (ω′ k ′) = 2(ω k ). For the average power to be the same, we must have Fkω = F ′k ′ω′, so kω = 4k ′ω′ and k ′ω′ = kω 4 . Multiplying the first and second equations together gives ω′ 2 = ω 2 2, so ω′ = ω 2. Thus, the frequency must decrease by a factor of the first equation gives k ′2 = k 2 8, so k ′ = k 8. 2. Dividing the second equation by (a) (b) The wave moves in the + x direction with speed v, so in the expression for y (x,0) replace x with x − vt : for ( x − vt ) < − L 0 h ( L + x − vt ) L for − L < ( x − vt ) < 0  y ( x, t ) =  h ( L − x + vt ) L for 0 < ( x − vt ) < L 0 for ( x − vt ) > L (c) From Eq. (15.21): for( x − vt ) < − L − F (0)(0) = 0  2 ∂y ( x, t ) ∂y ( x, t ) − F (h L)(− hv L) = Fv(h L) for − L < ( x − vt ) < 0 P ( x, t ) = − F = 2 ∂x ∂t − F (− h L)(hv L) = Fv(h L) for 0 < ( x − vt ) < L − F (0)(0) = 0 for ( x − vt ) > L  Thus the instantaneous power is zero except for − L < ( x − vt ) < L, where it has the constant value Fv(h L) 2 . a) Pav = v= F F ω 2 A2 1 2 F =v so ω = 2πf = 2π (v λ) Using these two expressions to replace F and ω gives Pav = 2 π 2v 3 A2 λ2 ; = (6.00 × 10−3 kg) (8.00 m) 1  2λ2 P  A =  2 3av  = 7.07 cm  4π v  b) Pav ~ v 3 so doubling v increases Pav by a factor of 8. Pav = 8(50.0 W) = 400.0 W 2 a) , d) b)The power is a maximum where the displacement is zero, and the power is a minimum of zero when the magnitude of the displacement is a maximum. c) The direction of the energy flow is always in the same direction. d) In this case, ∂y = −kΑ sin( kx + ωt ), and so Eq. (15.22) becomes ∂x P( x, t ) = − FkωA2sin 2 (kx + ωt ). The power is now negative (energy flows in the − x 1direction), but the qualitative relations of part (b) are unchanged. v12 = F1 , v22 = F2 = F1 − YΑαLΤ Solving for α, α= v12 − v22 v 2 − v22 = 1 ⋅ Y ( Α ) Τ (Y ρ) Τ ⋅ 1 min, so (a) The string vibrates through 1 2 cycle in 4 × 5000 1 4 T= min → T = 1.6 × 10−3 min = 9.6 × 10− 2 s 2 5000 f = 1 T = 1 9.6 × 10−2 s = 10.4 Hz λ = L = 50.0 cm = 0.50 m (b) Second harmonic (c) v = fλ = (10.4 Hz)(0.50 m) = 5.2 m s (d) (i)Maximum displacement, so v = 0 (ii) v y = ∂y ∂t = ∂∂t (1.5 cm sin kx sin ωt ) Speed = v y = ω(1.5 cm)sinkx sin ωt at maximum speed, sin kx = sin ωt = 1 v y = ω(1.5 cm) = 2πf (1.5 cm) = 2π (10.4 Hz)(1.5 cm) = 98 cm s = 0.98 m s (e) v = F → = F v2 M = L= F (1.00 N)(0.500 m) L= = 1.85 × 10− 2 kg 2 v (5.2 m s) 2 =18.5 g There is a node at the post and there must be a node at the clothespin. There could be additional nodes in between. The distance between adjacent nodes is λ 2, so the distance between any two nodes is n (λ 2) for n = 1, 2, 3, ... 45.0 cm = n(λ 2), λ = v f , so f = n[v (90.0 cm)] = (0.800 Hz)n, n = 1, 2, 3, ... (a) The displacement of the string at any point is y ( x, t ) = ( ASW sin kx) sin ωt. For the fundamental mode λ = 2 L, so at the midpoint of the string sin kx = sin(2π λ )( L 2) = 1, and y = ASW sin ωt. Taking derivatives gives v y = ∂y ∂t = ωASW cos ωt , with maximum value ∂v y v y max = ωASW , and a y = ∂t = −ω2 ASW sin ωt , with maximum value a y max = ω2 ASW . Dividing these gives ω = a y max v y max = (8.40 × 103 m s 2 ) (3.80 m s) = 2.21 × 103 rad s, and then ASW = v y max ω = (3.80 m s) (2.21 × 103 rad s) = 1.72 × 10−3 m. (b) v = λf = (2 L)(ω 2π ) = Lω π = (0.386 m) (2.21 × 103 rad s) π = 272 m s. a) To show this relationship is valid, take the second time derivative: ∂ 2 y ( x, t ) ∂ 2 = 2 [( ASW sin kx) cos ωt ], ∂t ∂t 2 2 ∂ ∂ y ( x, t ) = −ω [( ASW sin kx) sin ωt ] 2 ∂t ∂t 2 ∂ y ( x, t) = −ω2 [( Αsw sin kx) cos ωt ], 2 ∂t ∂ 2 y ( x, t ) = −ω2 y ( x, t ), Q.E.D. 2 ∂t The displacement of the harmonic oscillator is periodic in both time and space. b) Yes, the travelling wave is also a solution of the wave equation. a) The wave moving to the left is inverted and reflected; the reflection means that the wave moving to the left is the same function of − x, and the inversion means that the function is − f (− x). More rigorously, the wave moving to the left in Fig. (15.17) is obtained from the wave moving to the right by a rotation of 180° , so both the coordinates ( f and x) have their signs changed. b). The wave that is the sum is f ( x) − f (− x) (an inherently odd function), and for any f , f (0) − f (−0) = 0. c) The wave is reflected but not inverted (see the discussion in part (a) above), so the wave moving to the left in Fig. (15.18) is + f (− x). d) dy d df ( x) df (− x) df ( x) df (− x) d (− x) = + = + ( f ( x) + f (− x)) = dx dx dx dx dx d ( − x) dx = df df − dx dx . x=− x At x = 0 , the terms are the same and the derivatives is zero. (See Exercise 2012 for a situation where the derivative of f is not finite, so the string is not always horizontal at the boundary.) y ( x, t ) = y1 ( x, t ) + y2 ( x, t ) = Α[cos(kx + ωt ) + cos(kx − ωt )] = Α[cosωt cos kx − sin ωt sin kx + cos ωt cos kx + sin ωt sin kx] = (2Α) cosωt cos kx. b) At x = 0, y (0, t ) = (2 Α)cosωt , and so x = 0 is an antinode. c) The maximum displacement is, front part (b), ΑSW = 2 Α, the maximum speed is ωΑSW = 2ωΑ and the a) magnitude of the maximum acceleration is ΑSW ω 2 Αsw = 2ω 2 Α. a) λ = v f = (192.0 m s) (240.0 Hz) = 0.800 m , and the wave amplitude is = 0.400 cm. The amplitude of the motion at the given points is (i) (0.400 cm)sin (π ) = 0 (a node), (ii) (0.400 cm) sin(π 2) = 0.004 cm (an antinode) and (iii) (0.400 cm) sin(π 4) = 0.283 cm. b). The time is half of the period, or 1 (2 f ) = 2.08 × 10−3 s. c) In each case, the maximum velocity is the amplitude multiplied by ω = 2πf and the maximum acceleration is the amplitude multiplied by ω2 = 4π 2 f 2 , or (i) 0, 0; (ii) 6.03 m s, 9.10 × 103 m s 2 ; (iii) 4.27 m s , 6.43 × 103 m s 2 . The plank is oscillating in its fundamental mode, so λ = 2 L = 10.0 m, with a frequency of 2.00 Hz. a) v = fλ = 20.0 m s. b) The plank would be its first overtone, with twice the frequency, or 4 jumps s. (a) The breaking stress is F πr 2 = 7.0 × 108 N m 2 , and the maximum tension is F = 900 N, so solving for r gives the minimum radius r = The mass and density are fixed, ρ = length L = M πr 2 ρ = 4.0×10 −3 kg π (6.4×10 − 4 m) 2 ( 7800 kg m 3 ) M πr 2 L 900 N π (7.0×10 8 N m 2 ) = 6.4 × 10 −4 m. , so the minimum radius gives the maximum = 0.40 m. (b) The fundamental frequency is f 1 = 1 2L F , = 1 2L F M L = 1 2 F ML . Assuming the maximum length of the string is free to vibrate, the highest fundamental frequency occurs when F = 900 N, f1 = 1 2 900 N (4.0×10 −3 kg)(0.40 m) = 376 Hz. a) The fundamental has nodes only at the ends, x = 0 and x = L. b) For the second harmonic, the wavelength is the length of the string, and the nodes are at x = 0, x = L 2 and x = L. b) d) No; no part of the string except for x = L 2, oscillates with a single frequency. a) The new tension F ′ in the wire is F′ = F − B = w −  1 ρwater  (1 3w) ρwater  = w1 − ρA1  3 ρ A1   (1.00 × 103 kg m3 )   = (0.8765) w = (0.87645) F . = w 1 − 3 3   3(2.7 × 10 kg m )  The frequency will be proportional to the square root of the tension, and so f ′ = (200 Hz) 0.8765 = 187 Hz. b) The water does not offer much resistance to the transverse waves in the wire, and hencethe node will be located a the point where the wire attaches to the sculpture and not at the surface of the water. a) Solving Eq. (15.35) for the tension F, F = 4 L2 f12 = 4mLf12 = 4(14.4 × 10−3 kg)(0.600 m)(65.4 Hz) 2 = 148 N. .4 2 ) , and the percent increase is b) The tension must increase by a factor of ( 73 65.4 (73.4 65.4) 2 − 1 = 26.0%. a) Consider the derivation of the speed of a longitudinal wave in Section 16.2. Instead of the bulk modulus B, the quantity of interest is the change in force per fractional length change. The force constant k ′ is the change in force force per length change, so the force change per fractional length change is k ′L, the applied force at one end is F = (k ′L)(v y v) and the longitudinal impulse when this force is applied for a time t is k ′Lt v y v . The change in longitudinal momentum is ((vt ) m L)v y and equating the expressions, canceling a factor of t and solving for v gives v 2 = L2 k ′ m. An equivalent method is to use the result of Problem 11.82(a), which relates the force constant k ′ and the “Young’s modulus” of the Slinky TM , k ′ = YA L , or Y = k ′ L A. The mass density is ρ = m ( AL), and Eq. (16.8) gives the result immediately. b) (2.00 m) (1.50 N m) (0.250 kg) = 4.90 m s. 2 2 K (1 2) mv y 1  ∂y  = = a) uk =  . x m 2  ∂t  b) ∂y ∂t = ωA sin(kx − ωt ) and so uk = c) The piece has width 1 ω2 A2 sin 2 (kx − ωt ). 2 x and height x ∂∂yx , and so the length of the piece is 12 2    ( x) 2 +  x ∂y     ∂x    12   ∂y 2  = x1 +      ∂x      1  ∂y  2  ≈ x 1 +   ,  2  ∂x   where the relation given in the hint has been used. [ ] x 1 + 12 ( ∂∂yx ) 2 − x 1  ∂y  = F  . 2  ∂x  x = −kA sin( kx − ωt ), and so d) up = F e) ∂y ∂x up = 2 1 2 2 Fk A sin 2 (kx − ωt ) 2 and f) comparison with the result of part (c)with k 2 = ω 2 v 2 = ω 2 sinusoidal wave u (f). k = uv p F , shows that for a . g) In this graph, uk and up coincide, as shown in part a) The tension is the difference between the diver’s weight and the buoyant force, F = (m − ρwaterV ) g = (120 kg − (1000 kg m3 )(0.0800 m 3 )(9.80 m s 2 )) = 392 N. b) The increase in tension will be the weight of the cable between the diver and the point at x, minus the buoyant force. This increase in tension is then (,x − ρ( Ax))g = (1.10 kg m − (1000 kg m3 )π (1.00 × 10 −2 m) 2 )(9.80 m s ) x = (7.70 N m) x The tension as a function of x is then F ( x) = (392 N) + (7.70 N m) x. c) Denote the tension as F ( x) = F0 + ax, where F0 = 392 N and a = 7.70 N m. Then, the speed of 2 transverse waves as a function of x is v = dxdt = ( F 0 + ax ) needed for a wave to reach the surface is found from t = ∫ dt = ∫ dx dx. = dx dt ∫ F0 + ax Let the length of the cable be L, so ∫ t= = = L 0 2 a ( 2 F0 + ax a dx = F0 + ax F0 + aL − F0 L 0 ) 2 1.10 kg m ( 392 N + (7.70 N m)(100 m ) − 392 N ) = 3.98 s. 7.70 N m and the time t The tension in the rope will vary with radius r. The tension at a distance r from the center must supply the force to keep the mass of the rope that is further out than r accelerating inward. The mass of this piece in m LL− r , and its center of mass moves in a circle of radius L2+ r , and so 2  L − r  2  L + r  mω 2 T (r ) =  m ω = ( L − r 2 ). L   L  2L  An equivalent method is to consider the net force on a piece of the rope with length dr and mass dm = dr m L . The tension must vary in such a way that T (r ) − T (r + dr ) = −ω2 r dm, or dT = −(mω2 L)r dr. This is integrated to obtained dr T (r ) = −(mω2 2 L)r 2 + C , where C is a constant of integration. The tension must vanish at r = L, from which C = (mω2 L 2) and the previous result is obtained. The speed of propagation as a function of distance is dr T (r ) TL ω v(r ) = = = = L2 − r 2 , dt m 2 where drdt > 0 has been chosen for a wave traveling from the center to the edge. Separating variables and integrating, the time t is 2 L dr t = ∫ dt = . ω ∫0 L2 − r 2 The integral may be found in a table, or in Appendix B. The integral is done explicitly by letting r = L sin θ , dr = L cos θ dθ , L2 − r 2 = L cos θ , so that dr r ∫ L2 − r 2 = θ = arcsin L , and t= 2 π arcsin(1) = . ω ω 2 a) ∂∂yx = kAS W coskx sinωt, ∂∂yt = −ωASW ωsinkx cosωt , and so the instantaneous power is P = FA2 SW ωk (sin kx cos kx)(sin ωt cos ωt ) 1 = FA2 SW ωk sin( 2kx) sin(2ωt ). 4 b) The average value of P is proportional to the average value of sin( 2ωt ), and the average of the sine function is zero; Pav = 0. c) The waveform is the solid line, and the power is the dashed line. At time t = π 2ω , y = 0 and P = 0 and the graphs coincide. d) When the standing wave is at its maximum displacement at all points, all of the energy is potential, and is concentrated at the places where the slope is steepest (the nodes). When the standing wave has zero displacement, all of the energy is kinetic, concentrated where the particles are moving the fastest (the antinodes). Thus, the energy must be transferred from the nodes to the antinodes, and back again, twice in each cycle. Note that P is greatest midway between adjacent nodes and antinodes, and that P vanishes at the nodes and antinodes. a) For a string, f n = n 2L F and in this case, n = 1. Rearranging this and solving for F gives F = 4 L2 f 2 . Note that = πr 2 ρ, so = π (.203 × 10−3 m)2 (7800 kg m 3 ) = 1.01 × 10−3 kg m. Substituting values, F = (1.01 × 10 −3 kg m)4(.635 m) 2 (247.0 Hz) 2 = 99.4 N. b) To find the fractional change in the frequency we must take the ration of f to f : f = 1 2L F ,  1 F =    2L   1 1 f = F2 2L 1 1 F f = . 2 F 2L (f )= ( )  1   2L  1  F 2 ,  Now divide both sides by the original equation for f and cancel terms: f = f 1 2L 1 2L 1 2 F F F f 1 LF = f 2 F F = − YαA T , so c) From Section 17.4, 11 F = − 2.00 × 10 Pa 1.20 × 10−5 C° × (π (.203 × 10−3 m) 2 )(11°C) = 3.4 N. Then, F F = − 0.034, f f = −0.017, and finally, f = −4.2 Hz, or the pitch falls. This also explains the constant the constant tuning in the string sections of symphonic orchestras. ( )( ) Capítulo 16 a) λ = v f = (344 m s) (100 Hz) = 0.344m. b) if p → 1000 p 0 , then A → 1000A0 Therefore, the amplitude is 1.2 × 10−5 m. c) Since pmax = BkA, increasing pmax while keeping A constant requires decreasing k, and increasing π , by the same factor. 344 m s Therefore the new wavelength is (0.688 m)(20) = 6.9 m, f new = 6.9 m = 50 Hz. A= p max v ( 3.0×10 −2 Pa) (1480 m s ) = 2 πBf 2 π ( 2.2×10 9 Pa) (1000 Hz) , or A = 3.21 × 10−12 m. The much higher bulk modulus increases both the needed pressure amplitude and the speed, but the speed is proportional to the square root of the bulk modulus. The overall effect is that for such a large bulk modulus, large pressure amplitudes are needed to produce a given displacement. From Eq. (16.5), pmax = BkA = 2π BA λ = 2πBA f v. a) 2π (1.42 × 105 Pa) (2.00 × 10−5 m) (150 Hz) (344 m s) = 7.78 Pa. b) 10 × 7.78 Pa = 77.8 Pa. c) 100 × 7.78 Pa = 778 Pa. The amplitude at 1500 Hz exceeds the pain threshold, and at 15,000 Hz the sound would be unbearable. The values from Example 16.8 are B = 3.16 × 104 Pa, f = 1000 Hz, A = 1.2 × 10−8 m. Using Example 16.5, v = 344 m s amplitude of this wave is pmax = BkA = B 216 K 293K = 295 m s , so the pressure 2πf A = (3.16 × 104 Pa). v 2π (1000 Hz) (1.2 × 10−8 m) = 8.1 × 10− 3 Pa. This is (8.1 × 10−3 Pa) (3.0 × 10−2 Pa) = 0.27 295 m s times smaller than the pressure amplitude at sea level (Example 1621), so pressure amplitude decreases with altitude for constant frequency and displacement amplitude. a) Using Equation (16.7), B = v 2 ρ = (λf ) 2 , so B = [(8 m)(400 s)] × (1300 kg m 3 ) = 1.33 × 1010 Pa. 2 [ b) Using Equation (16.8), Y = v 2 ρ = ( L t ) 2 ρ = (1.5 m) (3.9 × 10− 4 s × (6400 kg m3 ) = 9.47 × 1010 Pa. ] 2 a) The time for the wave to travel to Caracas was 9 min 39 s = 579 s and the speed was 1.085 ×10 4 m s (keeping an extra figure). Similarly, the time for the wave to travel to Kevo was 680 s for a speed of 1.278 ×10 4 m s, and the time to travel to Vienna was 767 s for a speed of 1.258 × 10 4 m s. The average speed for these three measurements is 1.21 × 10 4 m s. Due to variations in density, or reflections (a subject addressed in later chapters), not all waves travel in straight lines with constant speeds. b) From Eq. (16.7), B = v 2 ρ , and using the given value of ρ = 3.3 × 10 3 kg m 3 and the speeds found in part (a), the values for the bulk modulus are, respectively, 3.9 × 1011 Pa, 5.4 × 1011 Pa and 5.2 × 1011 Pa. These are larger, by a factor of 2 or 3, than the largest values in Table (1121). Use vwater = 1482 m s at 20°C, as given in Table (16.1) The sound wave travels in water for the same time as the wave travels a distance 22.0 m − 1.20 m = 20.8 m in air, and so the depth of the diver is (20.8 m ) vwater = (20.8 m )1482 m s = 89.6 m. vair 344 m s This is the depth of the diver; the distance from the horn is 90.8 m. a), b), c) Using Eq. (16.10), vH 2 = vH e = vAr = (1.41)(8.3145 J mol ⋅ K )(300.15 K ) = 1.32 × 103 m (2.02 × 10 (1.67 )(8.3145 J −3 kg mol) s mol ⋅ K )(300.15 K ) = 1.02 × 103 m s −3 4.00 × 10 kg mol ( ) (1.67 )(8.3145 J mol ⋅ K )(300.15 K ) = 323 m s . (39.9 × 10−3 kg mol) d) Repeating the calculation of Example 16.5 at T = 300.15 K gives vair = 348 m s , and so vH 2 = 3.80 vair , vHe = 2.94 vair and vAr = 0.928 vair . Solving Eq. (16.10) for the temperature, Mv 2 T= = γR (28.8 × 10 −3   850 km h   1 m s   kg mol)       0.85   3.6 km hr   (1.40)(8.3145 J mol ⋅ K ) 2 = 191 K, or − 82°C. b) See the results of Problem 18.88, the variation of atmospheric pressure with altitude, assuming a non2constant temperature. If we know the altitude we can use  Mg     αy  Rα   the result of Problem 18.88, p = p0 1 − . Since T = To − αy, T0   for T = 191 K, α = .6 × 10−2 °C m, and T0 = 273 K, y = 13,667 m (44,840 ft.). Although a very high altitude for commercial aircraft, some military aircraft fly this high. This result assumes a uniform decrease in temperature that is solely due to the increasing altitude. Then, if we use this altitude, the pressure can be found: −2  (.6 × 10 ° C m) (13,667 m)   p = p o 1 − 273 K    ( 28.8×10 −3 kg mol)(9.8 m s 2 )     ( 8.315 J mol⋅ K)(.6×10 − 2 ° C m)    , p = p o (.70) 5.66 = .13p o , or about .13 atm. Using an altitude of 13,667 m in the equation derived in Example 18.4 gives p = .18p o , which overestimates the pressure due to the assumption of an isothermal atmosphere. and As in Example 16 2 5, with T = 21°C = 294.15 K, v= γRT (1.04)(8.3145 J mol ⋅ K)(294.15 K) = = 344.80 m s. M 28.8 × 10− 3 kg mol The same calculation with T = 283.15 Κ gives 344.22 m s, so the increase is 0.58 m s. Table 16.1 suggests that the speed of longitudinal waves in brass is much higher than in air, and so the sound that travels through the metal arrives first. The time difference is t= 80.0 m 80.0 m L L − = − = 0.208 s. 11 vair vBrass 344 m s (0.90 × 10 Pa) (8600 kg m 3 ) (1.40)(8.3145 J mol ⋅ K)(300.15 K) (1.40)(8.3145 J mol ⋅ K)(260.15 K) − −3 (28.8 × 10 kg mol) (28.8 × 10 − 3 kg mol) = 24 m s. (The result is known to only two figures, being the difference of quantities known to three figures.) The mass per unit length is related to the density (assumed uniform) and the cross2section area A by = Aρ, so combining Eq. (15.13) and Eq. (16.8) with the given relations between the speeds, Υ F Υ = 900 so F A = ⋅ ρ Αρ 900 a) λ = Υ ρ (11.0 × 1010 Pa) (8.9 × 103 kg m3 ) v = = = 16.0 m. f f 220 Hz b) Solving for the amplitude A (as opposed to the area a = πr 2 ) in terms of the average power Pav = Ιa, A= = (2 Pav a) ρΥω2 2(6.50 × 10−6 ) W) (π (0.800 × 1022 m)2 ) 3 (8.9 × 10 kg m )(11.0 × 10 Pa) (2π (220 Hz)) 3 10 2 c) ωΑ = 2π f Α = 2π (220 Hz)(3.289 × 10−8 m) = 4.55 × 10−5 m s. = 3.29 × 10−8 m. a) See Exercise 16.14. The amplitude is 2Ι Α= ρΒω2 2(3.00 × 10−6 W m 2 ) = (1000 kg m )(2.18 × 10 Pa) (2π (3400 Hz)) 3 9 2 = 9.44 × 10−11 m. The wavelength is B ρ (2.18 × 109 Pa) (1000 kg m 3 ) = = 0.434 m. f 3400 Hz v = f λ= b) Repeating the above with B = γp = 1.40 × 105 Pa and the density of air gives A = 5.66 × 10−9 m and λ = 0.100 m. c) The amplitude is larger in air, by a factor of about 60. For a given frequency, the much less dense air molecules must have a larger amplitude to transfer the same amount of energy. 2 2 B, and from Eq. (19.21), v 2 = B ρ . Using From Eq. (16.13), I = vp max Eq. (16.7) to eliminate v, I = eliminate B, I = vp 2 max ( ) 2 2 B ρ pmax 2 B = pmax 2 ρB . Using Eq. (16.7) to 2( v ρ ) = p 2 a) pmax = BkA = 2 πBfA v = 2 max 2 ρv. 2 π (1.42×10 5 Pa) (150 Hz) (5.00×10 −6 m) (344 m s) = 1.95 Pa. b) From Eq. (16.14), 2 I = pmax 2 ρv = (1.95 Pa) 2 (2 × (1.2 kg m 3 )(344 m s)) = 4.58 × 10−3 W m 2 . c) 10 × log ( 4.58×10 −3 10 −12 ) = 96.6 dB. (a) The sound level is W m2 β = (10 dB) log II0 , so β = (10 dB) log 0.500 , or β = 57 dB. 10 −12 W m 2 b) First find v, the speed of sound at 20.0 °C, from Table 16.1, v = 344 m s. The density of air at that temperature is 1.20 kg m 3 . Using Equation (16.14), I= 2 pmax (0.150 N m 2 ) 2 = , or I = 2.73 × 10−5 W m 2 . Using this in Equation 3 2 ρv 2(1.20 kg m )(344 m s) (16.15), β = (10 dB) log 2.73 × 10−5 W m 2 , or β = 74.4 dB. 10−12 W m 2 a) As in Example 16.6, I = ( 6.0×10 −5 Pa) 2 2 (1.20 kg m 3 )( 344 m s) = 4.4 × 10−12 W m 2 . β = 6.40 dB. a) 10 × log ( 4II ) = 6.0 dB. b) The number must be multiplied by four, for an increase of 12 kids. Mom is five times further away than Dad, and so the intensity she hears is = 5 −2 of the intensity that he hears, and the difference in sound intensity levels is 10 × log(25) = 14 dB. 1 25 (Sound level) = 75 dB − 90 dB = −25 dB I (Sound level) = 10 log I 0f − 10 log I 0i = 10 log Ifi I I Therefore I − 25 dB = 10 log Ifi If Ii = 10−2.5 = 3.2 × 10−3 β = (10 dB)log II0 , or 13 dB = (10 dB)log II0 . Thus, I I 0 = 20.0, or the intensity has increased a factor of 20.0. Open Pipe: λ1 = 2 L = v v = f1 594 Hz Closed at one end: λ1 = 4L = v f Taking ratios: 2 L v 594 Hz = 4L v f f = L 2 594 Hz = 297 Hz 2 a) Refer to Fig. (16.18). i) The fundamental has a displacement node at = 0.600 m , the first overtone mode has displacement nodes at L4 = 0.300 m and 34L = 0.900 m and the second overtone mode has displacement nodes at L = 0.200 m, L2 = 0.600 m and 56L = 1.000 m . ii) Fundamental: 0, L = 1.200 m. First : 0, 6 L = 0.600 m, L = 1.200 m. Second : 0, L3 = 0.400 m, 23L = 0.800 m, L = 1.200 m. 2 b) Refer to Fig. (16.19); distances are measured from the right end of the pipe in the figure. Pressure nodes at: Fundamental: L = 1.200 m . First overtone: L 3 = 0.400 m, L = 1.200 m. Second overtone: L 5 = 0.240 m, 3L 5 = 0.720 m , L = 1.200 m. Displacement nodes at Fundamental: 0. First overtone: 0, 2 L 3 = 0.800 m. Second overtone: 0, 2 L 5 = 0.480 m, 4 L 5 = 0.960 m a) f1 = v 2L = ( 344 m s ) 2 ( 0.450 m) = 382 Hz, 2 f1 = 764 Hz, f3 = 3 f1 = 1147 Hz, f 4 = 4 f1 = 1529 Hz. b) f1 = 4vL = 191 Hz, f 3 = 3 f1 = 573 Hz, f 5 = 5 f1 = 956 Hz, f 7 = 7 f1 = 1338 Hz. Note that the symbol “ f1 ” denotes different frequencies in the two parts. The frequencies are not always exact multiples of the fundamental, due to rounding. = 52.3, so the 52nd harmonic is heard. Stopped; 20,f000 = 104.7, so 103 rd c) Open: 20,f000 1 1 highest harmonic heard. f1 = ( 344 m/s) 4(0.17 m) = 506 Hz, f 2 = 3 f1 = 1517 Hz, f 3 = 5 f1 = 2529 Hz. a) The fundamental frequency is proportional to the square root of the ratio (see Eq. (16.10)), so f He = f air = γ M γ He M air (5 3) 28.8 ⋅ = (262 Hz) ⋅ = 767 Hz, γ air M He (7 5) 4.00 b) No; for a fixed wavelength , the frequency is proportional to the speed of sound in the gas. a) For a stopped pipe, the wavelength of the fundamental standing wave is 4 L = 0.56 m, and so the frequency is f1 = (344 m s ) (0.56 m) = 0.614 kHz. b) The length of the column is half of the original length, and so the frequency of the fundamental mode is twice the result of part (a), or 1.23 kHz. For a string fixed at both ends, Equation (15.33), f n = nv 2L , is useful. It is important to remember the second overtone#is the third harmonic. Solving for v, v = and inserting the data, v = ( 2 )(.635 m )(588 /s ) 3 2 fnL n , , and v = 249 m s . a) For constructive interference, the path difference d = 2.00 m must be equal to an integer multiple of the wavelength, so λ n = d n, fn = v vn 344 m s v = n(172 Hz ). = = n  = n λn d 2.00 m d  Therefore, the lowest frequency is 172 Hz. b) Repeating the above with the path difference an odd multiple of half a wavelength, f n = (n + 12 )(172 Hz ). Therefore, the lowest frequency is 86 Hz (n = 0 ). The difference in path length is x = (L − x ) − x = L − 2 x, or x = (L − x ) 2 . For destructive interference, x = (n + (1 2))λ ,and for constructive interference, x = nλ. The wavelength is λ = v f = (344 m s ) (206 Hz) = 1.670 m (keeping an extra figure), and so to have 0 ≤ x ≤ L, − 4 ≤ n ≤ 3 for destructive interference and − 4 ≤ n ≤ 4 for constructive interference. Note that neither speaker is at a point of constructive or destructive interference. a) The points of destructive interference would be at x = 0.58 m, 1.42 m. b) Constructive interference would be at the points x = 0.17 m, 1.00 m, 1.83 m. c) The positions are very sensitive to frequency, the amplitudes of the waves will not be the same (except possibly at the middle), and exact cancellation at any frequency is not likely. Also, treating the speakers as point sources is a poor approximation for these dimensions, and sound reaches these points after reflecting from the walls , ceiling, and floor. λ = v f = (344 m s ) (688 Hz ) = 0.500 m To move from constructive interference to destructive interference, the path difference must change by λ 2. If you move a distance x toward speaker B, the distance to B gets shorter by x and the difference to A gets longer by x so the path difference changes by 2x. 2 x = λ 2 and x = λ 4 = 0.125 m We are to assume v = 344 m s , so λ = v f = (344 m/s) (172 Hz ) = 2.00 m. If rA = 8.00 m and rB are the distances of the person from each speaker, the condition for destructive interference is rB − rA = (n + 12 )λ, where n is any integer. Requiring rB = rA + (n + 12 )λ > 0 gives n + 12 > − rA λ = − (8.00 m ) (2.00 m ) = −4, so the smallest value of rB occurs when n = −4, and the closest distance to B is rB = 8.00 m + (2 4 + 12 )(2.00 m ) = 1.00 m. λ = v f = (344 m s ) (860 Hz ) = 0.400 m The path difference is 13.4 m − 12.0 m = 1.4 m. path difference = 3.5 λ The path difference is a half2integer number of wavelengths, so the interference is destructive. a) Since f beat = f a − fb , the possible frequencies are 440.0 Hz ± 1.5 Hz = 438.5 Hz or 441.5 Hz b) The tension is proportional to the square of the frequency. 2 f (1.5 Hz ) = 6.82 × 10−3. Therefore T ∝ f 2 and T ∝ 2 f f . So TT = f . i) TT = 2440 Hz ii) T T = 2 ( −1.5 Hz ) 440 Hz = −6.82 × 10−3. a) A frequency of 12 (108 Hz + 112 Hz ) = 110 Hz will be heard, with a beat frequency of 112 Hz–108 Hz = 4 beats per second. b) The maximum amplitude is the sum of the amplitudes of the individual waves, 2 1.5 × 10 −8 m = 3.0 × 10 −8 m. The minimum amplitude is the difference, zero. ( ) Solving Eq. (16.17) for v, with vL = 0, gives fL 1240 Hz   v= vS =  (− 25.0 m s ) = 775 m s , fS − f L  1200 Hz − 1240Hz  or 780 m s to two figures (the difference in frequency is known to only two figures). Note that vS < 0, since the source is moving toward the listener. Redoing the calculation with +20.0 m s for vS and − 20.0 m/s for vL gives 267 Hz. a) From Eq. (16.17 ), with vS = 0, vL = −15.0 m s , f A′ = 375 Hz. b) With vS = 35.0 m s , vL = 15.0 m s , f B′ = 371 Hz. c) f A′ − f B′ = 4 Hz (keeping an extra figure in f A′ ) . The difference between the frequencies is known to only one figure. In terms of wavelength, Eq. (16.29) is v + vs λL = λS ⋅ v + vL ) (344 m s) (400 Hz ) = 0.798 m. This is, of a) vL = 0, vS = −25.0 m and λ L = ( 319 344 course, the same result as obtained directly from Eq. (16.27). vS = 25.0 m s and vL = (369 m s ) (400 Hz ) = 0.922 m. The frequencies corresponding to these wavelengths are c) 431 Hz and d) 373 Hz. a) In terms of the period of the source, Eq. (16.27) becomes vS = v − 0.12 m λ = 0.32 m s − = 0.25 m s . TS 1.6 s b) Using the result of part (a) in Eq. (16.18), or solving Eq. (16.27) for v S and substituting into Eq. (16.28) (making sure to distinguish the symbols for the different wavelengths) gives λ = 0.91 m.  v + vL   fS f L =   v + vS  a) The direction from the listener to source is positive, so vS = − v 2 and vL = 0.  v   fS = 2 fS = 2.00 kHz f L =  v−v 2 b) vS = 0, vL = + v 2 v +v 2 fL =   fS = 32 fS = 1.50 kHz  v  This is less than the answer in part (a). The waves travel in air and what matters is the velocity of the listener or source relative to the air, not relative to each other. For a stationary source, vS = 0, so f L = which gives vL = v ( fL fS ) v + vL v + vS fS = (1 + vL v ) fS , 490 Hz − 1 = (344 m s )( 520 − 1) = −19.8 m/s. Hz This is negative because the listener is moving away from the source. )(262 Hz ) a) vL = 18.0 m/s, vS = −30.0 m s , and Eq. (16.29 ) gives f L = ( 362 314 = 302 Hz. b) vL = −18.0 m s , vS = 30.0 m/s and f L = 228 Hz. a) In Eq. (16.31), v vS = 1 1.70 = 0.588 and α = arcsin(0.588) = 36.0°. b) As in Example16.20, (950 m ) t= = 2.23 s. (1.70) (344 m s) ( tan(36.0°)) a) Mathematically, the waves given by Eq. (16.1) and Eq. (16.4) are out of phase. Physically, at a displacement node, the air is most compressed or rarefied on either side of the node, and the pressure gradient is zero. Thus, displacement nodes are pressure antinodes. b) (This is the same as Fig. (16.3).) The solid curve is the pressure and the dashed curve is the displacement. c) The pressure amplitude is not the same. The pressure gradient is either zero or undefined. At the places where the pressure gradient is undefined mathematically (the “cusps” of the y 2 x plot), the particles go from moving at uniform speed in one direction to moving at the same speed in the other direction. In the limit that Fig. (16.43) is an accurate depiction, this would happen in a vanishing small time, hence requiring a very large force, which would result from a very large pressure gradient. d) The statement is true, but incomplete. The pressure is indeed greatest where the displacement is zero, but the pressure is equal to its largest value at points other than those where the displacement is zero. The altitude of the plane when it passes over the end of the runway is (1740 m − 1200 m)tan 15° = 145 m , and so the sound intensity is 1 (1.45) 2 of what the intensity would be at 100 m. The intensity level is then [ ] 100.0 dB − 10 × log (1.45) 2 = 96.8 dB, so the airliner is not in violation of the ordinance. a) Combining Eq. (16.14) and Eq. (16.15), pmax = 2 ρvI 010( β 10 ) = 2(1.20 kg m3 )(344 m s)(10−12 W m )105.20 2 = 1.144 × 10 −2 Pa, or = 1.14 × 10−2 Pa, to three figures. A= b) From Eq. (16.5), and as in Example 16.1, pmax pmax v (1.144 × 10−2 Pa) (344 m s) = = = 7.51 × 10−9 m. Bk B 2πf 2π (1.42 × 105 Pa)(587 Hz ) c) The distance is proportional to the reciprocal of the square root of the intensity, and hence to 10 raised to half of the sound intensity levels divided by 10. Specifically, (5.00 m)10(5.20 − 3.00) 2 = 62.9 m. # ( ) a) p = IA = I 010( β 10 dB) A. b) 1.00 × 10−12 W m 2 (105.50 )(1.20 m 2 ) −7 = 3.79 × 10 W. For the flute, the fundamental frequency is 344.0 m s v f1f = = = 800.0 Hz 4 L 4(0.1075 m) For the flute and string to be in resonance, nf f1f = ns f1s , where f1s = 600.0 Hz is the fundamental frequency for the string. ns = nf ( f1 f f1s ) = 43 nf ns is an integer when nf = 3- , - = 1, 3, 5... (the flute has only odd harmonics) nf = 3 - gives ns = 4 Flute harmonic 3- resonates with string harmonic 4 - , - = 1,3,5,... (a) The length of the string is d = L 10, so its third harmonic has frequency = 3 21d F . The stopped pipe has length L, so its first harmonic has frequency v 1 f1pipe = s . Equating these and using d = L 10 gives F = vs2 . 4L 3600 (b) If the tension is doubled, all the frequencies of the string will increase by a factor of 2 . In particular, the third harmonic of the string will no longer be in resonance with the first harmonic of the pipe because the frequencies will no longer match, so the sound produced by the instrument will be diminished. (c) The string will be in resonance with a standing wave in the pipe when their frequencies are equal. Using f1pipe = 3 f1string , the frequencies of the pipe are f string 3 nf1pipe = 3nf1string , (where n=1, 3, 5…). Setting this equal to the frequencies of the string n′f1string , the harmonics of the string are n′ = 3n = 3, 9, 15,... a) For an open pipe, the difference between successive frequencies is the fundamental, in this case 392 Hz, and all frequencies would be integer multiples of this frequency. This is not the case, so the pipe cannot be an open pipe. For a stopped pipe, the difference between successive frequencies is twice the fundamental, and each frequency is an odd integer multiple of the fundamental. In this case, f1 = 196 Hz, and 1372 Hz = 7 f1 , 1764 Hz = 9 f1. b) n = 7 for 1372 Hz, n = 9 for 1764 Hz. c) f1 = v 4 L, so L = v 4 f1 = (344 m/s ) (784 Hz ) = 0.439 m. The steel rod has standing waves much like a pipe open at both ends, as shown in Figure (16.18). An integral number of half wavelengths must fit on the rod, that is, nv fn = . 2L a) The ends of the rod are antinodes because the ends of the rod are free to ocsillate. b) The fundamental can be produced when the rod is held at the middle because a node is located there. (1)(5941 m s ) = 1980 Hz. c) f1 = 2(1.50 m ) d) The next harmonic is n = 2, or f 2 = 3961 Hz. We would need to hold the rod at an n = 2 node, which is located at L 4 from either end, or at 0.375 m from either end. The shower stall can be modeled as a pipe closed at both ends, and hence there are nodes at the two end walls. Figure (15.23) shows standing waves on a string fixed at both ends but the sequence of harmonics is the same, namely that an integral number of half wavelengths must fit in the stall. a) The condition for standing waves is f n = 2nvL , so the first three harmonics are n = 1, 2, 3. b) A particular physics professor’s shower has a length of L = 1.48 m. Using f n = 2nvL , the resonant frequencies can be found when v = 344 m s . Hz 1 116 2 232 3 349 Note that the fundamental and second harmonic, which would have the greatest amplitude, are frequencies typically in the normal range of male singers. Hence, men do sing better in the shower! (For a further discussion of resonance and the human voice, see Thomas D. Rossing , The#Science#of#Sound, Second Edition, Addison2Wesley, 1990, especially Chapters 4 and 17.) a) The cross2section area of the string would be a = (900 N) (7.0 × 108 Pa) = 1.29 × 10−6 m 2 , corresponding to a radius of 0.640 mm (keeping extra figures). The length is the volume divided by the area, V m ρ (4.00 × 10−3 kg) L= = = = 0.40 m. a a (7.8 × 103 kg m 3 )(1.29 × 10− 6 m 2 ) b) Using the above result in Eq. (16.35) gives f1 = 377 Hz, or 380 Hz to two figures. a) The second distance is midway between the first and third, and if there are no other distances for which resonance occurs, the difference between the first and third positions is the wavelength λ = 0.750 m. (This would give the first distance as λ 4 = 18.75 cm, but at the end of the pipe, where the air is not longer constrained to move along the tube axis, the pressure node and displacement antinode will not coincide exactly with the end). The speed of sound in the air is then v = fλ = (500 Hz)(0.750 m) = 375 m s. b) Solving Eq. (16.10) for γ , γ= Mv 2 (28.8 × 10−3 kg mol)(375 m s) 2 = 1.39. = RT (8.3145 J mol ⋅ K)(350.15 K) c) Since the first resonance should occur at τ 4 = 0.875 m but actually occurs at 0.18 m, the difference is 0.0075 m. a) Considering the ear as a stopped pipe with the given length, the frequency of the fundamental is f1 = v 4 L = (344 m s) (0.10 m) = 3440 Hz; 3500 Hz is near the resonant frequency, and the ear will be sensitive to this frequency. b) The next resonant frequency would be 10,500 Hz and the ear would be sensitive to sounds with frequencies close to this value. But 7000 Hz is not a resonant frequency for an open pipe and the ear is not sensitive at this frequency. a) From Eq. (15.35), with m the mass of the string and M the suspended mass, f1 = F Mg = = 4mL πd 2 L2 ρ (420.0 × 10 −3 kg)(9.80 m s 2 ) = 77.3 Hz π (225 × 10− 6 m)2 (0.45 m)2 (21.4 × 103 kg m 3 ) and the tuning fork frequencies for which the fork would vibrate are integer multiples of 77.3 Hz. b) The ratio m M ≈ 9 × 10 −4 , so the tension does not vary appreciably along the string. a) L = λ 4 = v 4 f = (344 m s) (4(349 Hz)) = 0.246 m. b) The frequency will be proportional to the speed, and hence to the square root of the Kelvin temperature. The temperature necessary to have the frequency be higher is (293.15 K)(1.060) 2 = 329.5 K, which is 56.3° C. The wavelength is twice the separation of the nodes, so γRT v = λf = 2 Lf = . M Solving for γ , γ= M (16.0 × 10−3 kg) (2 Lf ) 2 = (2(0.200 m)(1100 Hz))2 = 1.27. RT (8.3145 J mol ⋅ K ) (293.15 K) If the separation of the speakers is denoted h, the condition for destructive interference is x 2 + h 2 − x = βλ , where β is an odd multiple of one2half. Adding x to both sides, squaring, cancelling the x 2 term from both sides and solving for x gives x= Using λ = v f h2 β − λ. 2 βλ 2 and h from the given data yields 9.01 m ( β = 12 ), 2.71 m( β = 32 ), 1.27 m (β = 52 ), 0.53 m (β = 72 ) and 0.026 m (β = 92 ) . These are the only allowable values of β that give positive solutions for x . (Negative values of x may be physical, depending on speaker design, but in that case the difference between path lengths is x 2 + h 2 + x.) b) Repeating the above for integral values of β , constructive interference occurs at 4.34 m, 1.84 m, 0.86 m, 0.26 m . Note that these are between, but not midway between, the answers to part (a). c) If h = λ 2 , there will be destructive interference at speaker B. If λ 2 > h, the path difference can never be as large as λ 2 . (This is also obtained from the above expression for x , with x = 0 and β = 12 .) The minimum frequency is then v 2h = (344 m s) (4.0 m) = 86 Hz. a) The wall serves as the listener, want f L = 600 Hz.  v + vS   f L fS =   v + vL  vL = 0, vS = −30 m s, v = 344 m s fS = 548 Hz b) Now the wall serves as a stationary source with f s = 600 Hz  v + vL   fS f L =   v + vS  vS = 0, vL = +30 m s, v = 344 m s f L = 652 Hz To produce a 10.0 Hz beat, the bat hears 2000 Hz from its own sound plus 2010 Hz coming from the wall. Call v the magnitude of the bat’s speed, f w the frequency the wall receives (and reflects), and V the speed of sound. Bat is moving source and wall is stationary observer: V V −v = f w 2000 Hz (1) Bat is moving observer and wall is stationary source: V +v V = 2010 Hz f w (2) Solve (1) and (2) together: v = 0.858 m s a) A = R ⋅ pmax = BkA = pmax = 2π ρB f R, I = 2πBA 2πBAf = . In air v = λ v B . Therefore ρ p 2 max = 2π 2 ρB f 2 ( R) 2 . 2 ρB b) PTot = 4πR 2 I = 8π3 ρB f 2 R 2 ( R) 2 c) I = PTot 4 πd 2 = 2π 2 ρB f 2 R 2 ( R ) 2 d2 p max = (2 ρB I )1 2 = , 2π ρB fR ( R ) p max R( R ) = . ,A= d d 2π ρB f (See also Problems 16.70 and 16.74). Let f 0 = 2.00 MHz be the frequency of the generated wave. The frequency with which the heart wall receives this wave is v+v f H = v H f 0 , and this is also the frequency with which the heart wall re2emits the wave. The detected frequency of this reflected wave is f ′ v −vvH , f H , with the minus sign indicating that the heart wall, acting now as a source of v+v + waves, is moving toward the receiver. Combining, f ′ v − v HH f 0 , and the beat frequency is  v + vH  2vH f beat = f ′ − f 0 =  f0 . − 1 f 0 = v − vH  v − vH  Solving for vH ,    f beat  85 Hz  = (1500 m s )   vH = v 6  2(2.00 × 10 Hz) + (85 Hz)   2 f 0 + f beat  = 3.19 × 10 −2 m s. Note that in the denominator in the final calculation, f beat is negligible compared to f 0 . a) λ = v f = (1482 m s) (22.0 × 103 Hz) = 6.74 × 10−2 m. or Problem 16.70; the difference in frequencies is b) See Problem 16.66  2vW  2(4.95 m s )  = 22.0 × 103 Hz f = fS  = 147 Hz. (1482 m s ) − (4.95 m s )  v − vW  ( ) The reflected waves have higher frequency. a) The maximum velocity of the siren is ωP AP = 2πf P AP . You hear a sound with frequency f L = f siren v (v + vS ), where vS varies between + 2πf P AP and − 2πf P AP . So f L − max = f siren v (v − 2πf P AP ) and f L − min = f siren v (v + 2πf P AP ). b) The maximum (minimum) frequency is heard when the platform is passing through equilibrium and moving up (down). a) Let vb be the speed of the bat, vi the speed of the insect and f i the frequency with which the sound waves both strike and are reflected from the insect. The frequencies at which the bat sends and receives the signals are related by  v + vi  v + vb   v + vb   .  = f s  f L = f i   v − vi   v − vb  v − vi  Solving for vi ,  fS  v + vb    1 −  f L  v − vb    f (v − vb ) − f S (v + vb )   vi = v = v L .  f L (v − vb ) + fS (v + vb )  f S  v + vb     1 +  f L  v − vb    Letting f L = f refl and fS = f bat gives the result. b) If f bat = 80.7 kHz, f refl = 83.5 kHz, and vbat = 3.9 m s , vinsect = 2.0 m s . (See Problems 16.66, 16.74, 16.67). a) In a time t, the wall has moved a distance v1t and the wavefront that hits the wall at time t has traveled a distance vt, where v = f 0 λ0 , and the number of wavecrests in the total distance is ( v + v1 ) t λ0 . b) The reflected wave has traveled vt and the wall has moved v1t , so the wall and the wavefront are separated by (v − v1 )t. c) The distance found in part (b) must contain the number of reflected waves found in part (a), and the ratio of the quantities is the wavelength of the v−v reflected wave, λ 0 v + v11 . d) The speed v divided by the result of part (c), expressed in v −v terms of f 0 is f 0 v + v11 . This is what is predicted by the problem2solving strategy. v+v e) f 0 v − v11 − f 0 = f 0 v − v11 . 2v a) fR = fL 12 −1 2 1 − vc c−v  v  v = fS = f S 1 −  1 +  . c+v 1 + vc  c  c b) For small x, the binomial theorem (see Appendix B) gives (1 − x ) 12 (1 + x ) −1 2 ≈ 1 − x 2, ≈ 1 − x 2, so 2 v    v f L ≈ f S 1 −  ≈ f S 1 −   2c   c where the binomial theorem has been used to approximate (1 − x 2) ≈ 1 − x. 2 The above result may be obtained without resort to the binomial theorem by expressing f R in terms of f S as fR = fS 1 − (v c ) 1 − ( v c ) 1 + (v c ) 1 − ( v c ) = fS 1 − (v / c) 1 − (v c ) 2 . To first order in v c , the square root in the denominator is 1, and the previous result is obtained. c) For an airplane, the approximation v << c is certainly valid, and solving the expression found in part (b) for v, v=c f 46.0 Hz fS − f R = c beat = (3.00 × 108 m s) = 56.8 m s , fS 2.43 × 108 Hz fS and the approximation v << c is seen to be valid. Note that in this case, the frequency difference is known to three figures, so the speed of the plane is known to three figures. a) As in Problem 16.71, v=c fS − f R − 0.018 × 1014 Hz = (3.00 × 108 m s) = −1.2 × 106 m s , 8 fS 4.568 × 10 Hz with the minus sign indicating that the gas is approaching the earth, as is expected since f R > f S . b) The radius is (952 yr) (3.156 × 107 s yr)(1.2 × 106 m s) = 3.6 × 1016 m = 3.8 ly . This may also be obtained from (952 yr) f R − fS fS . c) The ratio of the width of the nebula to 2π times the distance from the earth is the ratio of the angular width (taken as 5 arc minutes) to an entire circle, which is 60 × 360 arc minutes. The distance to the nebula is then (keeping an extra figure in the intermediate calculation) 2 × 3.75 ly (60) × (360) = 5.2 × 10 3 ly, 5 so the explosion actually took place about 4100 B.C a) The frequency is greater than 2800 MHz; the thunderclouds, moving toward the installation, encounter more wavefronts per time than would a stationary cloud, and so an observer in the frame of the storm would detect a higher frequency. Using the result of Problem 16.71, with v = −42.0 Km h, f R − f S = fS (42.0 km h) (3.6 km h/1 m s) −v = (2800 × 106 Hz) = 109 Hz. c (3.00 × 108 m s) b) The waves are being sent at a higher frequency than 2800 MHz from an approaching source, and so are received at a higher frequency. Repeating the above calculation gives the result that the waves are detected at the installation with a frequency 109 Hz greater than the frequency with which the cloud received the waves, or 218 Hz higher than the frequency at which the waves were originally transmitted at the receiver. Note that in doing the second calculation, fS = 2800 MHz + 109 Hz is the same as 2800 MHz to three figures. a) (See also Example 16.19 and Problem 16.66.) The wall will receive and v f 0 , and the woman will hear this reflected wave at a reflect pulses at a frequency v − vw frequency v + vw v + vw v ⋅ f0 ; f0 = v − vw v v − vw The beat frequency is  v + vw   2vw  . − 1 = f 0  f beat = f 0   v − vw   v − vw  b) In this case, the sound reflected from the wall will have a lower frequency, and using f 0 (v − vw ) (v + vw ) as the detected frequency (see Example 21212; vw is replaced by − vw in the calculation of part (a)),  v − vw   2vw   = f 0  . f beat = f 0 1 −  v + vw   v + vw  Refer to Equation (16.31) and Figure (16.38). The sound travels a distance vT and the plane travels a distance vsT before the boom is found. So, h 2 = (vT ) 2 + (vsT ) 2 , or hv v vsT = h 2 − v 2T 2 . From Equation (16.31), sin α = . Then, vs = . 2 vs h − v 2T 2 a) b) From Eq. (16.4), the function that has the given p ( x, 0) at t = 0 is given graphically as shown. Each section is a parabola, not a portion of a sine curve. The period is λ v = (0.200 m) (344 m s) = 5.81 × 10−4 s and the amplitude is equal to the area under the p − x curve between x = 0 and x = 0.0500 m divided by B, or 7.04 × 10 −6 m. c) Assuming a wave moving in the + x 2direction, y (0, t ) is as shown. d) The maximum velocity of a particle occurs when a particle is moving throughout the origin, and the particle speed is v y = − ∂∂yx v = pvB . The maximum velocity is found from the maximum pressure, and v ymax = (40 Pa)(344 m s) (1.42 × 105 Pa) = 9.69 cm s . The maximum acceleration is the maximum pressure gradient divided by the density, a max = (80.0 Pa) (0.100 m) = 6.67 × 10 2 m s 2 . (1.20 kg m 3 ) e) The speaker cone moves with the displacement as found in part (c ); the speaker cone alternates between moving forward and backward with constant magnitude of acceleration (but changing sign). The acceleration as a function of time is a square wave with amplitude 667 m s 2 and frequency f = v λ = (344 m s) (0.200 m) = 1.72 kHz. Taking the speed of sound to be 344 m s, the wavelength of the waves emitted by each speaker is 2.00 m. a) Point C is two wavelengths from speaker A and one and one2half from speaker B , and so the phase difference is 180° = π rad. b) I= P 8.00 × 10−4 W = = 3.98 × 10− 6 W m 2 , 2 2 4πr 4π (4.00 m) and the sound intensity level is (10 dB) log(3.98 × 106 ) = 66.0 dB. Repeating with P = 6.00 × 10−5 W and r = 3.00 m gives I = 5.31 × 10−7 W and β = 57.2dB. c) With the result of part (a), the amplitudes, either displacement or pressure, must be subtracted. That is, the intensity is found by taking the square roots of the intensities found in part (b), subtracting, and squaring the difference. The result is that I = 1.60 × 10−6 W and β = 62.1 dB. Capítulo 17 From Eq. (17.1), a) (9 5)(− 62.8) + 32 = −81.0°F. b) (9 5)(56.7 ) + 32 = 134.1°F. c) (9 5)(31.1) + 32 = 88.0°F. From Eq. (17.2 ), a) (5 9)(41.0 − 32) = 5.0°C. b) (5 9)(107 − 32) = 41.7°C. c) (5 / 9)(− 18 − 32) = −27.8°C. 1 C° = 95 F°, so 40.0 = 72.0 F° T2 = T1 + 70.0 F° = 140.2°F a) (5 9) (45.0 − (−4.0)) = 27.2° C. b) (5 9) (−56.0 − 44) = −55.6° C. a) From Eq. (17.1), (9 5)(40.2 ) + 32 = 104.4°F, which is cause for worry. b) (9 5)(12) + 32 = 53.6°F, or 54°F to two figures. (9 5)(11.8) = 21.2 F° 1 K = 1 C° = 95 F° , so a temperature increase of 10 K corresponds to an increase of 18 F° . Beaker B has the higher temperature. For (b), TC = TK = −10.0 C°. Then for (a), TF = = −18.0 F°. 9 5 TC = 9 5 (− 10.0 C°) Combining Eq. (17.2) and Eq. (17.3), 5 TK = (TF − 32°) + 273.15, 9 and substitution of the given Fahrenheit temperatures gives a) 216.5 K, b) 325.9 K, c) 205.4 K. (In these calculations, extra figures were kept in the intermediate calculations to arrive at the numerical results.) a) TC = 400 − 273.15 = 127°C, TF = (9 / 5)(126.85) + 32 = 260°F. b) TC = 95 − 273.15 = −178°C, TF = (9 / 5)(−178.15) + 32 = −289°F. c) TC = 1.55 × 107 − 273.15 = 1.55 × 107°C, TF = (9 / 5)(1.55 × 107 ) + 32 = 2.79 × 107°F. From Eq. (17.3), TK = (−245.92°C) + 273.15 = 27.23 K. From Eq. (17.4), (7.476)(273.16 K) = 2042.14 K − 273.15 = 1769°C. .15 K ) = 444 mm. From Eq. (17.4), (325.0 mm)( 373 273.16 K On the Kelvin scale, the triple point is 273.16 K, so °R = (9/5)273.15 K = 491.69°R. One could also look at Figure 17.7 and note that the Fahrenheit scale extends from − 460°F to + 32°F and conclude that the triple point is about 492 °R. From the point2slope formula for a straight line (or linear regression, which, while perhaps not appropriate, may be convenient for some calculators), 4.80 × 10 4 Pa (0.01°C) − (100.0°C) = −282.33°C, 6.50 × 10 4 Pa − 4.80 × 10 4 Pa which is − 282°C to three figures. b) Equation (17.4) was not obeyed precisely. If it were, the pressure at the triple ×104 Pa = 4.76 ×10 4 Pa. point would be P = (273.16) 6.50373 .15 ( ( ) T = ( L ) (αL0 ) = 25 × 10 −2 m so the temperature is 183° C ) ((2.4 × 10 (C°) (62.1 m )) = 168° C, −5 −1 αL0 T = (1.2 × 10−5 (C°) −1 )(1410 m)(18.0° C − (−5.0)°C) = +0.39 m. d + d = d (1 + α T ) = (0.4500 cm)(1 + (2.4 × 10−5 (C°) −1 )(23.0° C − (−78.0°C))) = 0.4511 cm = 4.511 mm. a) αD0 T = (2.6 × 10 −5 (C°) −1 )(1.90 cm) (28.0°C) = 1.4 × 10 −3 cm, so the diameter is 1.9014 cm. b) αD0 T = −3.6 × 10−3 cm, so the diameter is 1.8964 cm. α T = (2.0 × 10 −5 (C°) −1 )(5.00° C − 19.5° C) = −2.9 × 10 −4. α = ( L) ( L0 T ) = (2.3 × 10−4 m ) ((40.125 × 10−2 m )(25.0 C°)) = 2.3 × 10 −5 (C°) . −1 From Eq. (17.8), ( T= V V0 β β V0 T = 75 × 10 −5 (C°) −1 = 1.50×10 −3 5.1×10 −5 K −1 = 29.4°C, so T = 49.4°C. )(1700 L)(− 9.0°C) = −11 L, so there is 11 L of air. The temperature change is T = 18.0° C − 32.0° C = −14.0 C°. The volume of ethanol contracts more than the volume of the steel tank does, so the additional amount of ethanol that can be put into the tank is Vsteel − Vethanol = ( βsteel − βethanol )V0 T ( )( ) = 3.6 × 10 −5 (C°) − 75 × 10 −5 (C°) −1 2.80 m3 (− 14.0 C°) = 0.0280 m 3 −1 The amount of mercury that overflows is the difference between the volume change of the mercury and that of the glass; 8.95 cm3 −1 βglass = 18.0 × 10− 5 K −1 − = 1.7 × 10− 5 (C°) . 3 1000 cm (55.0°C ) ( a) A = L2 , A = 2 L L = 2 A = 2α TA0 = (2α )A0 T . b) L L ( L2 = 2 ) L L ) A0 . But L L = α T , and so A = (2α )A T = (2 ) (2.4 × 10−5 (C°) −1 ) (π × (.275 m ) ) (12.5°C ) = 1.4 × 10−4 m 2 . 2 a) A0 = πD 2 π 2 = (1.350 cm ) = 1.431 cm 2 . 4 4 b) A = A0 (1 + 2α T ) = (1.431 cm 2 )(1 + (2 )(1.20 × 10−5°C )(150°C )) = 1.437 cm 2 . (a ) No, the brass expands more than the steel. (b) call D the inside diameter of the steel cylinder at 20°C At 150°C : DST = DBR D + DST = 25.000 cm + D BR D + αST D° T = 25 cm + αBR (25 cm) T D = 25 cm(1 + αBR T ) 1 + αST T [ (25 cm) 1 + (2.0 × 10− 5 (C°) −1 )(130C°) = 1 + (1.2 × 10− 5 (C°) −1 )(130 C°) = 25.026 cm ] The aluminum ruler expands to a new length of L = L0 (1 + α T ) = (20.0 cm)[1 + (2.4 × 10−5 (C°) −1 )(100 C°)] = 20.048 cm The brass ruler expands to a new length of L = L0 (1 + α T ) = (20.0 cm)[1 + (2.0 × 10−5 (C°) −1 )(100 C°)] = 20.040 cm The section of the aluminum ruler will be longer by 0.008 cm From Eq. (17.12), F = −Yα TA = −(0.9 × 1011 Pa)(2.0 × 10−5 (C°) −1 )(−110°C)(2.01 × 10− 4 m 2 ) = 4.0 × 104 N. a) α = ( L L0 T ) = (1.9 × 10 −2 m) ((1.50 m)(400 C°) ) = 3.2 × 10 −5 (C°) −1. b) Yα T = Y L L0 = (2.0 × 1011 Pa)(1.9 × 10−2 m) (1.50 m) = 2.5 × 109 Pa. a) L = α TL = (1.2 × 10−5 K −1 )(35.0 K)(12.0 m) = 5.0 × 10−3 m. b) Using absolute values in Eq. (17.12), F = Yα T = (2.0 × 1011 Pa)(1.2 × 10−5 K −1 )(35.0 K) = 8.4 × 107 Pa. A a) (37°C − (−20°C))(0.50 L)(1.3 × 10−3 kg L) (1020 J kg ⋅ K ) = 38 J b) There will be 1200 breaths per hour, so the heat lost is (1200)(38 J) = 4.6 × ×10 4 J. t= Q mc T (70 kg)(3480 J kg ⋅ K )(7 C°) = = = 1.4 × 103 s, about 24 min. P P (1200 W ) Using Q=mgh in Eq. (17.13) and solving for T= Τ gives gh (9.80 m s 2 )(225 m) = = 0.53 C°. c (4190 J kg.K ) a) The work done by friction is the loss of mechanical energy, 1 1   mgh + m(v12 − v22 ) = (35.0 kg ) (9.80 m s 2 )(8.00 m) sin 36.9° − (2.50 m s) 2  2 2   3 = 1.54 × 10 J. b) Using the result of part (a) for Q in Eq. (17.13) gives ( T = 1.54 × 103 J ) ((35.0 kg )(3650 J kg ⋅ K )) = 1.21× 10 (210° C − 20°C)((1.60 kg )(910 J −2 C°. kg ⋅ K ) + (0.30 kg )(470 J kg ⋅ K )) = 3.03 × 105 J. Assuming Q = (0.60) × 10 × K , 1 (6) 12 (1.80 kg )(7.80 m s ) = 45.1 C°. MV 2 K T = (0.60) × 10 × =62 = mc mc 8.00 × 10− 3 kg (910 J kg ⋅ K ) 2 ( (85.0° C − 20.0° C)((1.50 kg )(910 J = 5.79 × 10 J. 5 ) kg ⋅ K ) + (1.80 kg )(4190 J kg ⋅ K )) a) Q = mc T = (0.320 kg )(4190 J kg ⋅ K )(60.0 K ) = 8.05 × 10 4 J. b) t = Q P = a) c = 8.05×10 4 J 200 W = 402 s. (120 s )(65.0 W ) Q = = 2.51 × 103 J kg ⋅ K. m T (0.780 kg )(22.54° C − 18.55° C ) b) An overstimate; the heat Q is in reality less than the power times the time interval. The temperature change is c= T = 18.0 K, so ( )( ) Q gQ 9.80 m s 2 1.25 × 104 J = = = 240 J kg ⋅ K. m T w T (28.4 N )(18.0 K ) a) Q = mc T , c = 470 J kg ⋅ K We need to find the mass of 3.00 mol: ( ) m = nM = (3.00 mol) 55.845 × 10 −3 kg mol = 0.1675 kg T = Q mc = (8950 J ) [(0.1675 kg )(470 J kg ⋅ K )] = 114 K = 114 C° b) For m = 3.00 kg, T = Q mc = 6.35 C° c) The result of part (a) is much larger; 3.00 kg is more material than 3.00 mol. Qmelt (10,000 J min )(1.5 min ) = = 30,000 J kg m 0.50 kg Q (b) Liquid : Q = mc T → c = m T ( 10,000 J min )(1.5 min ) c= = 1,000 J kg ⋅ C° (0.50 kg )(30C°) (a) Solid : c = LF = (10,000 J min )(1.0 min ) = 1300 J kg ⋅ C° Q = (0.50 kg )(15 C°) m T a) Qwater + Qmetal = 0 mwater c water Twater + mmetal cmetal Tmetal = 0 (1.00kg )(4190 J kg ⋅ K )(2.0 C°) + (0.500 kg )(cmetal )(− 78.0 C°) = 0 cmetal = 215 J/kg ⋅ K b) Water has a larger specific heat capacity so stores more heat per degree of temperature change. c) If some heat went into the styrofoam then Qmetal should actually be larger than in part (a), so the true c metal is larger than we calculated; the value we calculated would be smaller than the true value. a) Let the man be designated by the subscript m and the “‘water” by w, and T is the final equilibrium temperature. − mm C m Tm = m w C w Tw − mmCm (T − Tm ) = mw Cw (T − Tw ) mm C m (Tm − T ) = mw C w (T − Tw ) Or solving for T, T = m m C m Tm + m w C w Tw mm C m + mw C w . Inserting numbers, and realizing we can change K to °C , and the mass of water is .355 kg, we get (70.0 kg) (3480 J kg ⋅ K ) (37.0°C ) + (0.355 kg) (4190 J kg ⋅ °C ) (12.0°C ) T= (70.0 kg)(3480 J kg. °C ) + (0.355 kg) (4190 J kg ⋅ °C ) Thus, T = 36.85°C. b) It is possible a sensitive digital thermometer could measure this change since they can read to .1°C. It is best to refrain from drinking cold fluids prior to orally measuring a body temperature due to cooling of the mouth. The rate of heat loss is Q t= ( 70.355 kg)(3480 J kg.° C)(0.15° C) 7×10 6 J day ( )= t. Q t mC T t , or t = mC t Q ( t) . Interesting numbers, = 0.005 d, or t = 7.6 minutes. This may acount for mothers taking the temperature of a sick child several minutes after the child has something to drink. Q = m(c T + Lf ) ( = (0.350 kg) (4190 J kg ⋅ K )(18.0 K) + 334 × 103 J kg = 1.43 × 105 J = 34.2 kcal = 136 Btu. ) Q = m(cice Tice + Lf + cwater Twater + LV )  (2100 J kg ⋅ K)(10.0 C°) + 334 × 103 J kg   = (12.0 × 10 kg) 3  ( 100 C )(4190 J kg K) 2256 10 J kg + ° ⋅ + ×   4 = 3.64 × 10 J = 8.69 kcal = 34.5 Btu. −3 a) t = b) mLf P = Q mc T (0.550 kg)(2100 J kg ⋅ K)(15.0 K) = = = 21.7 min . P P (800 J min) ( 0.550 kg)(334×10 3 J kg ) (800 J min ) = 230 min, so the time until the ice has melted is 21.7 min + 230 min = 252 min. ((4000 lb) 2.205 lb kg) )(334 × 103 J kg) = 7.01 kW = 2.40 × 104 Btu hr. (86,400 s) a) m(c T + Lv ) = (25.0 × 10−3 kg)((4190 J kg ⋅ K)(66.0 K) + 2256 × 103 J kg ) = 6.33 × 104 J. b) mc T = (25.0 × 10−3 kg)(4190 J kg ⋅ K)(66.0 K) = 6.91 × 103 J. c) Steam burns are far more severe than hot2water burns. With Q = m(c T + Lf ) and K = (1 / 2)mv 2 , setting Q = K and solving for v gives v = 2((130 J Kg ⋅ K)(302.3 C°) + 24.5 × 103 J kg) = 357 m s . a) msweat = Mc T (70.0 kg)(3480 J kg ⋅ K)(1.00 K) = = 101 g. Lv (2.42 × 106 J kg) b) This much water has a volume of 101 cm 3 , about a third of a can of soda. The mass of water that the camel saves is Mc T (400 kg)(3480 J kg ⋅ K)(6.0 K) = = 3.45 kg, Lv (2.42 × 106 J kg) which is a volume of 3.45 L. For this case, the algebra reduces to  ((200)(3.00 × 10 −3 kg ))(390 J kg ⋅ K )(100.0 C°)      ( 0 . 240 kg )( 4190 J kg )( 20 . 0 C ) + °  = 35.1°C. T=  ((200)(3.00 × 10 −3 kg )(390 J kg ⋅ K )     + (0.240 kg )(4190 J kg ⋅ .K )   The algebra reduces to  ((0.500 kg)(390 J kg ⋅ K ) + (0.170 kg)(4190 J kg ⋅ K ))(20.0° C  + (0.250 kg)(470 J kg ⋅ K )(85.0°C) T=  ((0.500 kg)(390 J kg ⋅ K ) + (0.170 kg)(4190 J kg ⋅ K ))    + (0.250 kg)(470 J kg ⋅ K )      = 27.5°C The heat lost by the sample is the heat gained by the calorimeter and water, and the heat capacity of the sample is c= Q ((0.200 kg)(4190 J kg ⋅ K ) + (0.150 kg)(390 J kg ⋅ K ))(7.1 C°) = m T (0.0850 kg)(73.9 C°) =1010 J kg ⋅ K, or 1000 J kg ⋅ K to the two figures to which the temperature change is known. The heat lost by the original water is − Q = (0.250 kg)(4190 J kg ⋅ K )(45.0 C°) = 4.714 × 104 J, and the mass of the ice needed is mice = cice = −Q Tice + Lf + c water Twater (4.714 × 104 ) J (2100 J kg ⋅ K) (20.0 C°) + (334 × 103 J kg) + (4190 J kg ⋅ K )(30.0 C°) = 9.40 × 10 −2 kg = 94.0 g. The heat lost by the sample (and vial) melts a mass m, where Q ((16.0 g)(2250 J kg ⋅ K ) + (6.0 g)(2800 J kg ⋅ K ))(19.5K) m= = = 3.08 g. Lf (334 × 103 J kg) Since this is less than the mass of ice, not all of the ice melts, and the sample is indeed cooled to 0°C. Note that conversion from grams to kilograms was not necessary. (4.00 kg)(234 J kg ⋅ K)(750 C°) = 2.10 kg. (334 × 103 J kg) Equating the heat lost by the lead to the heat gained by the calorimeter (including the water2ice mixtue), mP b c Pb (200°C − T ) = (mw + mice )c wT + mcu ccu T + mice Lf . Solving for the final temperature T and using numerical values,  (0.750 kg)(130 J kg ⋅ K)(255 C°)    3   − ( 0 . 018 kg)(334 × 10 J kg)  = 21.4°C. T=  (0.750 kg)(130 ⋅ J kgK)     + (0.178 kg)(4190 J kg ⋅ K)   + (0.100 kg)(390 J kg ⋅ K)    (The fact that a positive Celsius temperature was obtained indicates that all of the ice does indeed melt.) The steam both condenses and cools, and the ice melts and heats up along with the original water; the mass of steam needed is (0.450 kg)(334 × 103 J kg) + (2.85 kg)(4190 J kg ⋅ K)(28.0 C°) msteam = 2256 × 103 J kg + (4190 J kg )(72.0 C°) = 0.190 kg. 2 The SI units of H&and& dQ dt are both watts, the units of area are m , temperature difference is in K, length in meters, so the SI units for thermal conductivity are [ W][m] W = . 2 [m ][K] m ⋅ K 100 K a) 0.450 = 222 K m. b)(385 W m ⋅ K)(1.25 × 1024 m 2 )(400 K m) = 10.7 W. m c)100.0°C − (222 K m)(12.00 × 10−2 m) = 73.3°C. dm Using the chain rule, H = dQ dt = Lf dt and solving Eq. (17.21) for k, dm L k = Lf dt A T (8.50 × 10− 3 kg) (60.0 × 10− 2 s) = (334 × 103 J kg ) & (600 s) (1.250 × 10− 4 m 2 )(100 K) = 227 W m ⋅ K. (Although it may be easier for some to solve for the heat flow per unit area, part (b), first the method presented here follows the order in the text.) a) See Example 17.13; as in that example, the area may be divided out, and solving for temperature T at the boundary, T= (k foam Lfoam )Tin + (k wood Lwood )Tout ) (k foam Lfoam ) + (k wood Lwood ) ((0.010 W m ⋅ K ) (2.2 cm))(19.0° C) + ((0.080 W m ⋅ K ) (3.0 cm ))(− 10.0° C) ((0.010 W m ⋅ K ) (2.2 cm )) + ((0.080 W m ⋅ K ) (3.0 cm)) = −5.8°C. Note that the conversion of the thickness to meters was not necessary. b) Keeping extra figures for the result of part, (a), and using that result in the temperature difference across either the wood or the foam gives (19.0° C − (− 5.767° C)) H foam H wood = = (0.010 W m ⋅ K ) A A 2.2 × 10− 2 m = (0.080 W m ⋅ K ) (− 5.767°C − (− 10.0°C )) 3.0 × 10 −2 m = 11 W m 2 . a) From Eq. (17.21), ( H = (0.040 W m ⋅ K ) 1040 m 2 K) ) (4.0(140 = 196 W, × 10 m ) −2 or 200 W to two figures. b) The result of part (a) is the needed power input. From Eq. (17.23), the energy that flows in time H t= ( ) t is A T 125 ft 2 (34F°) t= (5.0 h ) = 708 Btu = 7.5 × 105 J. 2 R 30 ft ⋅ F° ⋅ h Btu ( ) a) The heat current will be the same in both metals; since the length of the copper rod is known, ( H = (385.0 W m ⋅ K ) 400 × 10 −4 m 2 ) ((135.00.0 mK )) = 5.39 W. b) The length of the steel rod may be found by using the above value of H in Eq. (17.21) and solving for L2 , or, since H and A are the same for the rods, L2 = L (50.2 W m ⋅ K )(65.0 K ) = 0.242 m. k2 T2 = (1.00 m ) k T (385.0 W m ⋅ K )(35.0 K ) (see Problem 17.66) in Eq. (17.21), Using H = Lv dm dt dm L T = Lv dt kA (0.390 kg) (0.85 × 10−2 m) = (2256 × 103 J kg ) = 5.5 C°, (180 s) (50.2 W m ⋅ K )(0.150 m 2 ) and the temperature of the bottom of the pot is 100° C + 6 C° = 106° C. Q T = kA t L W   300 K   150 J s =  50.2  A  m. K   0.500 m   A = 4.98 × 10− 3 m 2 D A = πR 2 = π   2 2 D = 4A π = 4(4.98 × 10−3 m 2 ) π = 8.0 × 10− 2 m = 8.0 cm H a = H b (a = aluminum, b = brass) A(150.0°C − T ) A(T − 0° C) H a = ka , H b = kb La Lb (It has been assumed that the two sections have the same cross2sectional area.) A(T − 0°C) A(150.0°C − T ) ka = kb Lb La (2050 W m ⋅ K)(150.0° C − T ) (109.0 W m ⋅ K)(T − 0°C) = 0.800 m 0.500 m Solving for T gives T = 90.2°C From Eq. (17.25), with e = 1, a) (5.67 × 10−8 W m 2 ⋅ K 4 )(273 K) 4 = 315 W m 2 . b) A factor of ten increase in temperature results in a factor of 10 4 increase in the output; 3.15 × 10 6 W m 2 . Repeating the calculation with Ts = 273 K + 5.0 C° = 278 K gives H = 167 W. The power input will be equal to H net as given in Eq. (17.26); P = Aeσ (T 4 − Ts4 ) = (4π (1.50 × 10− 2 m) 2 )(0.35)(5.67 × 10−8 W m 2 ⋅ K 4 )((3000 K) 4 − (290 K) 4 ) = 4.54 × 103 W. A= H 150 W = = 2.10 cm 2 4 4 −8 2 4 eσ T (0.35) 5.67 × 10 W m ⋅ K (2450K ) ( ) The radius is found from R= A = 4π ( ) H σT 2 = 4π H 1 . 4πσ T 2 Using the numerical values, the radius for parts (a ) and (b ) are Ra = Rb = (2.7 × 10 32 ) W 1 = 1.61 × 1011 m 2 4 4π 5.67 × 10 W m ⋅ K (11,000 K )2 ( −8 (2.10 × 10 23 ) ) W 1 = 5.43 × 106 m 2 4 4π 5.67 × 10 W m ⋅ K (10,000 K )2 ( −8 ) c) The radius of Procyon B is comparable to that of the earth, and the radius of Rigel is comparable to the earth2sun distance. a) normal melting point of mercury: − 30° C = 0.0°Μ normal boiling point of mercury: 357° C = 100.0 °Μ 100 Μ° = 396 C° so 1 Μ° = 3.96 C° Zero on the M scale is − 39 on the C scale, so to obtain TC multiple TΜ by 3.96 and then subtract 39° : TC = 3.96TM − 39° 1 (TC + 39°) Solving for TM gives TM = 3.96 1 (100° + 39°) = 35.1°Μ The normal boiling point of water is 100°C; TM = 3.96 b)10.0 Μ ° = 39.6° C° All linear dimensions of the hoop are increased by the same factor of α T , so the increase in the radius of the hoop would be Rα T = (6.38 × 106 m )(1.2 × 10−5 K −1 )(0.5 K ) = 38 m. The tube is initially at temperature T0 , has sides of length L0 volume V0 , density ρ0 , and coefficient of volume expansion β. a) When the temperature increase to T0 + T , the volume changes by an amount m m V , where V = β V0 T . Then, ρ = , or eliminating V , ρ = . V0 + V V0 + βV0 T Divide the top and bottom by V0 and substitute ρ0 = m V0 . Then m V0 ρ0 ρ= or ρ = . This can be rewritten as ρ = ρ0 (1 + β T ) −1. Then V0 V0 + βV0 T V0 1+ β T using the expression (1 + x ) ≈ 1 + nx, where n = −1, ρ = ρ0 (1 − β T ). n b) The copper cube has sides of length 1.25 cm = .0125 m, and T = 70.0° C − 20.0° C = 50.0° C. V = βV0 T = (5.1 × 10 −5 ° C )(.0125 m ) (50.0° C ) = 5 × 10 −9 m3 . 3 Similarly, ρ = 8.9 × 103 kg m3 (1 − (5.1 × 10−5 ° C)(50.0° C)), or ρ = 8.877 × 103 kg m3 ; extra significant figures have been keep. So ρ = −23 kg m3 . (a) We can use differentials to find the frequency change because all length changes are small percents . Let m be the mass of the wire v = F 0 = F (m L) = FL m v f = and λ = 2 L(fundamental) λ FL m 1 F = f =v λ= 2L 2 mL ∂F f ≈ L (only L changes due to heating) ∂L f ≈ f 1 2 ( 12 )( F mL) −1 2 (− mLF2 ) L 1 2 F mL = 1 L 2 L 1 1 f ≈ (α T ) f = (1.7 × 10−5 (C°) −1 )(40C°)(440 C°)(440 Hz) = 0.15 Hz 2 2 The frequency decreases since the length increases (b) v = F 0 = FL m v 12 ( FL m) −1 2 ( F m) L L α T = = = v 2L 2 FL m 1 = (1.7 × 10 −5 (C°) −1 )(40C°) = 3.4 × 10 − 4 = 0.034% 2 (c) λ = 2 L → λ = 2 L → λλ = 22 LL = LL = α T λ = (1.7 × 10−5 C −1 )(40C°) = 6.8 × 10− 4 = 0.068% : λ it increases Both the volume of the cup and the volume of the olive oil increase when the temperature increases, but β is larger for the oil so it expands more. When the oil starts to overflow, Voil = Vglass + (1.00 × 10−3 m) A, where A is the cross2sectional area of the cup. Voil = V0, oil βoil T = (9.9 cm) Aβoil T Vglass = V0, glass β glass T = (10.0 cm) Aβglass T (9.9 cm) Aβoil T = (10.0 cm) Aβglass T + (1.00 × 10−3 m) A The area A divides out. Solving for T gives T = 15.5 C° T2 = T1 + T = 37.5°C Volume expansion: dV = βV dT dV dT Slope of graph β= = V V Construct the tangent to the graph at 2°C and 8°C and measure the slope of this line. 3 At 22°C : Slope ≈ − 0.103Ccm and V ≈ 1000 cm3 ° 0.10 cm3 3C° ≈ −3 × 10− 5 (C°) −1 1000 cm3 The slope in negative, as the water contracts or it is heated. At 3 3 8° C : slope ≈ 0.244Ccm ° and V ≈ 1000 cm β≈− 0.24 cm3 4C ° ≈ 6 × 10−5 (C°) −1 1000 cm3 The water now expands when heated. β≈ La + Ls = 0.40 cm (a = aluminum, s = steel) L = L0α T , so (24.8 cm)(2.4 × 10−5 (C °) −1 )) T + (34.8 cm)(1.2 × 10−5 (C °) −1 )) T = 0.40 cm T = 395 C° T2 = T1 + T = 415°C a) The change in height will be the difference between the changes in volume of the liquid and the glass, divided by the area. The liquid is free to expand along the column, but not across the diameter of the tube, so the increase in volume is reflected in the change in the length of the columns of liquid in the stem. b) h= V liquid − V glass = V ( β liquid − β glass ) T A A (100 × 10 m3 ) (8.00 × 10− 4 K −1 − 2.00 × 10 −5 K 21 )(30.0 K) = (50.0 × 10− 6 m 2 ) −6 = 4.68 × 10 −2 m. To save some intermediate calculation, let the third rod be made of fractions f1 and f 2 of the original rods; then f1 + f 2 = 1 and f1 (0.0650) + f 2 (0.0350) = 0.0580. These two equations in f1 and f 2 are solved for 0.0580 − 0.0350 f1 = , f 2 = 1 − f1 , 0.0650 − 0.0350 and the lengths are f 1 (30.0 cm) = 23.0 cm and f 2 (30.0 cm) = 7.00 cm a) The lost volume, 2.6 L, is the difference between the expanded volume of the fuel and the tanks, and the maximum temperature difference is V T= ( βfuel − βA1 )V0 (2.6 × 10− 3 m 3 ) (9.5 × 10− 4 (C°) −1 − 7.2 × 10− 5 (C°) −1 )(106.0 × 10− 3 m 3 ) = 2.78 C°, or 28°C to two figures; the maximum temperature was 32°C. b) No fuel can spill if the tanks are filled just before takeoff. = L L0 = F AY a) The change in length is due to the tension and heating + α T . Solving for F A,& FA = Y 2L − α2T . L0 ( ) b) The brass bar is given as “heavy” and the wires are given as “fine,” so it may be assumed that the stress in the bar due to the fine wires does not affect the amount by which the bar expands due to the temperature increase. This means that in the equation preceding Eq. (17.12), L is not zero, but is the amount αbrass L0 T that the brass expands, and so F = Ysteel (αbrass − αsteel ) T A = 20 × 1010 Pa)(2.0 × 10− 5 (C°) −1 − 1.2 × 10− 5 (C°) −1 )(120°C) = 1.92 × 108 Pa. In deriving Eq. (17.12), it was assumed that L = 0; if this is not the case when there are both thermal and tensile stresses, Eq. (17.12) becomes F   L = L0  α T + . AY   For the situation in this problem, there are two length changes which must sum to zero, and so Eq. (17.12) may be extended to two materials a&and b in the form   F  F   = 0.  + L0b  αb T + L0a  αa T + AY AY a  b    Note that in the above, T , F and A are the same for the two rods. Solving for the stress F A, F αa L0a + αb L0b T =− A (( Loa Ya ) + ( L0b Yb )) = (1.2 × 102 5 (C°) −1 )(0.350 m) + (2.4 × 10− 5 (C°) −1 )(0.250 m) (60.0 C°) ((0.350 m) 20 × 1010 Pa) + (0.250 m 7 × 1010 Pa)) = −1.2 × 108 Pa to two figures. a) T = ( 0.0020 in.) R = αR0 (1.2×10 − 5 (C ° ) −1 ( 2.5000 in.) = 67 C° to two figures, so the ring should be warmed to 87°C. b) the difference in the radii was initially 0.0020 in., and this must be the difference between the amounts the radii have shrunk. Taking R0 to be the same for both rings, the temperature must be lowered by an amount R T= (αbrass − αsteel )R0 (0.0020 in.) = = 100 C° −1 −1 −5 2.0 × 10 (C°) − 1.2 × 10 −5 (C°) (2.50 in.) to two figures, so the final temperature would be − 80°C. ( ) a) The change in volume due to the temperature increase is βV T , and the change in volume due to the pressure increase is − VB p (Eq. (11.13)). Setting the net change equal to zero, βV T = V p = (1.6 × 10 Pa )(3.0 × 10 K 11 −5 −1 p B , or p = Bβ V . b) From the above, )(15.0 K ) = 8.64 × 10 7 Pa. As the liquid is compressed, its volume changes by an amount V = − pkV0 . When cooled, the difference between the decrease in volume of the liquid and the decrease in volume of the metal must be this change in volume, or (α1 − αm )V0 T = V . Setting the expressions for V equal and solving for T gives pk (5.065 × 106 Pa )(8.50 × 10−10 Pa −1 ) = −9.76 C°, T= = α m − α1 (3.90 × 10− 5 K −1 − 4.8 × 10− 4 K −1 ) so the temperature is 20.2°C. Equating the heat lost be the soda and mug to the heat gained by the ice and solving for the final temperature T =  ((2.00 kg )(4190 J kg ⋅ K ) + (0.257 kg )(910 J kg ⋅ K ))(20.0C°)    − (0.120 kg )((2100 J kg ⋅ K )(15.0 C°) + 334 × 10 3 J kg )    (2.00 kg )(4190 J kg ⋅ K ) + (0.257 kg )(910 J kg ⋅ K ) + (0.120 kg )(4190 J kg ⋅ K ) = 14.1°C. Note that the mass of the ice (0.120 kg ) appears in the denominator of this expression multiplied by the heat capacity of water; after the ice melts, the mass of the melted ice must be raised further to T . (7700 m s ) K (1 2 )mv 2 v2 = = = = 54.3. Q cm T 2c T 2(910 J kg ⋅ K )(600C°) b) Unless the kinetic energy can be converted into forms other than the increased heat of the satellite cannot return intact. 2 a) a) The capstan is doing work on the rope at a rate 2π 2π P = τω = Fr = (520 N ) 5.0 × 10− 2 m = 182 W, T (0.90 s ) or 180 W to two figures. The net torque that the rope exerts on the capstan, and hence the net torque that the capstan exerts on the rope, is the difference between the forces of the ends times the radius. A larger number of turns might increase the force, but for given forces, the torque is independent of the number of turns. dT dQ dt P (182 W) b) = = = = 0.064 C° s. dt mc mc (6.00 kg)(470 J mol ⋅ K) ( ) a) Replacing m with nM and nMc with nC, nk T2 nk Q = ∫ dQ = 3 ∫ T 3 &dt = (T24 − T14 ). Θ T1 4Θ 3 For the given temperatues, (1.50 mol)(1940 J mol ⋅ K) Q= ((40.0 K) 4 − (10.0 K) 4 ) = 83.6 J. 4(281 K)3 b) ( 83.6 J) Q = (1.50 mol) (30.0 K) n2 T = 1.86 J mol ⋅ K. c) C = (1940 J mol ⋅ K) (40.0 K 281 K) 3 = 5.60 J mol ⋅ K. Setting the decrease in internal energy of the water equal to the final gravitational potential energy, Lf ρwVw + Cw ρwVw LT = mgh. Solving for h, and inserting numbers: ρ V ( L + Cw LT ) h= w w f mg = [ (1000 kg m3 )(1.9 × .8 × .1 m3 ) 334 × 103 J kg + (4190 J kg ⋅ ° C)(37° C) (70 kg)(9.8 m s 2 ) ] = 1.08 × 105 m = 108 km. a) (90)(100 W)(3000 s) = 2.7 × 10 7 J. Q Q 2.7 × 107 J = = = 6.89 C°, cm cρV (1020 J kg ⋅ K)(1.20 kg m 3 )(3200 m3 ) or 6.9 C° to the more apropriate two figures. c) The answers to both parts (a) and (b) are multiplied by 2.8, and the temperature rises by 19.3 C°. b) LT = See Problem 17.97. Denoting C by C = a + bT , a and b independent of temperature, integration gives. b   Q = n a(T2 − T1 ) + (T22 − T12 ) . 2   In this form, the temperatures for the linear part may be expressed in terms of Celsius temperatures, but the quadratic must be converted to Kelvin temperatures, T1 = 300 K and T2 = 500 K. Insertion of the given values yields Q = (3.00 mol)(29.5 J mol ⋅ K)(500 K 2 300 K) &&&&&&&&&&&&&&&&&&&&&&& + (4.10 × 10 −3 J mol ⋅ K 2 )((500 K) 2 − (300 K) 2 )) = 1.97 × 10 4 J. a) To heat the ice cube to 0.0°C, heat must be lost by the water, which means that some of the water will freeze. The mass of this water is m C T (0.075 kg)(2100 J kg ⋅ K)(10.0 C°) = 4.72 × 10−3 kg = 4.72 g. mwater = ice ice ice = Lf (334 × 103 J kg) b) In theory, yes, but it takes 16.7 kg of ice to freeze 1 kg of water, so this is impractical. The ratio of the masses is ms C w Tw (4190 J kg ⋅ K)(42.0 K) = 0.0696, = = mw C w Ts + Lv (4190 J kg ⋅ K)(65.0 K) + 2256 × 10 3 J kg so 0.0696 kg of steam supplies the same heat as 1.00 kg of water. Note the heat capacity of water is used to find the heat lost by the condensed steam. a) The possible final states are steam, water and copper at 100°C, water, ice and copper at 0.0°C or water and copper at an intermediate temperature. Assume the last possibility; the final temperature would be  (0.0350 kg)((4190 J kg ⋅ K)(100 C°) + 2256 × 103 J kg)    3   − × ( 0 . 0950 kg)(334 10 J kg)  = 86.1°C T=  (0.0350 kg)(4190 J kg ⋅ K) + (0.446 kg)(390 J kg ⋅ K)    + (0.0950 kg)(4190 J kg ⋅ K)   This is indeed a temperature intermediate between the freezing and boiling points, so the reasonable assumption was a valid one. b) There are 0.13 kg of water. a) The three possible final states are ice at a temperature below 0.0°C, an ice2 water mixture at 0.0° C or water at a temperature above 0.0°C. To make an educated guess at the final possibility, note that (0.140 kg)(2100 J kg ⋅ K)(15.0 C°) = 4.41 kJ are needed to heat the ice to 0.0°C, and (0.190 kg)(4190 J kg ⋅ K)(35.0 C°) = 27.9 kJ must removed to cool the water to 0.0°C, so the water will not freeze. Melting all of the ice would require an additional (0.140 kg)(334 × 103 J kg) = 46.8 kJ, so some of the ice melts but not all; the final temperature of the system is 0.0°C. Considering the other possibilities would lead to contradictions, as either water at a temperature below freezing or ice at a temperature above freezing. b) The ice will absorb 27.9 kJ of heat energy to cool the water to 0°C. Then, ( 27.9 kJ − 4.41 kJ ) m = 334×10 3 J kg = 0.070 kg will be converted to water. There will be 0.070 kg of ice and 0.260 kg of water. a) If all of the steam were to condense, the energy available to heat the water would be (0.0400 kg)(2256 × 10 3 J kg ) = 9.02 × 10 4 J. If all of the water were to be heated to 100.0°C, the needed heat would be (0.200 kg)(4190 J kg ⋅ K )(50.0 C°) = 4.19 × 104 J. Thus, the water heats to 100.0°C and some of the steam condenses; the temperature of the final state is 100°C. b) Because the steam has more energy to give up than it takes to raise the water temperature, we can assume that some of the steam is converted to water: 4.19 × 10 4 J m= = 0.019 kg. 2256 × 10 3 J kg Thus in the final state, there are 0.219 kg of water and 0.021 kg of steam. The mass of the steam condensed 0.525 kg − 0.490 kg = 0.035 kg. The heat lost by the steam as it condenses and cools is (0.035 kg) Lv + (0.035 kg)(4190 J kg ⋅ K )(29.0 K ), and the heat gained by the original water and calorimeter is ((0.150 kg)(420 J kg ⋅ K) + (0.340 kg)(4190 J kg ⋅ K ))(56.0 K) = 8.33 × 104 J. Setting the heat lost equal to the heat gained and solving for Lv gives 2.26 × 10 6 J kg, or 2.3 × 10 6 J kg to two figures (the mass of steam condensed is known to only two figures). a) The possible final states are in ice2water mix at 0.0°C, a water2steam mix at 100.0°C or water at an intermediate temperature. Due to the large latent heat of vaporization, it is reasonable to make an initial guess that the final state is at 100.0°C. To check this, the energy lost by the steam if all of it were to condense would be (0.0950 kg)(2256 × 103 J kg) = 2.14 × 105 J. The energy required to melt the ice and heat it to 100°C is (0.150 kg)(334 × 103 J kg + (4190 J kg ⋅ K) (100 C°)) = 1.13 × 105 J, and the energy required to heat the origianl water to 100°C is (0.200 kg)(4190 J kg.K) (50.0 C°) = 4.19 × 104 J. Thus, some of the steam will condense, and the final state of the system wil be a water2steam mixture at 100.0°C. b) All of the ice is converted to water, so it adds 0.150 kg to the mass of water. Some of the steam condenses giving up 1.55 × 103 J of energy to melt the ice and raise the temperature. Thus, m = 1.55×10 5 J 2256×10 3 J kg = 0.69 kg and the final mass of steam is 0.026 kg, and of the water, .150 kg + .069 kg + .20 kg = 0.419 kg. c) Due to the much larger quantity of ice, a reasonable initial guess is an ice2water mix at 0.0° C. The energy required to melt all of the ice would be (0.350 kg) (334 × 103 J kg) = 1.17 × 10 5 J . The maximum energy that could be transferred to the ice would be if all of the steam would condense and cool to 0.0°C and if all of the water would cool to 0.0 C , (0.0120) kg (2256 × 103 J kg + (4190 J kg ⋅ K)(100.0 C°)) + (0.200 kg)(4190 J kg ⋅ K)(40.0 C°) = 6.56 × 104 J. This is insufficient to melt all of the ice, so the final state of the system is an ice2water 6.56×104 J mixture at 0.0° C. 6.56 × 104 J of energy goes into melting the ice. So, m = 334 ×103 J kg = 0.196 kg. So there is 0.154 kg of ice, and 0.012 kg + 0.196 kg + 0.20 kg = 0.408 kg of water. Solving Eq. (17.21) for k, Τ (3.9 × 10−2 m) k=H = (180 W) = 5.0 × 10− 2 W m ⋅ K. 2 Α Τ (2.18 m )(65.0 K)   Τ 28.0 C°  = (0.120 J mol.K) (2.00 × 0.95 m 2 ) −2 −2 L  5.0 × 10 m + 1.8 × 10 m  = 93.9 W. b) The flow through the wood part of the door is reduced by a factor of ( 0.50 ) 2 1 − ( 2.00×0.95) = 0.868 to 81.5 W . The heat flow through the glass is a) H = kΑ  28.0 C°   = 45.0 W, H glass = (0.80 J mol ⋅ K) (0.50 m)2  −2  12.45 × 10 m  5 + 45.0 = 1.35. and so the ratio is 81.93 .9 R1 = L k1 , R2 = L k2 , H1 = H 2 , and so T1 = H A R1 , Τ 2 = H A R2 . The temperature difference across the combination is Τ = Τ1 + T2 = H H ( R1 + R2 ) = R, A A so, R = R1 + R2 . The ratio will be the inverse of the ratio of the total thermal resistance, as given by Eq. (17.24). With two panes of glass with the air trapped in between, compared to the single pane, the ratio of the heat flows is (2( Lglass kglass ) + R0 + ( Lair kair ) , ( Lglass kglass) + R0 where R0 is the thermal resistance of the air films. Numerically, the ratio is (2((4.2 × 10 −3 m) (0.80 W m ⋅ K)) + 0.15 m 2 ⋅ K W + ((7.0 × 10−3 m) (0.024 W m ⋅ K))) =2 (4.2 × 10− 3 m) (0.80 W m ⋅ K) + 0.15 m 2 ⋅ K W Denote the quantites for copper, brass and steel by 1, 2 and 3, respectively, and denote the temperature at the junction by T0 . a) H 1 = H 2 + H 3 , and using Eq. (17.21) and dividing by the common area, k1 (100°C − T0 ) = k 2 T0 + k 3 T0 . L1 L2 L3 Solving for T0 gives (k1 L1 ) (100°C ). T0 = (k1 L1 ) + (k 2 L2 ) + (k 3 L3 ) Substitution of numerical values gives T0 = 78.4°C. b) Using H = kAL T for each rod, with T1 = 21.6 C°, T2 = T3 = 78.4°C gives H1 = 12.8 W, H 2 = 9.50 W and H 3 = 3.30 W . If higher precision is kept, H 1 is seen to be the sum of H 2 and H 3 . a) See Figure 17.11. As the temperature approaches 0.0°C, the coldest water rises to the top and begins to freeze while the slightly warmer water, which is more dense, will be beneath the surface. b) (As in part (c), a constant temperature difference is assumed.) Let the thickness of the sheet be x, and the amount the ice thickens in time dt be dx. The mass of ice added per unit area is then ρic e dx, meaning a heat transfer of ρic e Lf dx. This must be the product of the heat flow per unit area times the time, (H A)dt = (k T x )dt. Equating these expressions, k T k T ρic e Lf dx = dt or xdx = dt. x ρice Lf This is a separable differential equation; integrating both sides, setting x = 0 at t = 0, gives 2k T x2 = t. ρice Lf The square of the thickness is propotional to the time, so the thickness is propotional to the square root of the time. c) Solving for the time in the above expression, 920 kg m 3 334 × 10 3 J kg (0.25 m )2 = 6.0 × 10 5 s. t= 2(1.6 J mol ⋅ K )(10°C ) ( )( ) d) Using x = 40 m in the above calculation gives t = 1.5 × 1010 s, about 500 y, a very long cold spell. Equation(17.21) becomes H = kA ∂∂Tx . a) H = (380 J kg ⋅ K)(2.50 × 10 −4 m 2 )(140 C° m) = 13.3 W. b) Denoting the points as 1 and 2, H 2 − H1 = dQ dt = mc ∂∂Tt . Solving for ∂T ∂x at 2, mc ∂T ∂T ∂T . + = ∂x 2 ∂x 1 kA ∂t The mass m is ρA x, so the factor multiplying ∂∂Tt in the above expression is cρ k x = 137 s m. Then, ∂T ∂x = 140 C° m + (137 s m)(0.250 C° s) = 174 C° m. 2 The mass of ice per unit area will be the product of the density and the thickness x, and the energy needed per unit area to melt the ice is product of the mass per unit area and the heat of fusion. The time is then ρxLf (920 kg m 3 )(2.50 × 10−2 m)(334 × 103 L kg) t= = P A (0.70)(600 W m 2 ) = 18.3 × 103 s = 305 min. a) Assuimg no substantial energy loss in the region between the earth and the sun, the power per unit area will be inversely proportional to the square of the distance from the center of the sun, and so the energy flux at the surface of the sun is (1.50 × 103 W m 2 ) ( 1.50×1011 m) 8 6.96×10 m) ) = 6.97 × 10 2 1 7 W m 2 . b) Solving Eq. (17.25) with e = 1, 1 7 2 4  H 1  4  6.97 × 10 W m T = = = 5920 K. −8 2 4  A σ   5.67 × 10 W m ⋅ K  : The rate at which the helium evaporates is the heat gained from the surroundings by radiation divided by the heat of vaporization. The heat gained from the surroundings come from both the side and the ends of the cylinder, and so the rate at which the mass is lost is 2 hπ d + 2π (d 2 ) σe(Ts4 − T 4 ) Lv ( = )   (0.250 m )π (0.090 m ) + 2π (0.045 m )2 (.200 )    × 5.67 × 10−8 W m 2 ⋅ K 4 (77.3 K )4 − (4.22 K )4  ( (2.09 × 10 )( 4 J kg )     ) −6 = 1.62 × 10 kg s, which is 5.82 g h. a) With p = 0, p V = nR T = pV T, T or V T 1 = , and β = . V T T βair 1 = = 67. βcopper (293 K)(5.1 × 102 5 K −1 ) b) a) At steady state, the input power all goes into heating the water, so c T and (1800 W) P = = 51.6 K, T= c(dm dt ) (4190 J kg ⋅ K) (0.500 kg min) (60 s min) and the output temperature is 18.0°C + 51.6°C. b) At steady state, the apparatus will neither remove heat from nor add heat to the water. P=H = dm dt a) The heat generated by the hamster is the heat added to the box; dT P = mc = (1.20 kg m 3 )(0.0500 m 3 )(1020 J mol ⋅ K)(1.60 C° h) = 97.9 J h. dt b) Taking the efficiency into account, M P0 P (10%) 979 J h = = = = 40.8 g h. t Lc Lc 24 J g For a spherical or cylindrical surface, the area A in Eq. (17.21) is not constant, and the material must be considered to consist of shells with thickness dr and a temperature difference between the inside and outside of the shell dT . The heat current will be a constant, and must be found by integrating a differential equation. a)Equation (17.21) becomes dT H dr H = k (4πr 2 ) or = k dT . dr 4πr 2 Integrating both sides between the appropriate limits, H 1 1  −  = k (T2 − T1 ). 4π  a b  In this case the “appropriate limits” have been chosen so that if the inner temperature T2 is at the higher temperature T1 , the heat flows outward; that is, dT dr < 0. Solving for the heat current, k 4πab(T2 − T1 ) H= . b−a b) Of the many ways to find the temperature, the one presented here avoids some intermediate calculations and avoids (or rather sidesteps) the sign ambiguity mentioned above. From the model of heat conduction used, the rate of changed of temperature with B radius is of the form dT dr = r 2 , with B a constant. Integrating from r = a to r and from r = a to r = b gives 1 1 1 1 T (r ) − T2 = B −  and T1 − R2 = B − . a b a r Using the second of these to eliminate B and solving T(r) (and rearranging to eliminate compound fractions) gives  r − a  b  T (r ) = T2 − (T2 − T1 )  .  b − a  r  There are, of course, many equivalent forms. As a check, note that at r = a, T = T2 and at r = b, T = T1. c) As in part (a), the expression for the heat current is dT H H = k (2πrL) or = kLdT , dr 2πr which integrates, with the same condition on the limits, to 2πkL(T2 − T1 ) H . ln(b a) = kL(T2 − T1 ) or H = ln(b a ) 2π d) A method similar (but slightly simpler) than that use in part (b) gives ln(r a ) . T (r ) = T2 + (T1 − T2 ) ln(b a ) e) For the sphere: Let b − a = l , and approximate b ~ a, with a the common radius. Then the surface area of the sphere is A = 4πa 2 , and the expression for H is that of Eq. (17.21) (with l instead of L, which has another use in this problem). For the cylinder: with the same notation, consider&& From the result of Problem 17.121, the heat current through each of the jackets T , where l is the length of the is related to the temperature difference by H = ln2(πlk b a) cylinder and b and a are the inner and outer radii of the cylinder. Let the temperature across the cork be T1 and the temperature across the styrofoam be T2 , with similar notation for the thermal conductivities and heat currents. Then, T1 + T2 = T = 125 C°. Setting H 1 = H 2 = H and canceling the common factors, T1k1 Tk = 2 2 1n 2 1n 1.5 −1  k ln 1.5   . Eliminating T2 and solving for T1 gives T1 = T 1 + 1  k2 ln 2  Substitution of numerical values gives T1 = 37 C°, and the temperature at the radius where the layers meet is 140°C − 37°C = 103°C. b) Substitution of this value for T1 into the above expression for H 1 = H gives 2π (2.00 m )(0.04 J mol ⋅ K ) (37 C°) = 27 W. H= ln 2 a) b) After a very long time, no heat will flow, and the entire rod will be at a uniform temperature which must be that of the ends, 0°C. c) d) ∂T ∂x = (100°C)(π L )cosπx L. At the ends, x = 0 and x = L, the cosine is ± 1 and the temperature gradient is ± (100°C)(π 0.100&m ) = ± 3.14 × 103 C° m. e) Taking the phrase “into the rod” to mean an absolute value, the heat current will be kA ∂∂Tx = (385.0 W m ⋅ K) (1.00 × 1024 m 2 )(3.14 × 103 C° m) = 121 W. f) Either by evaluating ∂∂Tx at the center of the rod, where πx L = π 2 and cos(π 2) = 0, or by checking the figure in part (a), the temperature gradient is zero, and no heat flows through the center; this is consistent with the symmetry of the situation. There will not be any heat current at the center of the rod at any later time. g) See Problem 17.114; k (385 W m ⋅ K) = = 1.1 × 10− 4 m 2 s. 3 3 ρc (8.9 × 10 kg m )(390 J kg ⋅ K) h) Although there is no net heat current, the temperature of the center of the rod is decreasing; by considering the heat current at points just to either side of the center, where there is a non2zero temperature gradient, there must be a net flow of heat out of the region around the center. Specifically, ∂T H (( L 2) + x) − H (( L 2) − x) = ρALxc ∂t   ∂T ∂T  − = kA  ∂x ( L 2) + x ∂x ( L 2 ) − x    2 ∂t = kA 2 x, ∂x a) In hot weather, the moment of inertia I and the length d in Eq. (13.39) will both increase by the same factor, and so the period will be longer and the clock will run slow (lose time). Similarly, the clock will run fast (gain time) in cold weather. (An ideal pendulum is a special case of physical pendulum.) b) LL0 = α Τ = (1.2 × 10−5 (C°) −1 × (10.0 C°) = 1.2 × 1024. c) See Problem 13.97; to avoid possible confusion, denote the pendulum period by τ . For this problem, ττ = 12 LL = 6.0 × 10−5 , so in one day the clock will gain (86,400 s)(6.0 × 10 −5 ) = 5.2 s so two figures. d) 2τ τ = (1 2)αLT < (86,400) −1 , so T < 2((1.2 × 10−5 (C°) −1 ) × (86,400)) −1 = 1.93 C°. The rate at which heat is aborbed at the blackened end is the heat current in the rod, kA (T2 − T1 ), L where T1 = 20.00 K and T2 is the temperature of the blackened end of the rod. If this were to be solved exactly, the equation would be a quartic, very likely not worth the trouble. Following the hint, approximate T2 on the left side of the above expression as T1 to obtain σL 2 T2 = T1 + (TS − T14 ) = T1 + (6.79 × 10−12 K − 3 )(Ts4 − T14 ) = T1 + 0.424 K. k This approximation for T2 is indeed only slightly than T1, and is a good estimate of the Aσ (TS4 − T24 ) = temperature. Using this for T 2 in the original expression to find a better value of T gives the same T to eight figures, and further, and further iterations are not worth – while. A numerical program used to find roots of the quartic equation returns a value for T that differed from that found above in the eighth place; this, of course, is more precision than is warranted in this problem. a) The rates are: (i) 280 W, (ii) (54 J h ⋅ C° ⋅ m 2 )(1.5 m 2 )(11 C°) (3600 s h) = 0.248 W, (iii) (1400 W m 2 )(1.5 m 2 ) = 2.10 × 10 3 W, (iv) (5.67 × 10−8 W m 2 ⋅ K 4 )(1.5 m 2 )(320 K) 4 − (309 K 4 ) = 116 W. The total is 2.50 kW, with the largest portion due to radiation from the sun. b) P 2.50 × 103 W = = 1.03 × 10− 6 m3 s = 3.72 L h. 3 6 ρLv (1000 kg m )(2.42 × 10 J kg ⋅ K) c) Redoing the above calculations with e = 0 and the decreased area gives a power of 945 W and a corresponding evaporation rate of 1.4 L h. Wearing reflective clothing helps a good deal. Large areas of loose weave clothing also facilitate evaporation. Capítulo 18 R= $ % # ⋅ = ! " ⋅ ⋅ & ' ( n=m M = * !! ) ( *% ,- * $( p= nRT * " $ = V × (= "$ ) ! " ⋅ *! (* = "/ ! −$ =" × " (+ ' ⋅ (*!0 ( ( . ( 1 ( 2!/$ 3 0 $°4 m = nM = ( 6 %*$ % × −$ ) 5 (* $ (*! " ( = !% × ! " ⋅ 5 ⋅ (*$ % ( MpV *% = RT * ,- * !/$ ) −% ) "( p! = p *V V! ( = *$ % (7 ' $0 ( = 0" (* ' ' 3 − / °4 4 5$*!0$ (1 ' ' (2 8 9 V =% $ pV = nRT =n ( 1 M =! × = V $ $ 8 pV * × n= = RT * $ #5 ! × $× !/ " =! × $ pV * M= RT * × −$ 0 * (1 = (* 0 ! " ⋅ (*! (* 0 ! " ⋅ (*% × −$ ) ⋅ (*!0 ( = /× −$ ) ( = %0 × −% ) ( × −$ ) ⋅ (*!0 ( "(  *! ! × ( ×  * pV  T! = T  ! !  = *$  pV  (m $ 5 ) pV * M= RT * ,- * (& . (* $ ( = %" ⋅ (*!0$ ( ) m = nM = !0 ! °4 $ !/ M =% ; 6 :! T = !! °4 = !0 −$ m = nM = * (6 :! = =! × A ! MpV ($! × = RT −$ ( ! m′ = !/ ) 0 ) " . (*%" ! . (*%00 )( ) % $× $ % # ⋅ )($ $× (  = //" (  $ $ . )( ) / = $ $°4 )= $/$ ) . ) m 6 ,- * T p ! = p  ! T  V   V! "(   = *  ×  %$ . ( $  /   % $ $   = $ $" ×  . n= ( −$ ( *$! × pV = RT * ! " ⋅ (* / × ) V! = V *T! T ( = * " ( nRT V = / ! × ! = ( /! × . / $× " (* % × $ ( ⋅ ( *!0 * (= $ ( = / × $ ) (*// $ !0! (= . ,- * /( ! = ( 8 nV 0 /× " . (1 < p ! = p *V V! ( = * 8 V ( V! = . p T! p! T = *$ (* !0" !// ( = $ /% (*" /( = ( > p !V = nR * ' ( T! = (1 ' > * (* (*$ ! " ⋅ ( ⋅ ( = $%$ = / $°4 ) * (1 F= nRT *$ = L T= ( 6 F = pA = *nRT V ( A = *nRT ( L A V = L 6 T = ! ° 4 = !0$ ' F= (* $ % # ! ⋅ ( *!0$ ( = $ "" × % ? ⋅ (*$/$ ( =%" × % ? °4 = $/$ nRT *$ = L , (* $ % # ! °4 % g p = p e − Mgy < < y=− *< 6 y RT Mg 5 @1 0 = e − Mgy 5 @1 p= 0 p * 0 (= M =! × −$ ) 5 ( M =! 1 Mgy *! = RT p = p e − Mgy 5 @1 y % × −$ ) * $ % # × −$ ) 5 (*0 ⋅ (*!/$ ! (* − p p = − e− Mgy! RT = − −e /= !%$ = / = /= 1 !%= = 0 %% !%$ ' ( ( !%$ = = !%$ !% = !%= 8 p = p e − Myg RT ρ = ρ e − Mgy 1 % ' p = * ρ M ( RT. , ρ ' ,- * ( T = !/$ p Mgy = RT ' 8 ( m = *" ! × 6 V= m = ρ $ $"% ) )( *0 $/( p $ % # ⋅ A m = nM = *$ = ! %× =n = , (p=* 0 × −0 . (* × * $ % # ⋅ (*$ %* = %% × )(!/$ (* % × $ = A −" ρ ' !$ 0 ") 0× $ ) = 0 !% != R M= ρ = ρ e− !% @ p = ρRT M = =!! × % . % RT y= 6 , $ ( ( =!! × pV RT − $ . ) − " =! % −$ (= ) $ A !$ *! × −$ ) 5 ( = %0 ) ) (*"$ %" × *" !$ × " −! % ( 0 ") ( ( nRT RT  = = V V  p= × ! " ⋅ *   *" !$ × $ − ! . ( ( − / = !× !× ⋅ (*/ !$ 1 ' , !% (B 9 ) ' C V = *!! % × −$ $ − " = ! !$ × ( $ A *! !$ × − " = $$ × −" ( =" × $ $ * " " (*" !$ × !$ (3 ! (1 V = nRT p RT = n A Ap = * $ % # *" !$ × !$ − !" =% 0 × ⋅ (*$ (* ( $× $ L ( L= % 0 × * ( ) − !" $ $ 9 =$% × −0 . ( a( V = m ρ = n M ρ = ( (C , ! > 5$ V      )(  V   =  n 8  5$ )5( 5  =  ( )= 0 − 0 × )(" !$ × =$ (1 $ 5 =0 $  )  $ !$ × 5 ,- * "( ) ' ' ) * ' 0( v v = ? ! $ = %0 v v @ = v v @ = ? $ !!! ! !!! = $ = " % (8 > = % *D ' 1 ( n 5$ − × ( ,- $ ! $%0 $ 9 (6 ( −    ( <   ∑ ni xi = % "  '  (   )  !  ∑ ni xi    ' 5! =" *, ( V A = VB E ) < ) ( p = nRT 5 V >  ( pV =     RT A  = ( pV = (m M )RT < TA > TB n pV A RT 8 A E < E ) ' ( ! m(v ! ) = $! kT T8 F TG ) ( vrms = $kT m < E ) ' ' T, ' vrms G A 5 ' A C B 1 ) necessarily could a( m = m. + m = $ $% × = $kT m = 0 × v ( T = mv ! 6 v 6 " > v ) > T =$ × " c = "%= T = / $× / pV = nRT ' 1 %=! −!/ $k =$ × v !*! E ) (= = $RT M % ' $ ! ( kT = *$ !(* $ −!$ × !K m = ( v = $RT $* $ % # = M *$! × −$ ) −$ # ( ⋅ (*$ ( = % %× ) ( - = ! $% × ! s! ! * ( M  *$! × −$ ) v = ( mv =  *" !$ × !$  A = ! / × −!$ ) ⋅ 1 −! (="! × (*$ !*" ! × −! #( ( *" !$ × !$ ( *$! × # ( ( *% % × ! ( ' !*" ! × −! #(*$! × *" !$ × !$ = !mK −$ ) ( ( (1 ' 1 * ( * ( ' F ( = !mv mv = !L v L p =F ( P P = ( =n  =  ( =!% × ( 1 ? !K !*" ! × = L * ! = L! = !% × ( 8 ' $× = ( pV RT ! " ⋅ . − / −! ( − 0 = !% × ? . )( !% × − / . = ) ×   " !$ × )  !$ ⋅ ) )($ ! 8 )( 5 !! * ( #( ' '   1 × p=$ , − $ 1 = % "× $ "0 λ = "× > T? ! = T; ! * M ? ! M ; ! ( = (!0$ = $ /0 × $ ( ⋅ ! !( °4 $kT $* $ × −!$ # (*$ ( = m $ × −") - (1 "( ! )R = $R = !% 0 #5 )*! $R M ( ) $( $ % # =( × −$ ⋅ ) = " %% × −$ ( ' ) ) = $0 × ' $ #5) ⋅ ( ' ( − (6 ) ( $$ #5) ⋅ °4)( ) 5 )= ( Cv = (C ) ( # ⋅ °4 °4 ( "% #5) ⋅ °4)( ) 5 ) = !0 #5) ⋅ °4 − " °4 (! " #5) ⋅ °4) × ) 5 ) = $/ #5 ⋅ °4 − °4 ( B $R ; !+ ( CV (& ,- * ,- * ! ( $ !"( Q = (! )(! /0 # * ( 0$" # ⋅ )($ )= " )# c= ( /% = % 0 CV ! /" #5 ⋅ = $ M ! × ) 5 ' // m?C ? T? = m C ( = /% #5) ⋅ m? = T m C C? * ( m? = [( pV = nRT " ) 1 )5 ( " ) 6 )]( ! ) 5 1 * ⋅ ! " !( 5 V = ⋅ nRT = p )(!0$ ) = % * " (v $kT $RT v! v != = = m M * " (! * , % ( T= Mv ! *! × −$ ) = $* " ( ! R $* " (! * $ % # ( v ! = *% $ ⋅ ( ( *% $ × −% ⋅ ! ! (* (! = 0 / ( *% $ × −% ⋅ ! ! (* ( ! = %$ ( *% $ × −% ⋅ ! ! (* × −% ⋅ ! (! = ε= D )  m  f * v ( = %π    !πkT  $! ! mv ! $! !ε − ε kT π m  − ε kT =  e  εe m  !πkT  m ! (v ! , ,- * f = Aε e − ε kT $$( A 1 df ε − ε kT  ε    e = Ae − ε kT  −  = Ae − ε kT −  de kT   kT   1 f ,- * ε = mv ! = kT ! 9 $%( k R = m M ? ) 8 8 = R M −$ ( !* $ % # ⋅ (*$ ( *%% × ( ⋅ (*$ ( *π *%% × * $ % # ⋅ (*$ ( $* $ % # ( *%% × T= ' −$ −$ ( = $ $/ × ) (( = $ ) ( = % !× ) ! × ! ! °4> y=! × =! ) $ p =" (1 p< p . * ( p! = p ( p! p! - p °4 - H °4 C D . ' 1 1 - = !! × 1 4+ ! - 6 - V = βV T = *$ " × ( V = − kV p = *" ! × − °4( * − ! m = nM = = $ ( *! × . (* V = C (*! °4( = $ − % / . (=− = % "0 (I > V MpV RT × *! = " 0% × −$ − " ) (*! !" × − . (*$ × * $ % # ⋅ (*!0 ( −" $ ( ) pVM RT " . (** (π * " ( ! (*%% * $ % # ⋅ (*!0 m = nM = = * × −$ × ( ) ( = ! $) h′ (1 ' h′ = h > p T′ p T′ =h p + ρgy T p′ T = *! $ ( * $× * $× . (+* $ ) . ( (*0 $ ! (*/$ !  (  $ = !" h = h − h′ = ! % p = / $/ × (1 ' . ρgy    1 ' 1 m = nM = ρ g hV PV M M = ; RT RT $ !  * $ "× $ ) (*0 (* " (     × ** 0 − "0 ((* "! × − % ! (  = * ! ⋅ * $ % # ( * !0$ (       −$ = !$ × ρ′ = ρ − ' ρ' 1 1 m ' m V ρ '  ρ ρ m   T′ = T = = T  − ρ′ ρ − *m V ( ρV   = *! ( −  *!0 ) ( ( − $ (*  * !$ )   $  ( − = % !/!°4 V T p ! = p  !  V!T   = *! /!  0! * ( * $ 0 $ (*$ (*!/ (  = ! 0% (  ' *? $× * )( $ (= . (*$ m ( × % . pV = nM =  !  M  RT!   * × =  * $ % # % (  *$ (  . (* ⋅ (*$ (1 " . 8 = $ × $ × × p! = p *T! T ( = ( (= % * ! × ' " + $× ' ' " ' * 0 $ / $ (** $ × " × ( * (( =$ (1 ' pM ; !  * ρ − ρ; ! (Vg =  ρ − RT   =  !$ )  × *0 = %! × ( @ / × $  Vg  * $ − ! (*/ × . (*! ! × −$ ) * $ % # ⋅ (*! $ $ ? ( ? M =% × −$ (. ) (  (  h  h   1 p  h− y p + * ρg ( y ,- ' y y y=h− p ρ =* 0 (− $× * * $ "× $ ) $ . ( (*0 ! ( = % ( 1 ) ) J 1 ' ! K ρv ! = ρg h + p   p  = ! *0 v = ! g h + ρ    = !" ! h =$ ( v p =%! × % p = p   % h=h− h=$ v= " *h − p =y= ρ 7 ! ! % h! − * 8 . (  $  (   − p @  h=! v = %% h ' −h −h (= p* p p − ρ ρg % ( ρg h = /$/ *$ ! × * ) (+ . ( y + *% h = $ %/ )( ! ! - ( − z! ( = 1 ' h = /% - & v! = ( C −h = z ! = ! %$ + y (h + **% * ( *! h=! * *1 ' * (( ' ( ( t = n A t = pV A * = RT t * = × * ( 1 ' *!0$ !/$ ( = //$ $ K = $n $ 8 ( (*$" ( [− $" pt ] ! * % ( (*" !$ × !$ ⋅ (*!0$ (* % ! " ⋅ ( " (* ! − ( = "$( (× * /! ' ' /$$ J = $*m M ( 8 =$ *" !$ × * 1 % V = π *! × $ − = × * . !$ × −$ (* ( ) ) ( !/ RT ! ( ($ = $ % × − !0 $ 8 7 f p= f J? K . 7 ⋅ (*$ ( (*$ % × RT * $ % # = f *" !$ × !$ 8V − !0 $ ( = * !× f H ' ) ' . (f ' f ' (7 ,- *  *0 × ( x=  * $ % (*% x= $ $× C ' ! $ !0 ( U = mgh = M A ( C ) ' ! 1 ( + RTV  x !  − *% !0 × (  [ * %% ( * $ % (*% x = $ $× - (1 <  ! × gh =  !$  " !$ × U= /( ' −$  *0  − !$ # # 1 (x ] 1 1 ) −!! − ! ' ! !× $ kT T = $ $ × ! (1 > = a ! (*% ( (= !× − !! # ( ( *C U *r ( (1 F *r ( 1 U= ( < r = !− r! r! = R ( R  r $ ! U !  R   = !    r  r < r! ? " r =R ! " F= C ( U *r! ( = U * R ( = −U " nRT = $! pV = R $ ! ( × x=r R U . )( × −$ $ )= / ,- * !"( U ) - # (1 MpV RT ! ( MpV RT $ ! v! ! Mv ! = = $ RT $ pV (6 ,- ( v = $* $ % # ( v $ = !00 " × −$ ) ( $% # )($ ⋅ )($ ) )! = ! %! × −% = 0) ⋅ ( *$ ( *! × −$ ) (= / !%!= ( ( !GM !*" "/$ × = R ( 1 $!) 1 ( $kT $* $ × −!$ # ( * = m * "/ × − !/ ) ( ( − ( = ! × ? ⋅ ! ) ! ( * 00 × *" 0" × ( $ ) ( =" × ' !) ( 6 ! ) ,- ' (1 U = − GmM R = −mgR ' - T= % ' −$ ! *! × $ *" !$ × ) K +U > K > mgR (C * ( ($ ! )kT = mgR (*0 !$ ( *" $ × " (* $ × !$ # ! 8 ) T = (! $)(mgR k ) 6 ( ( = % × × * !! ! ( ' T = " $" × $ ' T =% 0 ' % (6 (1 ' 8 E *C (# , ! ( A −$ v , = $* $ % # A ⋅ (* % ( *! ! × v = $* $ % # ⋅ (*!! ( **! ! × (, # ) ' = $* $ % # (v $ (= " × ) =* $ !! (v =* $ %"(v ' ⋅ (*! R = *$M %πρ( −$ (= $ × ) ( *$! × −$ ( = $0 ) 1 =%" × !GM R = %! (6 ,- * 0( −!$ $kT $* $ × m= ! = v * − % M = * !% × = % "× ( m A ) (*" !$ × $  $V   $m ρ  ( D = !r = !  = !   %π   %π  # (*$ (! ( = !% × !$ − % ) × ( * −$ ) $  $* !% × − % ) (   =!0 × = ! $  (   %π *0! ) $ x=A 6 U & * ! θ( =* ωt v = −ωA = ! ! kA! * θ( = ωt ! ! −" ωt ( K mω! = k = ! mω! A! * K =U ! ωt ( ave ( ( - ' ) 6 ,- * !/( ⋅ !R 3 " " # * ( ' (1 ' ' (1 ⋅ (*$ * $ % # ( ( = ! %0 × $ = − %" 0% × ( & −$ ω = A I K = nRT = * )   *"  × (! ) ⋅ * ( !K ' ! * ( !*! %0 × $ #( = * 0%× − %" ) ⋅ ! (*" !$× !$ = " !× ! ' 6 6 4+ ! C+! 4B R = ! /0 # ! ;!C ( #  " × !( ! = !  !$  " !$ × I = ! m* ) ⋅ CV − 4V " R = !% 0% # ! 4V ! ⋅ $0 ! R= !/ CV ( ∞  m  f *v( dv = %π    !πkT  ∫ (  m  = %π    !πkT  * ' ! 5 ! kT dv   π   =  %*m ! KT  m ! KT ( f *v(dv ' ( v + dv> f *v( dv a = m5!kT  m  %π    !πkT  ,- * $! v ( * n=! ! − mv ∫v e . ' < $ !∞ $!   $ π $kT  $  = !  m  ! *m !kT (  *m !kT ( "( ∞ ∫  m  f *v(dv = %π    !πkT  $ !∞ ∫v e $ − mv ! ! kT dv v ! = x !v dv = dx v $dv = * !( x dx D ) ∞ ,- *  m  ∫ vf *v(dv = !π !πkT  $  m  = !π    !πkT  ! = π $ $ ( ! ∞ ∫ xe − mx  !kT     m  ! KT = m ! ! kT dx ! KT πm (C . v G v + dv f *v(dv f *v(dv ' 1 d = f *v(dv ∫ vv + = = (v v + dv v v f *v(dv ! kT m $ ! %  m   !kT  − f *v ( = %π  e =   e πv  !πkT   m  6 ' ! −% v ' / f *v( v = ! 0% × / e ' ' * 0 ) ' ! ( −! (D MpV * = RT !0/ −! $ × ! * ( × ! (C ' ! (8 * . (= % × −$ ) * $ % # ' * ( ' ( $ (* % × $ . (* ⋅ (*!0$ ( ' 6 . 1 1 ) ' !> × % ( * " (*! $% × ( m= %! ! ' ! f *v ( v = ! v ( f =$0 × $ %! = %! "= $ ( = g * $ (*$ / × (1 $ . ( = $!$ × $ . 1 !°4 *$ °4 − !°4( * "° 4 ( = $) * ( !%°4 (1 ) (6 λ = * %π ! r ! * ,- * ! ( V (( − = *%π ! * × − × (! * −$ " (( − =% × ( ⋅ (*! $* $ % # *% × ( V (kT = * × ' p=* ( ve = (* $ × −!$ # ! ' 4 × 0 %" × . − ?⋅ ! ) ! (* × " V ( mR −$ (* "/ × T CD = T −!/ ) ( ( ( pV = kT v thermal - ' − % ( = %× (*! !G * m V (* %πR $( = * π $(G * R = * π $(*" "/$ × 1 −$ ( =/ $ ) ( = " %× ( */ $ " −$ $ !GM = R ×* =" × ( * ( kT CD * (<  *V (   *V ( CD CD V ( CD = kT   = *!  * * (  × (   *! × 1 > V( " −" $ *   = !× (  $ − ) ' (6 , % dP dy = − pM RT dp Mg dy =− p R T − αy  p  p - ( &  Mg  =  Rα  αy   −   T  , %( " /" = $ 0 x p= p %  * " × −! 4° (* "$ (   −  = % *! (   ! Mg *! × −$ (*0 ( = = " /" −! Rα * $ % # ⋅ (* " × 4° ( ( * p * Mg  αy  Rα p = p  −   T  αy α * − αy ' T ( ≈ − T ** " /"( * y %(( , * ( ( % ? A ∂P ∂V (8 ' (C 6 * * ( p ( 6 * < p V ! ∂ p ∂V ! p= ( ( ∂p ∂V = nRT an ! − ! V − nb V nRT !an ! ∂p =− + *V − nb( ! V$ ∂V !nRT "an ! ∂! p = − V% ∂V ! *V − nb($ C V $ nRT = !an ! *V − nb( ! ( 9 V % nRT = $an ! *V − nb( $ V = *$ !(*V − nb( 6 *V n( = $b C T = a !/ Rb * ( ( ( p = R* a ( RT a a a − = !/ Rb − ! = *V n( − b *V n( !b 0b !/b ! RT = *V n( ( ;! A $ ! a !/ b a !/ b ! $b = $ ? ! A $ %% ; !+ A % $ $ (< 1 < - ( v = ! *v + v! ( v! − v! = = = v 1 ( v′ ! = + = v v ! + v!! ! *v ! + v!! ( − *v ! + v!! + !v v! ( ! % % % *v ! + v !! − !v v ! ( *v − v ! ( ! ≥v - ' * v ! + u ! ( v′ = + ' * v + u( (1 ' * (> v′ ! = = * + (! + * ** v! + + ( − (v ! + ! v u + ** + ( * ! * − v ! + !v u − u ! ( + + (− + (u ! ( u! 1 v′ ! − v′ ! = = v * + ( + *v ! − v ! ( + *v ! − v ! ( + >v (1 ' ' = + > ' * * + ( + (! ! * v ! − !v u + u ! ( *v − u ( ! v′ > v′ ' Capítulo 19 p V = nR T = ⋅ ° = !" # !" # W = nRT " pV = nRT T " "$ % &' V = V ) "p " ( ⋅ $V × $ " # $   "  =(   × * " $ W = p V = nR T $ " T= " * " p V= , × W = nR + T = - $ ⋅ T = , ° + $ dV = " # × . " + = ° =+ ° W= − ( =− × " W =p V = . × . $W = p V = − =− " " $ × " W = W = p V −V $W = $W = p V −V +W +W +W = p − p V −V $ ) 0 0 " " 0 % " # $ # "! " ! # " " $ " ! # ) / " " Q= $ W = −, p V = × U = Q −W = - ) / " on 0 × " W=p V " − , × U = Q −W " " p1 V $ ) / U = Q −W = $ " = , × . ) / =,, × "0 0 , - 0 " 2 U = Q −W " Q # " ! " ! 3 Q=+ W # " ! 4 " 3W =+ U= 1 = −( U = n( R ) T " 2 " T= −( U = nR ,° − * " " # U = Q −W = Q 6 " !0 0 " p V= × Q = U +W = − 7 3 ) =− ⋅ T =T + T = × $ °= 0 5 =− × $ $Q # $ " Q =− × # ! + , / ! (/ ' (× ! "0 0 ! ) / $) # " "! ! " !0 ! " " ) ! ) / " "! " p1 V " %! ( $) 2> " 3 Q = U +W$ Q > ! ) /$ "! 2 "Q> $ - " × - V1 4 ! ° 0 0 1 + − 0" " ° ) / " ) " . "0 + ! + , / = / + /! = ( = ! ( / = + , / ! = " (/ v= $ K m= " " ) / "! " ) $" ) ) $ W< $ " " " ' ! " 0 " Q =W > ) / " * " % $) " 9 # # " :Q:;, ? 3 b " " ! # /) % " !# " 4 " " " Q = −, $ ! ) " ) /! " ) / U= $ " " pV " " /) " # ) / " ) /$ W > " W < <# " 0 $ " ) / " 0 0 # % W =Q % $W= Q= $ " ) ! "" "! " " "! " " " U = $ Q = W 3 "$ Q = W > " Q = W < " 0 " 1# $ $" ) / +× . p V = U = Q − W = mL# − W = a3 ! " "! b Q =W = , " 0 " " 8 "! " # " " ) / " on $ U = −W > "! $ Q = W < . W= − , /) /) " /) " /) " ! # " ) / W +W > $ " 0 $ U= " " 0 " * " $ > ) " 0 " " " $Q > $ " $ "$ Q< - a # # Q =, ) 1" $ ! - ) / " " " × − + ) 0 % ! " ) / 7 ! U = Q −W = Q < " × /! − +, × " # 0 / 0 ! $ − = +, × = . × + "! @ "! &' @ "! &' " nCV T = " ( ( $ T= $ T= Q nC p Q nCA = = + , ⋅ + ,+ ( , ⋅ = ° = ⋅ = +, ( $ (( $ (( T =( T;( n= % " T =+ " p$ Q = nC p T = Q> % , ! " " ° =+ ⋅ ° =+ , ⋅ ° =+ " V$ Q= , ! " ⋅ ! Q = nC v T = % ° " ! " p$ Q = nC p T = Q= ! +( " ! % " ! $ U = CV T $ " p V = nR T @ "! CV = R "   U = n R  T =   p V = × " p$ Q = nC p T 8" ! $ pV = nRT "  p V   C p  p V Q = nC p  =  nR   R  8" ! 4 " $ V > " % " " $ ! $ × . − − × − = + " " " p$ p V = nR T Q> Q> " ! ! " " $ W = p V = nR T % " ! $ U = CV T $ " @ "! CV = R " ! $ U =n R T = p V = W - " Q = U +W = W$ W Q = " % " $ W = nRT " V V = = −+ ) % " U = $ Q = W = −+ % " % " ⋅ " U= $ , =+ ⋅ " ( ⋅ " "0 U = nCV T = & ! - % Q = U +W ! # ) " ) / = ⋅ = $ + Q = nC P T ) ) CA = R γ − " C p = CV + R = , pV − pV = nR T − T = " U= " W =Q=− γ = C CA = + R CA $ " ⋅ - ) / 8 " &' T= ! $ 3+ $ ! " ⋅ Q=( " "! 0 " ) =, × !" ! # $ " " " " " ( − ( γ )) $ ( Q = nC T = Cp = R " )( &4 )( )= = *" 4 ! ) / ( +3 γ   =  V p = p  V °) − nCV T = nC P T γ = ( " 8 )( ⋅ p - (   =   .   ) × 0 " &' ,+ × ( +$ . $ "! # p $   V  γ− W= p V  −    V  γ−  = % ( &' " "8 " ( ! × . )(      ) $ (T T ) = (V V ) = ( ! " " "! B "!C " B "! C (     −    =− + ×   ) γ− $ ! = ($ " " " γ= @ $ " &4 T = T (V V " &' )γ − (+ % &' =( ( )( $ p = p (V V ( γ= ( )γ = ( $ ) = ,+ )( ) = ( ° = ( ! Q = U +W = U = −W = − ∫ PdV PV γ = @ = −∫ γ D .A D Aγ " = PiVi γ A = −. A γ =− = - " " % D D⋅ " !0 " " " ! U = nCV T $ T γ p −γ = T γ p −γ " T γ = T γ p p ! # T <T A−γ + −γ+ D 1 D D D 1 − − ] × " on " !0 ) / " " T V γ − = T V γ − " V = nRT p p < p " γ− = [ ( ) !" ! " U> U !" γ− T " ! # U " ! # 3 " !0 T &' " ( " ( 0 T V  =  T  V  a γ= $ p = = &' V  p $ =   p V  = , γ = ,$ p = % γ− 1 4 ( % " U = Q − W = −W = − = , γ $T = = +, $T = = ( $ W = nCV T = $ Q= ⋅ , " " ) ° T= pV nR = . × × − . = ⋅ E " 4 " isothermal$ " " " " $" " 0 E × pV = nR T =+ T= * E @ "! &' 8 &4 × − @ "! &' " ( × " ( +$ pV − p V W= γ− = " G × W= T T = V V ) " " "! − = " $T = T V γ− V γ− × × × × − $ " # " " "1 p −V " Va " V - × +, = +, = (K . . − (− − pV " $ " - ) / " # pb + pa Vb − Va = − K V V +, = p =p V V γ= ( = × × " − ⋅ ( 7 γ− × . . × 7 ( = − - F $ " ! ! $ " " " 0 5 = " W " A B " pV 1! × . + × . $ WH W =− U = nCV T pV PV p V − pV $T = T = T = T −T = nR nR nR U = CV R p V − p V U= G - " U = Q −W ! # Q = U +W = − Q " ! # $ × × − . − ( × ) =− ! - =− ( × + × × # . F=− × Qabc = U ac + Wabc = nCv Tac + Wabc Tac E PV = nRT → T = PV nR ! Tac = Tc − Ta = × Tac = Wabc = * PV ! " + Wabc = PcVc PaVa PcVc − PaVa − = nR nR nR − × . . = = − − × × ( . . ×   U = nCv Tac = n  R  Tac   Qabc  =  =       × + × U ac "    = ( )= " × ! ; Wac = =( − Qac = U ac + Wac = I ( " " abc ) + " " ac ( × = . )= " ) / ! " " abc. U = Q − W = (( ) − (+ ) = "0 ) "a " b + = W; "! abd$ " Q = U + W = * "! $ " Q = U + W = (− ) + (− ) = −+ 3 " ! # $ U =− " 0 db$ dV = " ) / " " ad " 3 Qad = (U d − U a ) + Wad = ( )+ ( )= " db$ W= " Qdb = U b − U d = − = " !" % $ Q = U +W 7 ) / Wbc = Wabc " Wad = Wadc $ " ) bc$ Q = + = ( E ad $ Q = + ; $ " " " " "! 2 ) " &' " ( " " ab " dc, " ab$ Q = ( E E E dc$ Q = Q ) " "! " "! U = 7 $ W = p(V − V ) " - ) / " 0 0 "! E * "! ab Wbc = p c (Vc − Va ) * "! ad $Wad = p a (Vc − Va ) - ) " U ab = U b − U a $ U bc = U c − U b $ U ad = U d − U a $ U dc = U c − U d $ % Wabc 0 "! ,1 $ U = Q −W cd, W; * "! bc, E Q = U +W Qab = U b − U a + Qbc = (U c − U b ) + pc (Vc − Va ) Qad = (U d − U a ) + pa (Vc − Va ) Qdc = (U c − U d ) + a c "! abc E = pc (Vc − Va ) Qabc = U b − U a + (U c − U b ) + pc (Vc − Va ) = (U c − U a ) + pc (Vc − Va ) % a c "! adc E Wadc = pa (Vc − Va ) Qadc = (U c − U a ) + pa (Vc − Va ) * "! pc > pa $ Qabc > Qadc $ " Wabc > Wadc Q =W + U - " " - " " " $ " " !0 "! U " " " " abc " adc - ) / " 0 0 under " pV1 " " not " W ) " $ ! abc. 8 " U "$Q " ) )Q " " W = Wab + Wbc + Wcd + Wda = + bc + + = , =− $ Q = U +W = + − $ > out ! 2 "J 3 on da . + −, ! " W< + . =− $ " QH ) " " bc. - 0 " 0 $ ) / " ! U= $ Q =W - " ) " Q < W < 8/ # " " 0 ) / 0 W =Q =− W = Wab + Wbc + Wca Wbc = V= " Wab = p V " p " " 6 " Wab = nR Tb − Ta = ++ Wca = W − Wab = − − ++ = − + ! $ " ! $ p V = nR T # % 0 $ . " ac " $ Wac = p V = nR T $ " Wac = nR T − Ta = Q= cb . CV = C p − R$ ⋅ ( − = , (× Wcb = Q − U = − U = − nCV T $ " $ "! Wcb = −n C p − R Tb − T =− . ba ⋅ ( − ⋅ " " # $ Wba = W = Wac + Wcb + Wba = , (× =− ( × Ta = Tc n = C pQ T = + ( , −+ , × ⋅ U = nCV T = Q CCVP = − 8 + ) / × + = × W = Q − U = −, × U dV = $ W = " Q = U = − ,( × ,+ ( , $ " = − ,( × − ( " = −+ , × U= $ " Q =W = p V " V= ) " ! * " " V =V p p # − W = ( p !" " = D p =p = × T =T p p =, × × "! ( . $ × . × . - =− . + $ "# ) pV " "$ =+ D " " # "! !# " " ! $ ) / # " " nRT " V V = p V " p p / "! " 4 = × = $ ! p V −V % " = P V −V =pV = - ) / " × − =( × × . . × × . .    $ − p p × .  "   − × .   −  − > " " #   = −  "! $ - " V =V β T = "! " # − × − × p V = F A V= × β TV = = 7 Q = mC p T = V ρC p T = =, × U = Q −W = , " " − − × × × − × − =+ ,( /! × @" " ! /! ⋅ "$ " ) " cV " c p − × ° − ° , − = +× /! ⋅ , ° × − p V= × − Q = mC T = ρV C T −+ = (× /! × = ( ! $ U =Q= ( " - % " /" m K " !0 @" " "$ 0$ "L 0 0 mL = mC T − mv $ L=C T − v = ( /! ⋅ 8 # "! &' ( " " ( ° − ( " p  T = T   p  p γ− T γ = p γ− T γ $ @ "! γ = , $T = , + × × + , = ( = ( $) # −γ ,+° " × /! $ " !" ! * "$ "$ " # )" "! 8 . T = × ) - ) / 7 " ! 3 V V γ ( ( " . L + × 0 ( ° " V −V + CV p R ) " , V −p V  C W = pV  + V R  " # pV =T T =T pV n= = . , " / 0 " "! ! (° $) " " p 8" " 0 " 3 " 1 # "!$ Q " " ( +E - " W=p p =p ) V  V    γ V  V −    ) 0 " !" " −γ     = T      γ =T " ( ) −γ pV pV $Q = (CV + R )( T − T ) = p V  CV +  RT RT  R  ! " % &' " " 1 L = L (p p ( ) " $ Lγ # " "  p   L − L = L  −    = (  p     = , M "! &' ( γ− $ # $ "! p  T = T   p  @ "! < −( L γ ) =( W = nCV T = ( ) " 4  )  ! ) / )( × × × × γ γ− . . " V    −( " "! &' ) / " $ ⋅ # . .        L γ " ) V " L " ) 0 " " &' ( $ "0  −    ) " "! $ " "  Lγ ) " &' γ γ− " # "! ) = ( = !# ! )( ,( °) = , " +° ) × $ ( - " " " - m= ρV - " − = p = p " ) " " × × −+ " . = . " " 0 " − × 3 - = 3 $ , × /! $ " 0 " " /! " ! " " " 0 " $ " 0 &' # !" $ " ( $ " 0 "  p m = ρ V   p  γ = = ( +× − = " " , × /! − −+    × × . .    /!$ (= ( % " ! $ T = " U = $ Q =W = % " $Q; $ " W T = nR - "$ Q = nC T U = −W = − % $ W = pdV = nR T $ " "! T!# W Q = nC p nR = C p WR $ Q = R WR = - $Q = , - " U$ U = nCV 8 "! T " CV $ U = n R W = W= nR - , $ " * !" # ) × . $ W = −nRT = − W = nRT " W = - $W= N T= 6 !" / p = nR T V= "! 4 " "$ W = p V = nR T = " ⋅ " = −, 40! " !" 3 = ⋅ #" ⋅ = ( " "! 3 dV = " W = $ " Q = U = nCV T = − T =− T= " U= ( , $ ) $ , +( " ⋅ " - =, $ Q = nC p T U = nCV T = Q − W = "! "$ * " "! 4 " " 4 " " = × " $ # "! # # ! "" "! 4 " " 4 " "$ " &' ( $ = " " N "! "! # " #" $ . , =+ × . W = p V = nR T = = nC p T = ( , % &' ( $ ⋅ "! #" - # " " " ⋅ = −+ − 4 $ " "$ " − =− , $ $ U = Q − W = − +, " " ". (+ $ W= = " U = Q − W = −W = − ". (+ " = nCV T = ⋅ $ " dV = $ W= − = @ "! $ "! " $ ,+ ⋅ 1 , = $ " U = Q −W = Q = W = nRT " 8 " !0 - V V ( . V = V$ !" # "# # * ( ) = nRT " 3 nCV T = (× − p V = pV = nRT = + = × × ) / - " " $ ) / " " ) / " ! " "# # " " $ " #" $ " " 0 "# # " " " $ " !" ! "! " " ' " 5 "! " " " " !0 7 "! nC p T Q= + T= 4 " U= " "$ # 0 γ− = $ " - 4 " #" ⋅ "3 "! = + × 4 " " T= $ % U = $ W =Q= + × N N "! 4 " "$ &' ( # "! ) $ " − × = " # " "! "$ $ 0 ) ! " 2 " ) " " - " " $ " h+ y 0 ! # 0 " p + $ mg A "$ = p + πrmg " ! mg  h     p + πr  h + y   " # h$ y # " ) $ h = h+ y (+ ) y − h ' ω = " " "! ) "$) ! $ " " "$ " " " " " # "! " " $ / "! "  mg  y F =  p +  −  − p πr  h    y = − ( p πr + mg ) h ) # h$ " " ' $ " " "! ' " 0 - O − hy - ( p πr   (πr  ) − mg )"3 ! " ) " " ) + mg ) h g  p πr   =  + m h mg  $ y O −h3 ! ! ) " 0 " $ " " " " # 0 # $ " " " 0 ! " " " " y == h ! " ) ! ) " ) " " 8 # "! " p " V " T " " ! "! ) V, nRT an − V − nb V V V − nb    + an  −  W = ∫ pdV = nRT "   V  V − nb  V V  2 " a = b = $ W = nRT "(V V )$ 4 @ "! 4 $ p= W =(  × "  + = nRT " " "5 a ( ( )( ⋅ × × ( )− ( )− ( )( − − ⋅ )( ) )(+ )(+ )   × × − × − − " " " ) ) L L − × −   × = ) × ! " $ ! ! " "5 % " b " " $ " " "5 a " ) / " 4 " " Capítulo 20 a) 2200 J + 4300 J = 6500 J. b) 2200 6500 = 0.338 = 33.8%. a) 9000 J − 6400 J = 2600 J. 2600 J b) 9000 = 0.289 = 28.9%. J a) 163700 ,100 = 0.230 = 23.0%. b) 16,100 J − 3700 J = 12,400 J. c) 16 ,100 J 4.60×10 4 J kg = 0.350 g. d) (3700 J)(60.0 s) = 222 kW = 298 hp. a) Q = 1e Pt = (180×103 W)(1.00 s) ( 0.280 ) = 6.43 × 10 5 J. b) Q − Pt = 6.43 × 10 5 J − (180 × 10 3 W)(1.00 s) = 4.63 × 10 5 J. 330 MW a) e = 1300 MW = 0.25 = 25%. b) 1300 MW − 330 MW = 970 MW. Solving Eq. (20.6) for r , (1 − γ) ln r = ln(1 − e) or 1 r = (1 − e) 1−γ = (0.350) − 2.5 = 13.8. If the first equation is used (for instance, using a calculator without the x y function), note that the symbol “e” is the ideal efficiency , not the base of natural logarithms. a) Tb = Ta r γ −1 = (295.15 K)(9.5)0.40 = 726 K = 453°C. b) pb = pa r γ = (8.50 × 10 4 Pa)(9.50) γ = 1.99 × 106 Pa. a) From Eq. (20.6), e = 1 − r 1− γ = 1 − (8.8) −0.40 = 0.58 = 58%. b) 1 − (9.6) −0.40 = 60%, an increase of 2%. If more figures are kept for the efficiencies, the difference is 1.4%. a) W = QC K = 3.40×10 4 J 2.10 = 1.62 × 104 J. b) QH = QC + W = QC (1 + K1 ) = 5.02 × 10 4 J. P= Q W 1  m = C =  ( Lf + c p T ) t K t K t  = 1  8.0 kg    (1.60 × 105 J kg) + (485 J kg ⋅ K)(2.5 K) = 128 W. 2.8  3600 s  ( ) 1.44 × 10 5 J − 9.80 × 10 4 J a) = 767 W. b) EER = H P, or 60.0 s EER = (9.8 × 104 J) (60 s) 1633 W (3.413) = (3.413) = 7.27. 5 4 [(1.44 × 10 J) (60 s) − (9.8 × 10 J) (60 s)] 767 W a) QC = m( L f + cice Tice + cwater Twater ) ( = (1.80 kg) 334 × 103 J kg + (2100 J kg ⋅ K)(5.0 K) + (4190 J kg ⋅ K)(25.0 K) = 8.90 × 10 J. |Q | 10 5 J = 3.37 × 105 J. b) W = KC = 8.082×.40 5 c) | QH |= W + | QC |= 3.37 × 105 J + 8.08 × 105 J = 1.14 × 106 J (note that | QH |= | QC | (1 + K1 ).) a) | QH | − | QC |= 550 J − 335 J = 215 J. b) TC = TH (| QC | | QH |) = (620 K)(335 J 550 J) = 378 K. c) 1 − (| QC | | QH |) = 1 − (335 J 550 J) = 39 %. 3 K a) From Eq. (20.13), the rejected heat is ( 300 520 K )(6450 J) = 3.72 × 10 J. b) 6450 J − 3.72 × 103 J = 2.73 × 103 J. c) From either Eq. (20.4) or Eq. (20.14), e=0.423=42.3%. a) | QH |=| QC | T TH = mLf H TC TC (287.15 K) = 3.088 × 107 J, (273.15 K) or 3.09 × 107 J to two figures. b) | W |=| QH | − | QC |=| QH | (1 − (TC TH )) = = (85.0 kg)(334 × 103 J kg) (3.09 × 107 J) × (1 − (273.15 297.15)) = 2.49 × 106 J. ) K a) From Eq. (20.13), ( 320 270 K )( 415 J) = 492 J. b) The work per cycle is 77 J 492 J − 415 J = 77 J, and P = (2.75) × 1.00 = 212 W, keeping an extra figure. s c) TC (TH − TC ) = (270 K) (50 K) = 5.4. For all cases, | W |=| QH | − | QC | . a) The heat is discarded at a higher temperature, and a refrigerator is required; | W |=| QC | ((TH TC ) − 1) = (5.00 × 103 J) × ((298.15 263.15) − 1) = 665 J. b) Again, the device is a refrigerator, and | W |=| QC | ((273.15 / 263.15) − 1) = 190 J. c) The device is an engine; the heat is taken form the hot reservoir, and the work done by the engine is | W |= (5.00 × 10 3 J) × ((248.15 263.15) − 1) = 285 J. For the smallest amount of electrical energy, use a Carnot cycle. Qin = QCool water to 0° C + Qfreeze water = mc T + mLF ( ) = (5.00 kg) 4190 kgJ⋅K (20 K) + (5.00 kg)(334 × 103 J K) Carnot cycle: Qin Tcold = 2.09 × 106 J Q 2.09 × 106 J Q = out → = out Thot 268 K 293 K Qout = 2.28 × 10 6 J(into the room) W = Qout − Qin = 2.28 × 10 6 J − 2.09 × 10 6 J W = 1.95 × 10 5 J(electrical energy) The total work that must be done is Wtot = mgy = (500 kg)(9.80 m s 2 )(100 m) = 4.90 × 105 J QH = 250 J Find QC so can calculate work W done each cycle: QC T =− C QH TH QC = −(TC TH )QH = −(250 J)[(373.15 K) (773.15 K)] = −120.7 J W = QC + QH = 129.3 J The number of cycles required is Wtot 4.09 × 10 5 J = = 3790 cycles. W 129.3 J For a heat engine, QH = −QC / (1 − e ) = − (−3000 J ) (1 − 0.600 ) = 7500 J, and then W = eQH = (0.600)(7500 J ) = 4500 J. This does not make use of the given value of TH . If TH is used, then for a Carnot engine, TC = TH (1 − e ) = (800 K )(1 − 0.600) = 320 K and QH = −QCTH / TC , which gives the same result. ( ) QC = − mLf = −(0.0400 kg ) 334 × 103 J/kg = −1.336 × 10 4 J QC T =− C QH TH ( ) QH = −(TH TC )QC = − − 1.336 × 104 J [(373.15 K ) (273.15 K )] = +1.825 × 104 J W = QC + QH = 4.89 × 103 J The claimed efficiency of the engine is 1.51×10 8 J 2.60×10 8 J = 58%. While the most efficient 250 K engine that can operate between those temperatures has efficiency eCarnot = 1 − 400 K = 38%. The proposed engine would violate the second law of thermodynamics, and is not likely to find a market among the prudent. a) Combining Eq. (20.14) and Eq. (20.15), K= TC / TH 1− e 1− e = = . 1 − (TC / TH ) (1 − (1 − e )) e b) As e → 1, K → 0; a perfect (e = 1) engine exhausts no heat (QC = 0), and this is useless as a refrigerator. As e → 0, K → ∞; a useless (e = 0) engine does no work (W = 0), and a refrigerator that requires no energy input is very good indeed. a) Q mLf (0.350 kg ) (334 × 103 J kg ) = = = 428 J K . TC TC (273.15 k ) − 1.17 × 105 J = −392 J K . 298.15 K c) S = 428 J K + (−392 J K ) = 36 J K. (If more figures are kept in the intermediate calculations, or if S = Q((1 273.15 K) − (1 298.15 K)) is used, S = 35.6 J K. b) a) Heat flows out of the 80.0° C water into the ocean water and the 80.0° C water cools to 20.0° C (the ocean warms, very, very slightly). Heat flow for an isolated system is always in this direction, from warmer objects into cooler objects, so this process is irreversible. b) 0.100 kg of water goes form 80.0°C to 20.0° C and the heat flow is Q = mc T = (0.100 kg)(4190 J kg ⋅ K)(−60.0C°) = −2.154 × 104 J This Q comes out of the 0.100 kg of water and goes into the ocean. For the 0.100 kg of water, S = mc ln(T2 T1 ) = (0.100 kg)(4190 J kg ⋅ K) ln(293.15 353.15) = −78.02 J K For the ocean the heat flow is Q = +2.154 × 104 J and occurs at constant T: Q 2.154 × 104 J = +85.76 J K S= = T 293.15 K S net = S water + S ocean = −78.02 J K + 85.76 J K = +7.7 J K (a) Irreversible because heat will not spontaneously flow out of 15 kg of water into a warm room to freeze the water. (b) S = S ice + S room = mLF mLF + Tice Troom = (15.0 kg)(334 × 103 J kg) − (15.0 kg)(334 × 103 J kg) + 273 K 293 K = + 1,250 J K This result is consistent with the answer in (a) because S > 0 for irreversible processes. The final temperature will be (1.00 kg)(20.0°C) + (2.00 kg)(80.0°C) = 60°C, (3.00 kg) and so the entropy change is   333.15 K   333.15 K    + (2.00 kg) ln    = 47.4 J K. (4190 J kg ⋅ K) (1.00 kg) ln   293.15 K   353.15 K    For an isothermal expansion, T = 0, U = 0 and Q = W . The change of entropy is The entropy change is S= Q = T 1850 J 293.15 K = 6.31 J K. Q , and Q = mLv . Thus, T − mLv − (0.13 kg)(2.09 × 104 J kg) S= = = −644 J K. T (4.216 K) a) S = Q T = mL v T = (1.00 kg)(2256 × 10 3 J kg ) ( 373.15 K) = 6.05 × 103 J K. Note that this is the change of entropy of the water as it changes to steam. b) The magnitude of the entropy change is roughly five times the value found in Example 20.5. Water is less ordered (more random) than ice, but water is far less random than steam; a consideration of the density changes indicates why this should be so. a) b) S= Q mLv (18.0 × 10−3 kg)(2256 × 103 J kg) = = = 109 J K. T T (373.15 K) N2 : (28.0 × 10 −3 kg)(201× 103 J kg) = 72.8 J K (77.34 K) Ag : (107.9 × 10−3 kg)(2336 × 103 J kg) = 102.2 J K (2466 K) Hg : (200.6 × 10−3 kg)(272 × 103 J kg) = 86.6 J K (630 K) c) The results are the same order or magnitude, all around 100 J K .The entropy change is a measure of the increase in randomness when a certain number (one mole) goes from the liquid to the vapor state. The entropy per particle for any substance in a vapor state is expected to be roughly the same, and since the randomness is much higher in the vapor state (see Exercise 20.30), the entropy change per molecule is roughly the same for these substances. a) The final temperature, found using the methods of Chapter 17, is T= (3.50 kg)(390 J kg ⋅ K)(100 C°) = 28.94°C, (3.50 kg)(390 J kg ⋅ K) + (0.800 kg)(4190 J kg ⋅ K) or 28.9°C to three figures. b) Using the result of Example 20.10, the total change in entropy is (making the conversion to Kelvin temperature)  302.09 K  S = (3.50 kg)(390 J kg ⋅ K) ln    373.15 K   302.09 K   + (0.800 kg)(4190 J kg ⋅ K) ln   273.15 K  = 49.2 J K. (This result was obtained by keeping even more figures in the intermediate calculation. Rounding the Kelvin temperature to the nearest 0.01 K gives the same result. As in Example 20.8,  0.0420 m3  V   = 6.74 J K. S = nR ln  2  = (2.00 mol)(8.3145 J mol ⋅ K) ln  3   0.0280 m   V1  a) On the average, each half of the box will contain half of each type of molecule, 250 of nitrogen and 50 of oxygen. b ) See Example 20.11. The total change in entropy is S =k 1 ln(2) + k 2 ln(2) = ( 1 + 2 )k ln(2) = (600)(1.381 × 10−23 J K) ln(2) = 5.74 × 10−21 J K. c) See also Exercise 20.36. The probability is (1 2) × (1 2 ) = (1 2 ) = 2.4 × 10 −181 , and is not likely to happen. The numerical result for part (c) above may not be obtained directly on some standard calculators. For such calculators, the result may be found by taking the log base ten of 0.5 and multiplying by 600, then adding 181 and then finding 10 to the power of the sum. The result is then 10 −181 × 10 0.87 = 2.4 × 10 −181. 500 100 600 a) No; the velocity distribution is a function of the mass of the particles, the number of particles and the temperature, none of which change during the isothermal expansion. b) As in Example 20.11, w1 = 13 w2 (the volume has increased, and w2 < w1 ); ln(w2 w1 ) = ln (3 ) = ln(3), and S = k ln(3) = kn A ln(3) = nR ln(3) = 18.3 J K. c) As in Example 20.8, S = nR ln(V2 V1 ) = nR ln(3), the same as the expression used in part (b), and S = 18.3 J K. For those with a knowledge of elementary probability, all of the results for this exercise are obtained from P(k ) = ( )p (1 − p) n k k 4 n−k 4! 1 =   , k!(4 − k )!  2  where P(k) is the probability of obtaining k heads, n = 4 and p = 1 − p = This is of course consistent with Fig. (20.18). a) 4! 4!0! 4 (1 2)4 = 4! 0!4! (1 2)4 = 161 for all heads or all tails. b) 4! 3!1! 1 2 for a fair coin. (1 2)4 = 14!3!! = 14 . 4! (1 2) = 83 . d) 2 × 161 + 2 × 14 + 83 = 1. The number of heads must be one of 0, 1, 2, 3 or c) 2!2! 4, and there must be unit probability of one and only one of these possibilities. a) QH = +400 J, W = +300 J W = QC + QH , so QC = W − QH = −100 J Q T Since it is a Carnot cycle, C = − C QH TH TC = −TH (QC QH ) = −(800.15 K)[(−100 J) (400 J)] = +200 K = −73°C ( ) b) Total QC required is − mLf = −(10.0 kg ) 334 × 103 J kg = −3.34 × 106 J QC for one cycle is − 100 J, so the number of cycles required is − 3.34 × 106 J = 3.34 × 104 cycles − 100 J cycle 1 so the temperature change 1 − e, 1  1   1  1 TH′ − TH = TC  − −  = (183.15 K )  = 27.8 K.  1 − e′ 1 − e   0.55 0.600  a) Solving Eq. (20.14) for TH , TH = TC b) Similarly, TC = TH (1 − e ), and if TH′ = TH, TC′ − TC = TC e′ − e  0.050  = (183.15 K )  = 15.3 K. 1− e  0.600  The initial volume is V1 = nRT1 p1 = 8.62 × 10−3 m3 . a) At point 1, the pressure is given as atmospheric, and p1 = 1.01 × 10 5 Pa, with the volume found above, V1 = 8.62 × 10−3 m 3 . V2 = V1 = 8.62 × 10−3 m3 , and p2 = T2 T1 p1 = 2 p1 = 2.03 × 105 Pa (using pa = 1.013 × 105 Pa). p3 = p1 = 1.01 × 10 5 Pa and V3 = V1 TT13 = 1.41 × 10 −2 m 3 . b) Process 1 F 2 is isochoric, V = 0 so W = 0. U = Q = nCV T = (0.350 mol)(5 2 ) × (8.3145 J/mol ⋅ K )(300K ) = 2.18 × 103 J. The process 2 F 3 is adiabatic, Q = 0, − W = nCV and U = T = (0.350 mol)(5 2)(8.3145 J mol ⋅ K )(−108 K ) = −786 J (W > 0). The process 3 F 1 is isobaric; W = p V = nR T = (0.350 mol)(8.3145 J mol ⋅ K )(−192 K ) = − 559 J, U = nCV T = n(5 2)(8.3145 J mol ⋅ K )(−192 K ) = −1397 J and Q = nC p T = (0.350 mol)(7 2)(8.3145 J mol ⋅ K )(−192 K ) = −1956 J = U + W . c) The net work done is 786 J − 559 J = 227 J. d) Keeping extra figures in the calculations for the process 1 F 2, the heat flow into the engine for one cycle is 2183 J − 1956 J = 227 J. e) e = 227 J 2183 J = 0.104 = 10.4%. For a Carnot F cycle engine operating between 300 K and 600 K, the thermal efficiency is 1 − 300 600 = 0.500 = 50%. (a) The temperature at point c is Tc = 1000 K since from pV = nRT , the maximum temperature occurs when the pressure and volume are both maximum. So ( )( ) pcVc 6.00 × 105 Pa 0.0300m3 = = 2.16 mol. RTc (8.3145 J mol ⋅ K )(1000 K ) (b) Heat enters the gas along paths ab and bc, so the heat input per cycle is QH = Qac = Wac + U ac . Path ab has constant volume and path bc has constant pressure, so n= Wac = Wab + Wbc = 0 + p c (Vc − Vb ) = (6.00 × 10 5 Pa )(0.0300 m 3 − 0.0100 m 3 ) = 1.20 × 10 4 J. For an ideal gas, U ac = nCV (Tc − Ta ) = CV ( pcVc − paVa ) R, using nT = pV R. For CO 2 , CV = 28.46 J mol.K, so 28.46 J mol ⋅ K U ac = ((6.00 × 105 Pa)(0.0300 m 3 ) − (2.00 × 105 Pa)(0.0100 m 3 )) = 5.48 × 8.3145 J mol ⋅ K Then QH = 1.20 × 104 J + 5.48 × 104 J = 6.68 × 104 J. (c) Heat is removed from the gas along paths cd and da, so the waste heat per cycle is QC = Qca = Wca + U ca . Path cd has constant volume and path da has constant pressure, so Wca = Wcd + Wda = 0 + pd (Va − Vd ) = (2.00 × 105 Pa)(0.0100 m3 − 0.0300 m 3 ) = −0.400 × 104 J. From (b), U ca = − U ac = −5.48 × 104 J, so QC = −0.400 × 104 J − 5.48 × 104 J = −5.88 × 104 J. (d) The work is the area enclosed by the rectangular path abcd, W = ( pc − pa )(Vc − Va ), or W = QH + QC = 6.68 × 104 J − 5.86 × 104 J = 8000 J. (e) e = W QH = (8000 J) (6.68 × 104 J) = 0.120. a) W = 1.00 J, TC = 268.15 K, TH = 290.15 K For the heat pump QC > 0 and Q H < 0 Q T W = QC + QH ; combining this with C = − C gives QH TH W 1.00 J QH = = = 13.2 J 1 − TC TH 1 − (268.15 290.15) b) Electrical energy is converted directly into heat, so an electrical energy input of 13.2 J would be required. c) From part (a), QH = W ⋅ QH decrease as TC decreases. 1 − TC TH The heat pump is less efficient as the temperature difference through which the heat has to be “pumped” increases. In an engine, heat flows from TH to TC and work is extracted. The engine is more efficient the larger the temperature difference through which the heat flows. (a) Qin = Qab + Qbc Qout =Qca Tmax = Tb = Tc = 327°C = 600 K P 1 PaVa PbVb = → Ta = a Tb = (600 K) = 200 K Pb 3 Ta Tb PbVb = nRTb → Vb = nRTb (2 moles)(8.31 moleJ K )(600 K) = = 0.0332 m 3 Pb 3.0 × 105 Pa PbVb PcVc P  3 = → Vc = Vb b = (0.0332 m3 )  = 0.0997 m 3 = Va Tb Tc Pc 1 Monatomic gas : CV = 32 R and CP = 52 R J   3  3 Qab = nCV Tab = (2 moles)  8.31 (400 K) = 9.97 × 10 J mole K 2    c c nRT V b dV = nRTb ln c = nRTb ln 3 Qbc = Wbc = ∫ PdV = ∫ b b V Vb J   4 = (2.00 moles)  8.31 (600 K) ln 3 = 1.10 × 10 J mole K   Qin = Qab + Qbc = 2.10 × 104 J J  5 4 Qout = Qca = nC p Tca = (2.00 moles)   8.31 (400 K) = 1.66 × 10 J mole K   2 (b) Q = U + w = 0 + W → W = Qin − Qout = 2.10 × 104 J − 1.66 × 104 J = 4.4 × 103 J 4.4 × 103 J = 0.21 = 21% 2.10 × 104 J T 200 K = 1 − TCh = 1 − 600 = 0.67 = 67% K e = W Qin = (c) emax = ecannot a) b) QH = +500 J W = mgy = (15.0 kg)(9.80 m s 2 )(2.00 m) = 294 J W = QC + QH , QC = W − QH = 294 J − 500 J = −206 J QC T =− C QH TH TC = −TH (QC QH ) = −(773 K)[(−206 J) (500 J)] = +318 K = 45°C c) e = W QH = (294 J) (500 J) = 58.8% d) QC = −206 J; wastes 206 J of heat each cycle e) From part (a), state a has the maximum pressure and minimum volume. pV = nRT , p = nRT (2.00 mol)(8.3145 J mol ⋅ K)(773 K) = = 2.57 × 106 Pa 3 −3 V 5.00 × 10 m 279.15 K a) e = 1 − 300.15 K = 7.0%. b) 1 ( e − 1)(210 kW) = 2.8 MW. c) pout e = 210 kW 0.070 = 3.0 MW, 3.0 MW − 210 kW = dm d QC dt (2.8 × 106 W) (3600 s hr) = = = 6 × 105 kg hr = 6 × 105 L hr. dt c T (4190 J kg ⋅ K) (4 K) There are many equivalent ways of finding the efficiency; the method presented here saves some steps. The temperature at point 3 is T3 = 4T0 , and so 5 19 QH = U13 + W13 = nCV (T3 − T0 ) + (2 p0 )(2V0 − V0 ) = nRT0 (3) + 2 p0V0 = p0V0 , 2 2 where nRT0 = p 0V0 has been used for an ideal gas. The work done by the gas during one cycle is the area enclosed by the blue square in Fig. (20.22), W = p 0V0 , and so the efficiency is e = W QH = 192 = 10.5%. a) p 2 = p1 = 2.00 atm, V2 = V T2 1 T1 p4 = p = (4.00 L)(3/2) = 6.00 L. V3 = V2 = 6.00 L, p3 = p 2 V3 3 V4 = p3 (3 / 2) = 1.67 atm. As a check, p1 = p T1 4 T4 T3 T2 = p 2 (5 / 9) = 1.111 atm, = p 4 (6 / 5) = 2.00 atm. To summarize, ( p1 , V1 ) = (2.00 atm, 4.00 L) ( p3 , V3 ) = (1.111 atm, 6.00 L) b) The number of moles of oxygen is n = ( p 2 , V2 ) = (2.00 atm, 6.00 L) ( p 4 , V4 ) = (1.67 atm, 4.00 L). p1V1 RT1 , and the heat capacities are those in Table (19.1). The product p1V1 has the value x = 810.4 J; using this and the ideal gas law, i: ii : iii : iv :  T x 2 − 1 = (3.508)(810.4 J)(1 2) = 1422 J,   T1 T  W = p1 V = x 2 − 1 = (810.4 J)(1 2) = 405 J.  T1  Q = nCP T = CP R T −T  x 3 2  = (2.508)(810.4 J)(− 2 3) = −1355 J, W = 0.  T1  V  T V  W = nRT3 ln  4  = x 3 ln  4  = (810.4 J)(5 6) ln (2 3) = −274 J, Q = W T1  V3   V3  Q = nCV T = CV R Q = nCV T = CV R  T  x1 − 4  = (2.508)(810.4 J)(1 6) = 339 J, W = 0.  T1  In the above, the terms are given to nearest integer number of joules to reduce roundoff error. c) The net work done in the cycle is 405 J F 274 J = 131 J. d) Heat is added in steps i and iv, and the added heat is 1422 J + 339 J = 1761 J and the efficiency is 131 J 1761 J = 0.075, or 7.5%. The efficiency of a CarnotFcycle engine operating between 250 K and 450 K is 1 − 250 450 = 0.44 = 44%. a) U = 1657 kJ − 1005 kJ = 6.52 × 105 J, W = p V = (363 × 103 Pa) × (0.4513 m3 − 0.2202 m3 ) = 8.39 × 104 J, and so Q = U + W = 7.36 × 10 5 J. b) Similarly, QH = U − p V = (1171 kJ − 1969 kJ) + (2305 × 103 Pa)(0.00946 m3 − 0.0682 m 3 ) = −9.33 × 105 J. c) The work done during the adiabatic processes must be found indirectly (the coolant is not ideal, and is not always a gas). For the entire cycle, U = 0, and so the net work done by the coolant is the sum of the results of parts (a) and (b), − 1.97 × 10 5 J. The work done by the motor is the negative of this, 1.97 × 105 J. d) K = Qc W = 7.36×10 5 J 1.97×10 5 J = 3.74. For a monatomic ideal gas, CP = 52 R and CV = 32 R. a) ab: The temperature changes by the same factor as the volume, and so C Q = nCP T = P pa (Va − Vb ) = (2.5)(3.00 × 105 Pa)(0.300 m 3 ) = 2.25 × 105 J. R The work p V is the same except for the factor of 52 , so W = 0.90 × 105 J. U = Q − W = 1.35 × 105 J. bc: The temperature now changes in proportion to the pressure change, and Q = 32 ( pc − pb )Vb = (1.5)(−2.00 × 105 Pa)(0.800 m 3 ) = −2.40 × 105 J, and the work is zero ( V = 0). U = Q − W = −2.40 × 105 J. ca: The easiest way to do this is to find the work done first; W will be the negative of area in the p)V plane bounded by the line representing the process ca and the verticals from points a and c. The area of this trapezoid is 12 (3.00 × 105 Pa + 1.00 × 105 Pa) × (0.800 m 3 − 0.500 m3 ) = 6.00 × 104 J, and so the work is − 0.60 × 105 J. U must be 1.05 × 105 J (since U = 0 for the cycle, anticipating part (b)), and so Q must be U + W = 0.45 × 10 5 J. b) See above; Q = W = 0.30 × 105 J, U = 0. c) The heat added, during process ab and ca, is 2.25 × 105 J + 0.45 × 105 J ×105 = 0.111 = 11.1%. = 2.70 × 105 J and the efficiency is QWH = 02..30 70×105 a) ab: For the isothermal process, T = 0 and U = 0. W = nRT1 ln(Vb Va ) = nRT1ln(1/r ) = −nRT1 ln(r ), and Q = W = − nRT1 ln(r ). bc: For the isochoric process, V = 0 and W = 0; Q = U = nCV T = nCV (T2 − T1 ). cd: As in the process ab, U = 0 and W = Q = nRT2ln(r ). da: As in process bc, V = 0 and W = 0; U = Q = nCV (T1 − T2 ). b) The values of Q for the processes are the negatives of each other. c) The net work for one cycle is Wnet = nR(T2 − T1 )ln(r ), and the heat added (neglecting the heat exchanged during the isochoric expansion and compression, as W mentioned in part (b)) is Qcd = nRT2 ln(r ), and the efficiency is Qnetcd = 1 − (T1 T2 ). This is the same as the efficiency of a CarnotFcycle engine operating between the two temperatures. The efficiency of the first engine is e1 = TH − T ′ TH and that of the second is e2 = T ′ − TC T′ , and the overall efficiency is  T − T ′   T ′ − TC  e = e1e2 =  H  .  TH   T ′  The first term in the product is necessarily less than the original efficiency since T ′ > TC , and the second term is less than 1, and so the overall efficiency has been reduced. a) The cylinder described contains a mass of air m = ρ(πd 2 4 )L, and so the total kinetic energy is K = ρ(π 8 )d 2 Lv 2. This mass of air will pass by the turbine in a time t = L v, and so the maximum power is K P = = ρ(π 8)d 2v 3 . t Numerically, the product ρair (π 8 ) ≈ 0.5 kg m3 = 0.5 W ⋅ s 4 m 5 . 1/ 3 1/ 3  (3.2 × 106 W) (0.25)   P e  = 14 m s = 50 km h. b) v =  2  =  4 5 2   kd   (0.5 W ⋅ s m )(97 m)  c) Wind speeds tend to be higher in mountain passes.  1 gal  1 mi  3.788 L   = 9.89 L h. a) (105 km h )    25 mi  1.609 km  1 gal  b) From Eq. (20.6), e = 1 − r1− γ = 1 − (8.5) −0.40 = 0.575 = 57.5%.  9.89 L h  (0.740 kg L )(4.60 × 107 J kg )(0.575) = 5.38 × 10 4 W = 72.1 hp. c)  3600 s hr   d) Repeating the calculation gives 1.4 × 10 4 W = 19 hp, about 8% of the maximum power. (Extra figures are given in the numerical answers for clarity.) a) The efficiency is e = 1 − r −0.40 = 0.611 , so the work done is QH e = 122 J and | QC |= 78 J. b) Denote the length of the cylinder when the piston is at point a by L0 and the stroke as s. Then, L0 L0 − s = r , L0 = L0 A = r r −1 s and volume is r 10.6 sA = (86.4 × 10 −3 m)π (41.25 × 10 −3 m) 2 = 51.0 × 10 −4 m 3 . r −1 9.6 c) The calculations are presented symbolically, with numerical values substituted at the end. At point a, the pressure is p a = 8.50 × 10 4 Pa, the volume is Va = 5.10 × 10 −4 m 3 as found in part (b) and the temperature is Ta = 300 K. At point b, the volume is Vb = Va r , the pressure after the adiabatic compression is pb = pa r γ and the temperature is Tb = Ta r γ −1 . During the burning of the fuel, from b to c, the volume remains constant and so Vc = Vb = Va r . The temperature has changed by an amount Q QH RQH Ta T= H = = nCV ( paVa RTa )CV paVaCV (8.3145 J mol ⋅ K )(200 J ) Pa )(5.10 × 10− 4 m3 ) (20.5 J Ta = f Ta , mol ⋅ K ) where f is a dimensionless constant equal to 1.871 to four figures. The temperature at c is then Tc = Tb + f Ta = Ta (r γ −1 + f ) . The pressure is found from the volume and = (8.50 × 10 4 temperature, pc = pa r (r γ −1 + f ) . Similarly, the temperature at point d is found by considering the temperature change in going from d to a, QC Q = (1 − e) H = (1 − e) f Ta , so Td = Ta (1 + (1 − e) f ). The process from d to a is nCV nCV isochoric, so Vd = Va , and pd = pa (1 + (1 − e) f ). As a check, note that pd = pc r − γ . To summarize, p V T a pa Va Ta b pa r γ Va r Ta r γ −1 c pa r (r γ−1 + f ) Va r Ta (r γ −1 + f ) Va Ta (1 + (1 − e) f ) d pa (1 + (1 − e ) f ) Using numerical values (and keeping all figures in the intermediate calculations), (a) Q T =k A for furnace and water t L S S S = furnace + water t t t kA T L kA T L =− + Tf Tw = 1  kA T  1  − +  L  Tf Tw  = 2 1 1  (79.5 W m ⋅ K )     2 1m 15 cm +  (210 K )  −    0.65m  100 cm    523 K 313 K   = +0.0494 J K ⋅ s (b) S > 0 means that this process is irreversible. Heat will not flow spontaneously from the cool water into the hot furnace. a) Consider an infinitesimal heat flow dQH that occurs when the temperature of the hot reservoir is T ′ : dQC = −(TC / T ′)dQH dQH T′ dQ | QC |= TC ∫ H = TC | S H | T′ ∫ dQ C = −TC ∫ b) The 1.00 kg of water (the highFtemperature reservoir) goes from 373 K to 273 K. QH = mc T = (1.00 kg)(4190 J kg ⋅ K )(100 K ) = 4.19 × 105 J Sh = mcln (T2 T1 ) = (1.00 kg)(4190 J kg ⋅ K )ln(273 373) = −1308 J/K The result of part (a) gives | QC |= (273 K )(1308 J K ) = 3.57 × 105 J QC comes out of the engine, so QC = −3.57 × 105 J Then W = QC + QH = −3.57 × 105 J + 4.19 × 105 J = 6.2 × 104 J. c) 2.00 kg of water goes from 323 K to 273 K QH = mc T = (2.00 kg)(4190 J kg ⋅ K )(50 K ) = 4.19 × 105 J S h = mc ln (T2 T1 ) = (2.00 kg)(4190 J kg ⋅ K )ln (273 / 323) = −1.41 × 103 J K QC = −TC | S h |= −3.85 × 105 J W = QC + QH = 3.4 × 10 4 J d) More work can be extracted from 1.00 kg of water at 373 K than from 2.00 kg of water at 323 K even though the energy that comes out of the water as it cools to 273 K is the same in both cases. The energy in the 323 K water is less available for conversion into mechanical work. See Figure (20.15(c)), and Example 20.8. a) For the isobaric expansion followed by the isochoric process, follow a path from T to 2T to T . Use dQ = nCV dT or dQ = nC p dT to get S = nC p ln 2 + nCV ln 12 = n(C p − CV ) ln2 = nR ln2. b) For the isochoric cooling followed by the isobaric expansion, follow a path from T to T / 2 to T . Then S = nCV ln 12 + nC p ln 2 = n(C p − CV ) ln = nR ln 2. The much larger mass of water suggests that the final state of the system will be water at a temperature between 0°C and 60.0°C. This temperature would be  (0.600 kg )(4190 J kg ⋅ K )(45.0C°)     − (0.0500 kg )((2100 J kg ⋅ K )(15.0C°)   + 334 × 103 J kg)  = 34.83°C, T= (0.650 kg)(4190 J kg ⋅ K ) keeping an extra figure. The entropy change of the system is then  307.98  S = (0.600 kg)(4190 J kg ⋅ K)ln   318.15    273.15   (2100 J kg ⋅ K) ln  258.15       3   334 × 10 J kg + (0.0500 kg)  +  = 10.5 J K. 273.15 K    307.98   + (4190 J kg ⋅ K) ln  273.15     (Some precision is lost in taking the logarithms of numbers close to unity.) a) For constantFvolume processes for an ideal gas, the result of Example 20.10 may be used; the entropy changes are nCV ln(Tc Tb ) and nCV ln(Ta Td ). b) The total entropy change for one cycle is the sum of the entropy changes found in part (a); the other processes in the cycle are adiabatic, with Q = 0 and S = 0. The total is then TT  T T S = nCV ln c + nCV ln a = nCV ln c a . Tb Td  TaTd  From the derivation of Eq. (20.6), Tb = r γ −1Ta and Tc = r γ −1Td , and so the argument of the logarithm in the expression for the net entropy change is 1 identically, and the net entropy change is zero. c) The system is not isolated, and a zero change of entropy for an irreversible system is certainly possible. a) b) From Eq. (20.17), dS = dQ T , and so dQ = T dS , and Q = ∫ dQ = ∫ T dS which is the area under the curve in the TS plane. c) QH is the area under the rectangle bounded by the horizontal part of the rectangle at TH and the verticals. | QC | is the area bounded by the horizontal part of the rectangle at TC and the verticals. The net work is then QH − | QC |, the area bounded by the rectangle that represents the process. The ratio of the areas is the ratio of the lengths of the vertical sides of the respective rectangles, and T −T the efficiency is e = QWH = HTH C . d) As explained in problem 20.49, the substance that mediates the heat exchange during the isochoric expansion and compression does not leave the system, and the diagram is the same as in part (a). As found in that problem, the ideal efficiency is the same as for a CarnotFcycle engine. a) S = b) JS = Q T Q T 3 (334×10 = − mLT f = − ( 0.160 kg) ( 373.15 K ) = mLf T = 3 ( 0.160 kg)(334 × 10 J kg ) ( 273.15 K) J kg) = −143 J K. = 196 J K. c) From the time equilbrium has been reached, there is no heat exchange between the rod and its surroundings (as much heat leaves the end of the rod in the ice as enters at the end of the rod in the boiling water), so the entropy change of the copper rod is zero. d) 196 J K − 143 J K = 53 J K. S = mcln(T2 T1 ) = (250 × 10−3 kg)(4190 J kg ⋅ K)ln(338.15 K 293.15 K) = 150 J K. a) b) S= − mc T Telement = − ( 250×10 −3 kg)(4190 J kg ⋅ K)(338.15 K − 293.15 K) 393.15 K = −120 J K. c) The sum of the result of parts (a) and (b) is S system = 30 J K. d) Heating a liquid is not reversible. Whatever the energy source for the heating element, heat is being delivered at a higher temperature than that of the water, and the entropy loss of the source will be less in magnitude than the entropy gain of the water. The net entropy change is positive. a) As in Example 20.10, the entropy change of the first object is m1c1ln(T T 1 ) and that of the second is m2c2ln(T ′ T 2 ) , and so the net entropy change is as given. Neglecting heat transfer to the surroundings, Q1 + Q2 = 0, m1c1 (T − T1 ) + m2c2 (T ′ − T2 ) = 0, which is the given expression. b) Solving the energyFconservation relation for T ′ and substituting into the expression for S gives  T  m c  T T  S = m1c1ln  + m2 c21n 1 − 1 1  − 1  .  T1   m2 c2  T2 T2   Differentiating with respect to T and setting the derivative equal to 0 gives (m2 c2 )(m1c1 m2 c2 )(− 1 T2 ) mc 0= 1 1 + . T   T T1   1 − (m1c1 m2 c2 )  −    T T   2   2  This may be solved for m1c1T1 + m2 c2T2 , m1c1 + m2 c 2 which is the same as T ′ when substituted into the expression representing conservation of energy. T= Those familiar with Lagrange multipliers can use that technique to obtain the relations ∂ ∂Q ∂ ∂Q S =λ S =λ , ∂T ∂T ∂T ′ ∂T ′ and so conclude that T = T ′ immediately; this is equivalent to treating the differentiation as a related rate problem, as m c m c dT ′ d =0 S= 11+ 2 2 T T ′ dT dT ′ and using dT ′ dT = − m 21c12 gives T = T ′ with a great savings of algebra. mc c) The final state of the system will be that for which no further entropy change is possible. If T < T ′, it is possible for the temperatures to approach each other while increasing the total entropy, but when T = T ′, no further spontaneous heat exchange is possible. a) For an ideal gas, CP = CV + R, and taking air to be diatomic, CP = R, CV = 52 R and γ = 75 . Referring to Fig. (20.6), QH = n 72 R (Tc − Tb ) = 72 ( p cVc − p bVb ). Similarly, QC = n 52 R( p aVa − p d Vd ). What needs to be done is to find the relations between the product of the pressure and the volume at the four points. pV pV T For an ideal gas, Tcc c = Tbb b , so p cVc = p aVa Tac . For a compression ratio r, and 7 2 ( ) given that for the Diesel cycle the process ab is adiabatic, V  pbVb = paVa  a   Vb  γ −1 = paVa r γ −1. γ −1 V  Similarly, pdVd = pcVc  c  . Note that the last result uses the fact that process da is  Va  T isochoric, and Vd = Va ; also, pc = pb (process bc is isobaric), and so Vc = Vb Tac . Then, ( ) Vc Tc Vb Tb Ta Va = ⋅ = ⋅ ⋅ Va Tb Va Ta Tb Vb T = c Ta =  T V γ −1  V  ⋅  a aγ −1  a   TbVb  Vb  −γ Tc γ r Ta Combining the above results, γ T  2 pdVd = paVa  c  r γ − γ  Ta  Subsitution of the above results into Eq. (20.4) gives    Tc γ γ −γ 2  − 1 5 T r e = 1−  a 7   Tc  γ −1     − r    Ta   1  (5.002)r −0.56 − 1 , =1− 1.4  (3.167) − r 0.40  ( ) where Tc Ta = 3.167, γ = 1.4 have been used. Substitution of r = 21.0 yields e = 0.708 = 70.8%. Capítulo 21 mlead = 8.00 g and charge = −3.20 × 10 −9 C − 3.20 × 10 −9 C = 2.0 × 1010. a) ne = −19 − 1.6 × 10 C n 8.00 g = 2.33 × 10 22 and e = 8.58 × 10 −13. b) nlead = A × 207 nlead current = 20,000 C s and t = 100 s = 10 −4 s Q = It = 2.00 C Q ne = = 1.25 × 1019. 1.60 × 10 −19 C The mass is primarily protons and neutrons of m = 1.67 × 10 −27 kg, so: 70.0 kg np and n = = 4.19 × 1028 − 27 1.67 × 10 kg About one(half are protons, so np = 2.10 × 10 28 = ne and the charge on the electrons is given by: Q = (1.60 × 10 −19 C) × (2.10 × 10 28 ) = 3.35 × 10 9 C. Mass of gold = 17.7 g and the atomic weight of gold is 197 g mol. So the number of atoms A × mol = (6.02 × 10 23 ) × ( 17.7 g 197 g mol ) = 5.41 × 10 22 . a) np = 79 × 5.41 × 1022 = 4.27 × 1024 q = np × 1.60 × 10 −19 C = 6.83 × 105 C b) ne = n p = 4.27 × 10 24. 1.80 mol = 1.80 × 6.02 × 1023 H atoms = 1.08 × 1024 electrons. charge = −1.08 × 10 24 × 1.60 × 10 −19 C = −1.73 × 105 C. First find the total charge on the spheres: 1 q2 F= ⇒ q = 4πε0 Fr 2 = 4πε0 (4.57 × 10− 21 )(0.2) 2 = 1.43 × 10−16 C 2 4πε0 r And therefore, the total number of electrons required is n = q e = 1.43 × 10−16 C 1.60 × 10−19 C = 890. a) Using Coulomb’s Law for equal charges, we find: 1 q2 F = 0.220 N = ⇒ q = 5.5 × 10−13 C 2 = 7.42 × 10− 7 C. 2 4πε0 (0.150 m) b) When one charge is four times the other, we have: 1 4q 2 F = 0.220 N = ⇒ q = 1.375 × 10−13 C 2 = 3.71 × 10− 7 C 4πε0 (0.150 m)2 So one charge is 3.71 × 10 −7 C, and the other is 1.484 × 10 −6 C. a) The total number of electrons on each sphere equals the number of protons. ne = np = 13 × A × 0.0250 kg = 7.25 × 1024. 0.026982 kg mol b) For a force of 1.00 × 10 4 N to act between the spheres, F = 104 N = 1 q2 ⇒ q = 4πε0 (104 N) (0.08 m)2 = 8.43 × 10− 4 C. 4πε0 r 2 ⇒ ne′ = q e = 5.27 × 1015 c) ne′ is 7.27 × 10 −10 of the total number. The force of gravity must equal the electric force. 1 q2 1 (1.60 × 10−19 C) 2 2 mg = ⇒ r = = 25.8 m 2 ⇒ r = 5.08 m. 4πε0 r 2 4πε0 (9.11 × 10− 31 kg)(9.8 m s) a) Rubbing the glass rod removes electrons from it, since it becomes positive. 7.50 nC = (7.50 × 10 −9 C) (6.25 × 1018 electrons C) = 4.69 × 1010 electrons (4.69 × 1010 electrons) (9.11 × 10 −31 kg electron) = 4.27 × 10 −20 kg. The rods mass decreases by 4.27 × 10 −20 kg. b) The number of electrons transferred is the same, but they are added to the mass of the plastic rod, which increases by 4.27 × 10 −20 kg. is in the + x ( direction, so q 2 q3 qq F1 = F2 , k 12 3 = k 2 r23 r13 2 1 must be in the − x ( direction and q1 is positive. q1 = (0.0200 0.0400) q 2 = 0.750 nC 2 F= a) 1 (0.550 × 10−6 C) q2 1 q1q2 ⇒ = 0 . 200 N 4πε0 (0.30 m)2 4πε0 r 2 ⇒ q 2 = + 3.64 × 10 −6 C. b) F = 0.200 N, and is attractive. Since the charges are equal in sign the force is repulsive and of magnitude: kq 2 (3.50 × 10−6 C) 2 = 0.172 N F= 2 = 4πε0 (0.800 m)2 r We only need the y(components, and each charge contributes equally. 1 (2.0 × 10−6 C) (4 × 10−6 C) F= sin α = 0.173 N (since sin α = 0.6). 4πε0 (0.500 m)2 Therefore, the total force is 2 F = 0.35 N , downward. 2 and F2 = k 3 are both in the + x(direction. q1q2 qq = 6.749 × 10− 5 N, F3 = k 1 2 3 = 1.124 × 10− 4 N 2 r12 r13 F = F2 + F3 = 1.8 × 10−4 N, in the + x(direction. F21 = (9 × 10 9 N ⋅ m 2 C 2 ) (20. × 10 −6 C) (2.0 × 10 −6 C) (0.60m )2 = 0.100 N FQ1 is equal and opposite to F1Q (Ex. 21.4), so (F ) (F ) Q1 x Q1 y = −0.23 N = 0.17 N Overall: Fx = −0.23 N Fy = 0.100 N + 0.17 N = 0.27 N The magnitude of the total force is (0.23 N ) + (0.27 N ) = 0.35 N. The direction of the force, as measured from the +y axis is 0.23 θ= tan −1 = 40 0.27 2 2 2 is in the + x − direction. F2 = k q1 q 2 r122 = 3.37 N, so F2 x = +3.37 N Fx = F2 x + F3 x and Fx = −7.00 N F3 x = Fx − F2 x = −7.00 N − 3.37 N = −10.37 N For F3 x to be negative, q3 must be on the –x(axis. F3 = k q1q3 , so x = x2 k q1q3 = 0.144 m, so x = −0.144 m F3 The charge q3 must be to the right of the origin; otherwise both q 2 and q3 would exert forces in the + x direction. Calculating the magnitude of the two forces: 1 q1q2 (9 × 109 N ⋅ m 2 C 2 )(3.00 × 10−6 C)(5.00 × 10−6 C) F21 = = 4πε0 r122 (0.200 m)2 = 3.375 N in the + x direction. F31 = (9 × 10 9 N ⋅ m 3 C 2 ) (3.00 × 10 −6 C) (8.00 × 10 −6 C) r132 0.216 N ⋅ m 2 in the − x direction r132 We need F21 − F31 = −7.00 N : = 3.375 N − 0.216 N ⋅ m 2 = −7.00 N r132 0.216 N ⋅ m 2 = 0.0208 m 2 3.375 N + 7.00 N r13 = 0.144 m to the right of the origin r132 = = 1+ y = − 0.400 m so, F= 2 and F = F2 + F1 since they are acting in the same direction at  1.50 × 10−9 C 3.20 × 10−9 C  1  = 2.59 × 10− 6 N downward. + (5.00 × 10− 9 C)  2 2  4πε0 ( 0 . 200 m) ( 0 . 400 m)   = 1 + 2 and F = F1 − F2 since they are acting in opposite directions at x = 0 so,  4.00 × 10−9 C 5.00 × 10−9 C  1  = 2.4 × 10− 6 N to the right. + F= (6.00 × 10− 9 C)  2 2  4πε0 ( 0 . 200 m) ( 0 . 300 m)   a) qQ 1 2qQa 1 sin θ 2 2 2 4πε 0 (a + x 2 ) 3 2 4πε 0 (a + x ) 1 2qQ in the + y direction. c) At x = 0, Fy = 4πε0 a 2 b) Fx = 0, Fy = 2 d) a) b) Fx = −2 − 2qQx 1 1 qQ cos θ = , Fy = 0 2 2 2 4πε 0 (a + x ) 4πε 0 (a + x 2 ) 3 / 2 c) At x = 0, F = 0. d) ( ) 1 q2 1 q2 1 q2 b) F = + 2 = 1+ 2 2 at an angle of 45° below the 4πε0 2 L2 4πε0 L2 4πε0 2 L2 positive x(axis a) E = 1 q 1 (3.00 × 10−9 C) = 432 N C , down toward the particle. 4πε0 r 2 4πε0 (0.250 m)2 b) E = 12.00 N C = 1 q 1 (3.00 × 10−9 C) ⇒ r = = 1.50 m. 4πε0 r 2 4πε0 (12.0 N C) Let +x(direction be to the right. Find a x : v0 x = +1.50 × 103 m s , vx = −1.50 × 103 m s , t = 2.65 × 10−6 s, a x = ? vx = v0 x + axt gives ax = −1.132 × 109 m s 2 Fx = max = −7.516 × 10−18 N is to the left (− x ( direction ), charge is positive, so E = F q = (7.516 × 10−18 N) [(2) (1.602 × 10 −19 ] is to the left. C) = 23.5 N C (a) x = 12 at 2 2(4.50 m) 2x 2 = 1.00 × 1012 m s a= 2 = (6 2 (3.00 × 10 s) t F ma (9.11 × 10 −31 kg) (1.00 × 1012 m s ) = = q q 1.6 × 10 −19 C 2 E= = 5.69 N C The force is up, so the electric field must be downward since the electron is negative. (b) The electron’s acceleration is ~ 1011 g, so gravity must be negligibly small compared to the electrical force. a) q E = mg ⇒ q = b) qE = mg ⇒ E = a) b) E= (0.00145 kg) (9.8 m s 2 ) = 2.19 × 10− 5 C, sign is negative. 650 N C (1.67 × 10−27 kg) (9.8 m s 2 ) = 1.02 × 10− 7 N / C, upward. −19 1.60 × 10 C 1 q 1 (26 × 1.60 × 10−19 C) = = 1.04 × 1011 N C . 4πε0 r 2 4πε0 (6.00 × 10−10 m)2 Eproton = 1 q 1 (1.60 × 10−19 C) = = 5.15 × 1011 N C . −11 2 2 4πε0 r 4πε0 (5.29 × 10 m) a) q = −55.0 × 10 −6 C, and F is downward with magnitude 6.20 × 10 −9 N. Therefore, E = F q = 1.13 × 10 −4 N C, upward. b) If a copper nucleus is placed at that point, it feels an upward force of magnitude F = qE = (29) ⋅ 1.6 × 10 −19 C ⋅ 1.13 × 10 −4 N C = 5.24 × 10 −22 N. a) The electric field of the Earth points toward the ground, so a NEGATIVE charge will hover above the surface. 2 (60.0 kg) (9.8 m s ) mg = qE ⇒ q = − = −3.92 C. 150 N C 1 q2 1 (3.92 C) 2 = = 1.38 × 107 N. The magnitude of the charge is 2 2 4πε0 r 4πε0 (100.00 m) too great for practical use. b) F = a) Passing between the charged plates the electron feels a force upward, and just misses the top plate. The distance it travels in the y(direction is 0.005 m. Time of flight m = t = 1.600.×0200 = 1.25 × 10 −8 s and initial y(velocity is zero. Now, 106 m s y = v0 yt + 12 at 2 so 0.005 m = 12 a(1.25 × 10−8 s) 2 ⇒ a = 6.40 × 1013 m s . But also 2 a= F m = eE me ⇒E= ( 9.11 × 10 −31 kg)( 6.40 × 1013 m s 2 ) 1.60 × 10 −19 C = 364 N C . b) Since the proton is more massive, it will accelerate less, and NOT hit the plates. To find the vertical displacement when it exits the plates, we use the kinematic equations again: 1 1 eE (1.25 × 10 −8 s) 2 = 2.73 × 10 −6 m. y = at 2 = 2 2 mp c) As mention in b), the proton will not hit one of the plates because although the electric force felt by the proton is the same as the electron felt, a smaller acceleration results for the more massive proton. d) The acceleration produced by the electric force is much greater than g; it is reasonable to ignore gravity. 1 = 2 a) q1 4πε 0 r1 = 2 9 2 2 −9 ˆ = (9 × 10 N ⋅ m C ) (−5.00 × 10 C) = (−2.813 × 10 4 N C) ˆ (0.0400 m )2 q 2 (9 × 10 9 N ⋅ m 2 C 2 ) (3.00 × 10 −9 C) = 1.08 × 10 4 N C = 2 2 2 r2 (0.0300m ) + (0.0400 m) The angle of 2 , measured from the x ( axis, is 180 − tan −1 ( 4.00 cm 3.00 cm ) = 126.9° Thus = (1.080 × 104 N C) ( ˆ cos 126.9° + ˆ sin 126.9°) = ( − 6.485 × 103 N C) ˆ + (8.64 × 103 N C) ˆ 2 b) The resultant field is 3 4 3 ˆ ˆ 1 + 2 = ( − 6.485 × 10 N C) + ( − 2.813 × 10 N / C + 8.64 × 10 N C) = ( − 6.485 × 103 N / C) ˆ − (1.95 × 104 N C) ˆ Let + x be to the right and + y be downward. Use the horizontal motion to find the time when the electron emerges from the field: x − x0 = 0.0200 m, a x = 0, v0 x = 1.60 × 10 6 m s , t = ? x − x0 = v0 x t + 12 a x t 2 gives t = 1.25 × 10 −8 s v x = 1.60 × 10 6 m s y − y 0 = 0.0050 m, v0y = 0, t = 1.25 × 10 −8 s, v y = ?  v0 y + v y y − y 0 =  2    t gives v y = 8.00 × 10 5 m s  v = v x2 + v y2 = 1.79 × 10 6 m s a) = −11 N C ˆ + 14 N Cˆ, so E = (−11) 2 + (14) 2 = 17.8 N C. θ = tan −1 ( − 14 11) = − 51.8°, so θ = 128° counterclockwise from the x(axis b) = q so F = (17.8 N C) (2.5 × 10−9 C) = 4.45 × 10−8 N, i) at − 52° (repulsive) ii) at + 128° (repulsive). a) Fg = me g = (9.11 × 10 −31 kg) (9.8 m s ) = 8.93 × 10 −30 N. Fe = eE = 2 (1.60 × 10−19 C) (1.00 × 104 N C) = 1.60 × 10−15 N. Yes, ok to neglect Fg because Fe >> Fg . b) E = 10 4 N C ⇒ Fe = 1.6 × 10 −15 N = mg ⇒ m = 1.63 × 10 −16 kg ⇒ m = 1.79 × 1014 me . c) No. The field is uniform. 2(0.0160 m) (1.67 × 10 −27 kg ) 1 2 1 eE 2 t ⇒E= at = = 148 N C . 2 2 mp (1.60 × 10 −19 C) (1.50 × 10 − 6 s ) 2 eE t = 2.13 × 10 4 m s . b) v = v0 + at = mp a) x = π 2ˆ 2 ˆ π  − 1.35  −1  12  + a) tan −1   = − , = − ˆ b) tan   = , ˆ = 2 2 2  .2  4  0   2.6   = 1.97 radians = 112.9°, ˆ = − 0.39 ˆ + 0.92 ˆ (Second quadrant). c) tan −1  1 . 10 +   a) E = 614 N C , F = qE = 9.82 × 10 −17 N. b) F = e 2 4πε0 (1.0 × 10−10 ) 2 = 2.3 × 10−8 N. c) Part (b) >> Part (a), so the electron hardly notices the electric field. A person in the electric field should notice nothing if physiological effects are based solely on magnitude. a) Let + x be east. is west and q is negative, so is east and the electron speeds up. −19 Fx =| q | E = (1.602 × 10 C) (1.50 V m) = 2.403 × 10−19 N ax = Fx m = (2.403 × 10−19 N) (9.109 × 10− 31kg ) = + 2.638 × 1011 m s 2 v0 x = + 4.50 × 105 m s , ax = + 2.638 × 1011 m s 2 , x − x0 = 0.375 m, vx = ? vx2 = v02x + 2ax ( x − x0 ) gives vx = 6.33 × 105 m s b) q > 0 so is west and the proton slows down. Fx = − | q | E = − (1.602 × 10−19 C) (1.50 V m) = − 2.403 × 10−19 N ax = Fx m = ( − 2.403 × 10−19 N) (1.673 × 10− 27 kg) = − 1.436 × 108 m s 2 2 v0 x = + 1.90 × 104 m s , ax = − 1.436 × 108 m s , x − x0 = 0.375 m, vx = ? v 2 x = v 2 0 x + 2a x ( x − x0 ) gives vx = 1.59 × 104 m s Point charges q1 (0.500 nC) and q 2 (8.00 nC) are separated by x = 1.20 m. The electric field is zero when E1 = E2 ⇒ kq1 r12 = kq 2 (1.20 − r1 ) 2 ⇒ q2 r12 = q1 (1.2 − r1 ) 2 = q1r12 − 2q1 (1.2)r1 + 1.2 2 q1 ⇒ (q2 − q1 )r12 + 2(1.2)q1r1 − (1.2) 2 q1 = 0 or 7.5r12 + 1.2r1 − 0.72 = r1 = + 0.24, − 0.4 r1 = 0.24 is the point between. Two positive charges, q , are on the x(axis a distance a from the origin. a) Halfway between them, E = 0.  1   q q  , | x|< a −  2 2  ( a − x)   4πε0  (a + x)  1   q q  , x > a + b) At any position x, E =  2 2  4 πε ( a x ) ( a x ) + −   0   −1   q q   , x < −a + 2 2  (a − x)   4πε0  (a + x) For graph, see below. The point where the two fields cancel each other will have to be closer to the negative charge, because it is smaller. Also, it cant’t be between the two, since the two fields would then act in the same direction. We could use Coulomb’s law to calculate the actual values, but a simpler way is to note that the 8.00 nC charge is twice as large as the –4.00 nC Charge. The zero point will therefore have to be a factor of 2 farther from the 8.00 nC charge for the two fields to have equal magnitude. Calling x the distance from the –4.00 nC charge: 1.20 + x = 2 x x = 2.90 m a) Point charge q1 (2.00 nC) is at the origin and q 2 ( − 5.00 nC ) is at x = 0.800 m. k | q1 | k | q2 | + = 575 N C right. i) At x = 0.200 m, E = 2 (0.200 m) (0.600 m) 2 k | q2 | k | q1 | + = 269 N C left. ii) At x = 1.20 m, E = 2 (0.400 m) (1.20 m) 2 k | q1 | k | q2 | + = 405 N C left. iii) At x = − 0.200 m, E = 2 (0.200 m) (1.00 m) 2 b) F = − eE i) F = 1.6 × 10−19 C ⋅ 575 N C = 9.2 × 10−17 N left, ii) F = 1.6 × 10−19 C ⋅ 269 N C = 4.3 × 10−17 N right, iii) F = 1.6 × 10−19 ⋅ 405 = 6.48 × 10−17 N right. A positive and negative charge, of equal magnitude q , are on the x(axis, a distance a from the origin. 1 2q a) Halfway between them, E = , to the left. 4πε0 a 2  1   4πε0  1 b) At any position x, E =   4πε0  1   4πε0 with “+” to the right. This is graphed below.  −q  q  , | x|< a − 2 2  ( a − x)   ( a + x)  −q  q  , x > a + 2 2  (a − x)   ( a + x)  −q  q  , x < −a − 2 2  ( a − x)   ( a + x) a) At the origin, E = 0. b) At x = 0.3 m, y = 0 : =  ˆ 1 1 1  = 2667 ˆ N C . + (6.00 × 10−9 C)  2 2  4πε0 ( 0 . 15 m ) ( 0 . 45 m )   c) At x = 0.15 m, y = −0.4 m :  −1 ˆ 1 1 0.3 ˆ 1 0.4 ˆ   (6.00 × 10−9 C)  + − 2 2 2 4πε0 (0.5 m) 0.5 (0.5 m) 0.5   (0.4 m) ⇒ = (129.6 ˆ − 510.3 ˆ) N C ⇒ E = 526.5 N C and θ = 75.7 down from the x(axis.  0.2  2(6.00 × 10− 9 C) ⋅   1  0.25  = 1382 ˆ N C d) x = 0, y = 0.2 m : = (0.25 m) 2 4πε0 = − Calculate in vector form the electric field for each charge, and add them. − 1 (6.00 × 10−9 C) ˆ = = −150 ˆ N C 4πε0 (0.6 m) 2   −1 1 1 (0.6) ˆ + (0.8) ˆ  = 21.6 ˆ + 28.8 ˆ N C (4.00 × 10 − 9 C) 2 2 4πε0 (1.00 m)  (1.00 m)   28.8  ⇒ E = (128.4) 2 + (28.8) 2 = 131.6 N C , at θ = tan −1   = 12.6 up from  128.4  − x axis. + = a) At the origin, =− 1 2(6.0 × 10−9 C) ˆ = −4800 ˆ N C . 4πε0 (0.15 m) 2 b) At x = 0.3 m, y = 0 :  ˆ 1 1 1  = 2133 ˆ N C . − (6.0 × 10− 9C)  2 2  ( 0 . 15 m ) ( 0 . 45 m ) 4πε0   c) At x = 0.15 m, y = −0.4 m : = = ⇒ + x ( axis.  −1 ˆ 1 1 0.3 ˆ 1 0.4 ˆ   (6.0 × 10− 9 C) − + 2 2 2 4πε0 (0.5 m) 0.5 (0.5 m) 0.5   (0.4 m) = (−129.6 ˆ − 164.5 ˆ) N C ⇒ E = 209 N C and θ = 232° clockwise from d) x = 0, y = 0.2 m : E y = 0, =− ) = − 1037 ˆ N C 1 2(6.00 × 10−9 C) ( 0.15 0.25 2 4πε0 (0.25 m) For a long straight wire, E = λ 1.5 × 10−10 C m ⇒r= = 1.08 m. 2πε0 r 2πε0 (2.5 N C) a) For a wire of length 2a centered at the origin and lying along the y(axis, the electric field is given by Eq. (21.10). 1 λ ˆ = 2 2πε0 x x a 2 + 1 b) For an infinite line of charge: λ ˆ 2πε0 x Graphs of electric field versus position for both are shown below. = For a ring of charge, the electric field is given by Eq. (21.8). 1 Qx ˆ so with = a) 2 4πε0 ( x + a 2 )3 2 Q = 0.125 × 10−9 C, a = 0.025 m and x = 0.4 m ⇒ = 7.0 ˆ N C . b) on ring =− on q =−q = − ( − 2.50 × 10−6 C) (7.0 ˆ N C) = 1.75 × 10(5 ˆ N. For a uniformly charged disk, the electric field is given by Eq. (21.11):  1 σ  ˆ = 1− 2 2  2 ε0  R x 1 +   The x (component of the electric field is shown below. The earth’s electric field is 150 N C , directly downward. So, σ 2 E = 150 = ⇒ σ = 300ε0 = 2.66 × 10− 9 C m , and is negative. 2ε0 For an infinite plane sheet, E is constant and is given by E = σ directed 2ε0 perpendicular to the surface. 2 e−  C C   100 cm   = − 4 × 10− 9 2 σ = 2.5 × 10 − 1.6 × 10−19 −  ⋅  2  m e   1m  cm  6 so E = 4 × 10 −9 2ε0 C m2 = 226 N C directed toward the surface. By superposition we can add the electric fields from two parallel sheets of charge. a) E = 0. b) E = 0. σ σ = , directed downward. c) E = 2 2 ε0 ε0 The field appears like that of a point charge a long way from the disk and an infinite plane close to the disk’s center. The field is symmetrical on the right and left (not shown). An infinite line of charge has a radial field in the plane through the wire, and constant in the plane of the wire, mirror(imaged about the wire: Cross section through the wire: Length of vector does not depend on angle. Plane of the wire: Length of vector gets shorter at points further away from wire. a) Since field lines pass from positive charges and toward negative charges, we can deduce that the top charge is positive, middle is negative, and bottom is positive. b) The electric field is the smallest on the horizontal line through the middle charge, at two positions on either side where the field lines are least dense. Here the y components of the field are cancelled between the positive charges and the negative charge cancels the x(component of the field from the two positive charges. a) p = qd ⇒ (4.5 × 10−9 C)(0.0031 m) = 1.4 × 10−11 C ⋅ m , in the direction from and towards q 2 . b) If is at 36.9°, and the torque τ = pE sin φ , then: E= τ 7.2 × 10−9 N ⋅ m = = 856.5 N C . p sin φ (1.4 × 10−11 C ⋅ m) sin 36.9° a) d = p q = (8.9 × 10−30 C ⋅ m) (1.6 × 10−19 C) = 5.56 × 10−11 m. b) τ max = pE = (8.9 × 10−30 C ⋅ m)(6.0 × 105 N C) = 5.34 × 10−24 N ⋅ m. Maximum torque: a) Changing the orientation of a dipole from parallel to perpendicular yields: U = U f − U i = − ( pE cos 90° − pE cos 0°) = + (5.0 × 10 −30 C ⋅ m)(1.6 × 106 N C) = + 8 × 10−24 J. b) 3 2(8 × 10−24 J) kT = 8 × 10− 24 J ⇒ T = = 0.384 K. 2 3(1.38 × 10− 23 J K ) Edipole ( x) = p 6.17 × 10−30 C ⋅ m −9 E = 4.11 ⇒ ( 3 . 00 × 10 m) = dipole 2πε0 (3.0 × 10− 9 m)3 2πε0 x3 × 106 N C . The electric force F = qE = (1.60 × 10−19 C)(4.11 × 106 N C) = 6.58 × 10−13 N and is toward the water molecule (negative x(direction). ( y + d 2) 2 − ( y − d 2) 2 1 1 2 yd − = = 2 2 2 2 2 2 ( y − d 2) ( y + d 2) ( y − d 4) ( y − d 2 4) 2 q 2 yd qd y p ⇒ Ey = = ≈ 2 2 2 2 2 2 4πε0 ( y − d 4 ) 2πε0 ( y − d 4) 2πε0 y 3⋅ b) This also gives the correct expression for E y since y appears in the full expression’s denominator squared, so the signs carry through correctly. a) a) The magnitude of the field the due to each charge is  q  1 q 1  , E= = 2 2 2  4πε0 r 4πε0  (d 2) + x )  where d is the distance between the two charges. The x(components of the forces due to the two charges are equal and oppositely directed and so cancel each other. The two fields have equal y(components, so:  2q  1   sin θ E = 2Ey = 2 2  4πε0  (d 2) + x  where θ is the angle below the x(axis for both fields. d 2 sin θ = ; ( d 2) 2 + x 2 thus  2q   1 d 2   Edipole =  2 2  2 2  4πε0  (d 2) + x  (d 2) + x The field is the − y directions.  qd =  4πε0 ((d 2) 2 + x 2 )3  2 b) At large x, x 2 >> (d 2) 2 , so the relationship reduces to the approximations qd Edipole ≈ 4πε0 x 3 The dipoles attract. Fx = F1x + F2 x = 0, Fy = F1 y + F2 y = 2 F1 y b) Opposite charges are closest so the dipoles attract. a) The torque is zero when is aligned either in the same direction as or in the opposite directions b) The stable orientation is when is aligned in the same direction as c) sin θ = 1.50 2.00 so θ = 48.6° Opposite charges attract and like charges repel. Fx = F1x + F2 x = 0 qq′ (5.00 × 10−6 C)(10.0 × 10−6 C) F1 = k 2 = k = 1.124 × 103 N r (0.0200 m)2 F1 y = − F1 sin θ = −842.6 N F2 y = −842.6 N so Fy = F1 y + F2 y = −1680 N (in the direction from the + 5.00 ( &C charge toward the − 5.00 ( &C charge). b) The y − components have zero moment arm and therefore zero torque. F1x and F2 x both produce clockwise torques. F1x = F1 cos θ = 743.1 N τ = 2( F1x )(0.0150 m) = 22.3 N ⋅ m, clockwise a) 13 =+ ⇒ 13 =+ 1 q1q3 1 q1q3 cos θ ˆ + sin θˆ 2 2 4πε0 r13 4πε0 r13 1 (5.00 nC)(6.00 nC) 4 ˆ 1 (5.00 nC)(6.00 nC) 3 ˆ i+ j −4 4πε0 ((9.00 + 16.0) × 10 m) 5 4πε0 ((9.00 + 16.0) × 10− 4 m) 5 = +(8.64 × 10−5 N) ˆ + (6.48 × 10−5 N) ˆ. ⇒ 13 Similarly for the force from the other charge: − 1 q2 q3 ˆ − 1 (2.00 nC)(6.00 nC) ˆ = = −(1.20 × 10− 4 ) ˆ 23 = 2 2 4πε0 (0.0300 m) 4πε0 r23 Therefore the two force components are: Fx = 8.64 × 10 −5 N Fy = 6.48 × 10−5 − 12.0 × 10−5 = −5.52 × 10−5 N b) Thus, F = Fx2 + Fy2 = (8.64 × 10−5 N) 2 + (−5.52 × 10−5 N)2 = 1.03 × 10−4 N, and the angle is θ = arctan( Fy Fx ) = 32.6, below the x axis  1 qQ 1 qQ 1 qQ  1 1   − = − 2 2 2  2 2  4πε0 (a + x) 4πε0 (a − x) 4πε0 a  (1 + x a ) (1 − x a )   qQ  1 qQ x x 1 qQ  x  x. But this is ⇒ Fq ≈ − 4  = − (1 − 2 . . . − (1 + 2 . . .)) = 2 2  3  4πε0 a a a 4πε0 a  a πε a  0  the equation of a simple harmonic oscillator, so: qQ 1 qQ kqQ ω = 2πf = ⇒ f = = . 3 3 mπ ε0 a 2π mπ ε0 a mπ 2 a 3 b) If the charge was placed on the y(axis there would be no restoring force if q and Q had the same sign. It would move straight out from the origin along the y(axis, since the x(components of force would cancel. a) Fq = Examining the forces: ∑ Fx = T sin θ − Fe = 0 and ∑ Fy = T cos θ − mg = 0. So mg sin θ cos θ = Fe = kq 2 d 2 But tan θ ≈ d 2L ⇒ d3 = 2 kq 2 L mg ⇒d = ( q2L 2 πε 0 mg ) 1 3 . a) b) Using the same force analysis as in problem q 2 = 4πε0 d 2 mg tan θ and we find: 2 d = 2 ⋅ (1.2)sin25 ⇒ q = 4πε0 (2 ⋅ (1.2) ⋅ sin 25°) 2 tan 25°(0.015 kg)(9.80 m s ) ⇒ q = 2.79 × 10−6 C. c) From Problem 21.70, mg tan θ = kq 2 d2 . d kq 2 (8.99 × 109 Nm 2 C)(2.79 × 10(6 C) 2 ⇒ tan θ = = 2L mg (2 L) 2 sin 2 θ 4(0.6m)2 (0.015 kg)(9.8 m s 2 ) sin − 2 θ 0.331 Therefore tan θ = sin . Numerical solution of this transcendental equation leads to 2 θ sin θ = θ = 39.5°. a) Free body diagram as in electric forces. Each charge still feels equal and opposite b) T = mg cos 20° = 0.0834 N so Fe = T sin 20° = 0.0285 N = kq1q2 . (Note: r 21 r1 = 2(0.500 m)sin20° = 0.342 m.) c) From part (b), q1 q 2 = 3.71 × 10 −13 C 2 . d) The charges on the spheres are made equal by connecting them with a wire, but 2 q +q we still have Fe = mg tan θ = 0.0453 N = 4πε1 0 Qr 2 where Q = 1 2 2 . But the separation r2 is 2 known: r2 = 2(0.500 m) sin 30° = 0.500 m. Hence: Q = q1 + q 2 2 = 4πε0 Fe r 2 2 = 1.12 × 10−6 C. This equation, along with that from part (b), gives us two equations in q1 and q 2 . q1 + q 2 = 2.24 × 10 −6 C and q1 q 2 = 3.70 × 10 −13 C 2 . By elimination, substitution and after solving the resulting quadratic equation, we find: q1 = 2.06 × 10 −6 C and q 2 = 1.80 × 10 −7 C . a) 0.100 mol NaCl ⇒ m Na = (0.100 mol)(22.99 g mol) = 2.30 g ⇒ m Cl = (0.100 mol)(35.45 g mol) = 3.55 g Also the number of ions is (0.100 mol) A = 6.02 × 10 22 so the charge is: q = (6.02 × 10 22 )(1.60 × 10 −19 C) = 9630 C. The force between two such charges is: 1 q2 1 (9630) 2 F= = = 2.09 × 1021 N. 2 2 4πε0 r 4πε0 (0.0200 m) b) a = F / m = (2.09 × 1021 N) (3.55 × 10−3 kg) = 5.89 × 1023 m s . c) With such a large force between them, it does not seem reasonable to think the sodium and chlorine ions could be separated in this way. 2 a) F3 = 4.0 × 10 −6  (2.5 × 10−9 C) 4.5 × 10−9 C  kq1q3 kq2 q3 ⇒ = kq3  + N= 2 + 2 r13 r232 (+0.2 m)2   (−0.3 m) 4.0 × 10−6 N q3 = = 3.2 nC. (1262 N C) b) The force acts on the middle charge to the right. c) The force equals zero if the two forces from the other charges cancel. Because of the magnitude and size of the charges, this can only occur to the left of the negative kq1 kq 2 charge q 2 . Then: F13 = F23 ⇒ = where x is the distance 2 (0.300 − x) (−0.200 − x) 2 from the origin. Solving for x we find: x = −1.76 m. The other value of x was between the two charges and is not allowed. 1 1 6q 2 q (3q) , toward the lower the left charge. The other = 4πε0 ( L 2 ) 2 4πε0 L2 two forces are equal and opposite. a) F = + b) The upper left charge and lower right charge have equal magnitude forces at right angles to each other, resulting in a total force of twice the force of one, directed toward the lower left charge. So, all the forces sum to: 3 1  q (3q ) 2 q (3q)  q2    = 3 2 +  N. F= + 2 2  2   2 4πε0  L ( 2 L)  4πε0 L  q q 1  2q   + − 2  . 2 2 y  4πε0  ( y − a ) ( y + a) 1 q ((1 − a y ) − 2 + (1 + a y ) − 2 − 2). Using the binomial expansion: b) E ( p ) = 2 4πε0 y a) E ( p ) =  1 q  2a 3a 2 2a 3a 2 1 6qa 2   + + + ⋅ ⋅ ⋅ + − + + ⋅ ⋅ ⋅ − 1 1 2 =  4πε y 4 . 4πε0 y 2  y y2 y y2  0 1 1 Note that a point charge drops off like 2 and a dipole like 3 . y y ⇒ E ( p) ≈ a) The field is all in the x (direction (the rest cancels). From the + q charges: q q x 1 1 1 qx E= = . ⇒ Ex = 2 2 2 2 2 2 2 4πε0 a + x 4πε0 a + x 4πε0 (a + x 2 )3 2 a +x (Each + q contributes this). From the − 2q : 1 2q 1  2qx 2q  1 2q  2 Ex = − ((a 2 x 2 + 1) − 3 ⇒ Etotal = − 2  = 2 3 2 2 4πε0 x 4πε0  (a + x ) x  4πε0 x 2  1 2q  3a 2 1 3qa 2   − + ⋅ ⋅ ⋅ − b) Etotal ≈ 1 1 =  4πε x 4 , for x >> a.. 4πε0 x 2  2 x 2  0 Note that a point charge drops off like a) 20.0 g carbon ⇒ 2 − 1). 1 1 and a dipole like 3 . 2 x x 20.0 g = 1.67 mol carbon ⇒ 6(1.67) = 10.0 mol 12.0 g mol electrons ⇒ q = (10.0) A (1.60 × 10−19C) = 0.963 × 106 C. This much charge is placed at the earth’s poles (negative at north, positive at south), leading to a force: 1 q2 1 (0.963 ×106C) 2 F= = = 5.13 × 107 N. 4πε0 (2 Rearth ) 2 4πε0 (1.276 ×107 m) 2 b) A positive charge at the equator of the same magnitude as above will feel a force in the south(to(north direction, perpendicular to the earth’s surface: q2 1 F =2 sin 45 4πε0 (2 Rearth ) 2 ⇒F =2 1 4 (0.963 × 106C) 2 = 1.44 × 108 N. 4πε0 2 (1.276 × 107 m)2 a) With the mass of the book about 1.0 kg, most of which is protons and neutrons, we find: #protons = 12 (1.0 kg) (1.67 × 10 −27 kg) = 3.0 × 10 26. Thus the charge difference present if the electron’s charge was 99.999% of the proton’s is q = (3.0 × 1026 )(0.00001)(1.6 × 10−19 C) = 480 C. b) F = k ( q) 2 r 2 = k (480 C) 2 (5.0 m) 2 = 8.3 × 1013 N − repulsive. The acceleration 2 a = F m = (8.3 × 1013 N ) (1 kg ) = 8.3 × 1013 m s . c) Thus even the slightest charge imbalance in matter would lead to explosive repulsion! (a) Fnet (on central charge) = 1 q2 1 q2 − 4πε0 (b − x) 2 4πε0 (b + x) 2  1 1   (b − x) 2 − (b + x) 2    2 2 2 4bx q (b + x) − (b − x) q2 = = 2 2 4πε0 (b − x) (b + x) 4πε0 (b − x) 2 (b + x) 2 For x << b, this expression becomes = Fnet ≈ q2 4πε0 q 2 bx q2 = x Direction is opposite to x. πε0 b 2b 2 πε0b3 (b) ∑ F = ma : − −q2 πε 0 b3 x = m ddt 2x 2  q2  d 2x x = −  3 dt 2 mπ ε b 0   ω= q2 1 = 2π f → f = 3 mπ ε0b 2π q2 mπ ε0b3 (c) q = e, b = 4.0 × 10−10 m, m = 12 amu = 12(1.66 × 10−27 kg) f = 1 (1.6 × 10−19 C) 2 = 4.28 × 1012 Hz 2π 12(1.66 × 10− 27 kg)π (8.85 × 10−12 C 2 Nm 2 )(4.0 × 10−10 m)3 a) m = ρV = ρ( 43 πr 3 ) = (8.9 × 103 kg m3 )( 43 π )(1.00 × 10 −3 m)3 = 3.728 × 10 −5 kg n = m M = (3.728 × 10 −5 kg)(63.546 × 10 −3 kg mol) = 5.867 × 10 −4 mol = n A = 3.5 × 1020 atoms (b) e = (29)(3.5 × 10 20 ) = 1.015 × 10 22 electrons and protons qnet = e F =k e − (0.99900)e e = (0.100 × 10−2 )(1.602 × 10−19 C)(1.015 × 1022 ) = 1.6 C (1.6 C) 2 q2 = k = 2.3 × 1010 N r2 (1.00 m) 2 First, the mass of the drop: −6 m) 3  3  4π (15.0 × 10  = 1.41 × 10−11 kg. m = ρV = (1000 kg m ) 3   Next, the time of flight: t = D v = 0.02 20 = 0.00100 s and the acceleration : d= 1 2 2d 2(3.00 × 10 −4 m) 2 = 600 m s . at ⇒ a = 2 = 2 2 (0.001 s) t So: (1.41 × 10 −11 kg )(600 m s ) = 1.06 × 10 −13 C. 4 8.00 × 10 N C 2 a = F m = qE m ⇒ q = ma E = Fx = 0 Fy = eE ay = a) Fy mp = v 2y = v02y + 2a y y = v02 sin 2 α + ⇒ hmax = b) eE ax = 0 mp 2eE y mp y = hmax when v y = 0 vo2 m p sin 2 α 2eE 1 2 y = v0 yt + a y t 2 t = torig when y = 0 so torig 1   ⇒ 0 =  − v0 sin α + a ytorig  torig 2   2v sin α = 0, 0 ay or 2v0 mp sin α eE 2v 2 m d = v0 xtorig = 0 p cos α sin α eE torig = c) d) hmax = (4 × 105 m s) 2 (1.67 × 10−27 kg) sin 2 30° = 0.42 m 2(1.6 × 10−19 C)(500 N C) 2(4 × 105 m s) 2 (1.67 × 1027 kg) cos 30° sin 30° d= = 2.89 m (1.6 × 10−19 C)(500 N C) a) E = 50 N C = 1 q1 1 q2 1  q1 q2  + = + 2  ⇒ q2 = 2 2 4πε0 r1 4πε0 r2 4πε0  r12 r2    q  (4.00 × 10−9 C)   = − 8 × 10− 9 C. r22  4πε0 E − 12  ⇒ q2 = (1.2 m) 2  4πε0 50.0 N C − 2 r (0.6 m) 1     b) E = −50 N C = 1 q1 1 q2 1  q1 q2  + = + ⇒ q2 = 2 2 2 4πε0 r1 4πε0 r2 4πε0  r1 r22   − 50 q1  r22  − 2  ⇒ q2 = (1.2 m)2 r1   k E = 12.0 N C =  (4.00 × 10−9 C)   4πε0 (−50.0) −  = − 24.0 × 10− 9 C. 2 (0.6 m)   − k (16.0 nC) k (12.0 nC) kq + + 2 2 (3.00 m) (8.00 m) (5.00 m) 2  12 1.60 × 10 −8 C 1.20 × 10 −8 C   = +7.31 × 10 −8 C = +73.1 nC. ⇒ q = 25.0 m 2  + − 2 2 k 9 . 0 m 64 . 00 m   a 1 dq 1 Qdx a) On the x(axis: dEx = ⇒ Ex = = 2 ∫ 4πε0 (a + r ) 4πε0 0 a(a + r − x) 2 1 Q1 1   − . And E y = 0 4πε0 a  r a + r  1 Q 1 1 1 qQ  1 1   −  ⇒ = q = −  ˆ. 4πε0 a  x − a x  4πε0 a  x − a x  kqQ kqQ kqQ ((1 − a x) −1 − 1) = (1 + a x + ⋅ ⋅ ⋅ − 1) ≈ 2 ≈ c) For x >> a, F = ax ax x 1 qQ . ( Note that for x >> a, r = x − a ≈ x.) Charge distribution looks like a point 4πε0 r 2 from far away. b) If a + r = x, then E = a) dE = k dq kQ dy kQx dy = with dE x = 2 2 2 ( x + y ) a( x + y ) a( x 2 + y 2 ) 3 2 2 and dE y = − KQy dy . Thus: a( x 2 + y 2 ) 3 2 a Ex = 1 Qx dy 2 ∫ 4πε0 a 0 ( x + y 2 )3 2 = 1 Qx 1 2 4πε0 a ( x + a 2 )1 2 a 1 Q = 2 2 x 4πε0 x( x + a 2 )1 2  1 −1 Q  1  − 2  2 1 2  4πε0 a  x ( x + a )  b) Fx = −qE x and Fy = −qE y where E x and E y are given in (a). a Ey = ydy −1 Q 2 ∫ 4πε0 a 0 ( x + y 2 )3 c) For x >> a, Fy = 2 = 1 qQ 1 qQ a 2 1 qQa (1 − (1 + a 2 / x 2 ) −1 2 ) ≈ = . 2 4πε0 ax 4πε0 ax 2 x 4πε0 2 x 3 1 qQ  a2    1 + Looks dipole(like in y(direction Fx = − 4πε0 x 2  x 2  −1 2 ≈ qQ . 4πε0 x 2 Looks point(like along x(direction a) From Eq. (22.9), ⇒ = = 1 Q ˆ 4πε0 x x 2 + a 2 1 (−9.00 × 10−9 C ) ˆ = (−1.29 × 106 N C) ˆ. − − 3 3 2 2 4πε0 (2.5 × 10 m) (2.5 × 10 m) + (0.025 m) (b) The electric field is less than that at the same distance from an infinite line of charge ( Ea →∞ = 1 2λ − 1 2Q = = − 1.30 × 106 N C). This is because in the 4πε0 x 4πε0 x 2a approximation, the terms left off were negative. 1 λ 2 2πε0 x 1 + ax 2 ( ) 12 ≈  λ  x2 1 − 2 + ⋅ ⋅ ⋅  = 2πε0 x  2a  E ∞ − (Higher order terms). Line c) For a 1% difference, we need the next highest term in the expansion that was left off to be less than 0.01: x2 < 0.01 ⇒ x < a 2(0.01) = 0.025m 2(0.01) ⇒ x < 0.35 cm. 2a 2 (a) From Eq. (22.9), ⇒ = = 1 Q ˆ 4πε0 x x 2 + a 2 1 (−9.0 × 10−9 C) = (−7858 N C) ˆ. 4πε0 (0.100 m) (0.100 m)2 + (0.025 m) 2 b) The electric field is less than that at the same distance from a point charge (8100 N C). Since E x →∞ =  1 Q a2  − + ⋅ ⋅ ⋅  = Epoint –(Higher order terms). 1 2  2 4πε0 x  2 x  c) For a 1% difference, we need the next highest term in the expansion that was left off to be less than 0.01: a2 ≈ 0.01 ⇒ x ≈ a 1 (2 (0.01)) = 0.025 1 0.02 ⇒ x ≈ 0.177 m. 2 x2 (a) On the axis, −1 2   4.00 p C π (0.025 m) 2 σ   R2 + 1  = 1 −  E= 2ε0   x 2 2 ε0    −1 2   (0.025 m) 2   + 1  1 −  2   (0.0020 m)   ⇒ E = 106 N C , in the + x(direction. b) The electric field is less than that of an infinite sheet E∞ = σ = 115 N C . 2 ε0 c) Finite disk electric field can be expanded using the binomial theorem since the expansion terms are small: ⇒ E ≈ σ 2 ε0 infinite sheet and finite disk goes like   x x3 − + 1  R 2 R 3 − ⋅ ⋅ ⋅ So the difference between the   x . Thus: R E ( x = 0.20 cm) ≈ 0.2 2.5 = 0.08 = 8% and E ( x = 0.40 cm) ≈ 0.4 2.5 = 0.16 = 16%. −1  σ   R2 + 1 E= 1 −  2ε0   x 2   (a) As in 2    −1  4.00 pC π (0.025 m)2   (0.025 m)2 = + 1 1 −  2 2 ε0   (0.200 m)  2  ⇒E  = 0.89 N C in the + x(direction. b) x >> R, E = σ [1 − (1 − R 2 2 x 2 + 3R 4 8 x 4 − ⋅ ⋅ ⋅)] 2 ε0 σ R2 σπR 2 Q ≈ = = . 2 2 2ε0 2 x 4πε0 x 4πε0 x 2 c) The electric field of (a) is less than that of the point charge (0.90 N C) since the correction term that was omitted was negative. d) From above x = 0.2 m (0.9 − 0.89) = 0.01 = 1%. For x = 0.1 m 0.89 Edisk = 3.43 N C Epoint = 3.6 N C so (3.6 − 3.43) = 0.047 ≈ 5%. 3.6 a 0 −a −a a) f ( x ) = f (− x) : ∫ f ( x)dx = ∫ ∫ a 0 a −a 0 0 f ( x)dx + ∫ f ( x)dx = ∫ f ( − x) d ( − x) + a a a −a 0 0 f ( x)dx. Now replace − x with y : ⇒ ∫ f ( x)dx = ∫ f ( y )d ( y ) + ∫ f ( x)dx = a 2 ∫ f ( x)dx. 0 a 0 a −a −a −a 0 0 b) g ( x ) = − g (− x) : ∫ g ( x )dx = ∫ g ( x )dx + ∫ g ( x )dx = − ∫ − g (− x)(−d (− x )) + ∫ a 0 a a a −a 0 0 g ( x)dx. Now replace − x with y : ⇒ ∫ g ( x)dx = − ∫ g ( y )d ( y ) + ∫ g ( x)dx = 0. c) The integrand in E y for Example 21.11 is odd, so E y =0. a) The y(components of the electric field cancel, and the x(components from both charges, as given in problem Ex = 1 − 2Q 4πε0 a If y >> a, ≈ is: 1 1  − 2 2 1  y (y + a ) 2   ⇒  = 1 − 2Qq  1 1  − 2 4πε0 a  y ( y + a 2 )1 2 ˆ  .  1 Qqa 1 − 2Qq . (1 − (1 − a 2 2 y 2 + ⋅ ⋅ ⋅)) ˆ = − 4πε0 y 3 4πε0 ay b) If the point charge is now on the x(axis the two charged parts of the rods provide different forces, though still along the x(axis (see problem + =q = + 1 Qq  1 1 −  ˆ and  4πε0 a  x − a x  − =q − =− ). 1 Qq  1 1 ˆ  −  4πε0 a  x x + a  So, = + + − For x >> a, = 1 Qq  1 2 1 ˆ − +   4πε0 a  x − a x x + a   a a2 ˆ  1 2Qqa ˆ 1 Qq   a a 2     =   1 . . . 2 1 . . . . − + − + − ≈ + + + 2 2      4πε0 x 3 4πε0 ax   x x  x x   The electric field in the x(direction cancels the left and right halves of the semicircle. The remaining y(component points in the negative y(direction. The charge per unit length of the semicircle is: λ= Q kλ dl kλ dθ kλ sin θ dθ and dE = 2 = but dE y = dE sin θ = . πa a a a So, E y = 2kλ π a ∫0 2 sin θ dθ = 2kλ [−cosθ]π0 a 2 = 2kλ 2kQ = 2 , downward. a πa By symmetry, E x = E y . For E y , compared to problem , the integral over the angle is halved but the charge density doubles─giving the same result. Thus, Ex = E y = 2kλ 2kQ = 2. a πa mg cos α qσ qσ ∑ Fy = 0 ⇒ T sin α = 2ε ⇒ T = 2ε sin α 0 0 ∑F x ⇒ = 0 ⇒ T cos α = mg ⇒ T = qσ mg qσ = ⇒ tan α = cos α 2ε0 sin α 2ε0 mg  qσ   ⇒ α = arctan 2 ε mg  0  a) b) 1429 1.4 × 10 5 N C m E qE = 10 mg ⇒ = = = 1429 kg C. q 10g 10(9.8 m s 2 ) kg 1 mol 6.02 × 1023 carbons 1.6 × 10−19 C carbons . . . = 1.15 × 1010 . −3 − C 12 × 10 kg mol excess e excess e − a) E x = E y , and Ex = 2 Elength of wire a , charge Q =2 Q 1 , where 4πε0 x x 2 + ( a )2 2 2Q 2Q a ⇒ Ex = − , Ey = − . 2 2 πε0 a 2 πε0 a b) If all edges of the square had equal charge, the electric fields would cancel by symmetry at the center of the square. x= a) E ( P) = − σ1 σ σ 0.0200 C m 2 0.0100 C m 2 0.0200 C m 2 − 2 + 3 =− − + 2 ε0 2ε0 2 ε0 2 ε0 2 ε0 2 ε0 ⇒ E ( P) = 0.0100 C m 2 = 5.65 × 108 N C, in the − x(direction. 2ε0 b) E ( R) = + ⇒ E ( R) = σ1 σ σ 0.0200 C m 2 0.0100 C m 2 0.0200 C m 2 − 2 + 3 =+ − + 2ε0 2 ε0 2ε 0 2 ε0 2 ε0 2 ε0 0.0300 C m 2 = 1.69 × 109 N C, in the + x(direction. 2ε0 E(S ) = + c) ⇒ E (S ) = σ1 σ σ 0.0200 C m 2 0.0100 C m 2 0.0200 C m 2 + 2 + 3 =+ + + 2 ε0 2ε0 2 ε0 2 ε0 2 ε0 2 ε0 0.0500 C m 2 = 2.82 × 109 N C, in the + x(direction. 2 ε0 d) E (T ) = + ⇒ E (S ) = σ1 σ σ 0.0200 C m 2 0.0100 C m 2 0.0200 C m 2 + 2 − 3 =+ + − 2 ε0 2 ε0 2 ε0 2 ε0 2 ε0 2ε0 0.0100 C m 2 = 5.65 × 108 N C, in the + x(direction. 2 ε0  − σ 2 + σ 3  2.00 × 10 − 4 C m 2  = = σ1  = +1.13 × 107 N m. 2 ε0 2 ε0    + σ1 + σ 3  4.00 × 10− 4 C m 2 Fon II qEat II  = = +2.26 × 107 N m = = σ 2  A A 2ε0 2 ε0    + σ1 + σ 2  − 6.00 × 10− 4 C m 2 Fon III qEat III  = = −3.39 × 107 N m = = σ 3  A A 2 ε 2 ε 0 0   (Note that “+” means toward the right, and “–” is toward the left.) Fon I qEat = A A I By inspection the fields in the different regions are as shown below:  σ  ˆ ˆ  σ  ˆ ˆ (− + ), EII =  (+ + ) EI =   2ε0   2 ε0   σ  ˆ ˆ  σ  ˆ ˆ (+ − ), EIV =  (− − ) EIII =  2 ε 2 ε  0  0 x ˆ z ˆ  σ  (− + ). ∴ =  x z  2 ε0  a) Q = Aσ = π ( R22 − R22 )σ b) Recall the electric field of a disk, Eq. (21.11): E = (1 ([ ][ σ 1 −1 2ε0 ( R2 x) 2 + 1 − 1 − 1 ( R2 x) 2 + 1 − 1 ( R1 x) 2 + 1 ( x) = c) Note that 1 ⇒ E ( x) = ( R1 x) 2 + 1 [ σ 1 −1 2 ε0 ] ( R x) 2 + 1 . So, ] ) xx ˆ ⇒ E( x) = 2−εσ × 0 ( R1 x) 2 + 1 = ) xx ˆ. x (1 + ( x R1 ) 2 ) −1 R1 2 ≈ x  ( x R1 ) 2  1 − + ... R1  2  σ  x x x σ 1 1  x ˆ  −   −  ˆ = , and sufficiently close means that 2ε0  R1 R2  x 2ε0  R1 R2  x 2 ( x R1 ) 2 << 1. d) F = qE ( x) = − 1  1 qσ  1 ω  −  x = mx ⇒ f = = 2ε0  R1 R2  2π 2π 1  qσ  1  − . 2ε0 m  R1 R2  a) The four possible force diagrams are: Only the last picture can result in an electric field in the –x direction. b) q1 = −2.00 & C, q3 = +4.00, &C, and q2 > 0. q1 1 1 q2 sin θ2 sin θ1 − c) E y = 0 = 2 4πε0 (0.0300 m)2 4πε0 (0.0400 m) 3 5 27 9 sin θ1 9 q1 q1 = 0.843 &C. ⇒ q2 = q1 = = 16 sin θ2 16 4 5 64 d) F3 = q3 E x = q3 1  q1 4 q2 3  +   = 56.2 N 4πε0  0.0016 5 0.0009 5  (a) The four possible diagrams are: The first diagram is the only one in which the electric field must point in the negative y direction. b) q1 = −3.00 &C, and q2 < 0. kq1 kq 2 kq 2 kq1 5 12 5 − ⇒ = c) E x = 0 = 2 2 2 2 (0.120 m) (0.050 m) 12 (0.050 m) 13 (0.120 m) 13 E = Ey = kq1 kq 2 kq1 5 12 + = 2 2 (0.050 m) 13 (0.120 m) 13 (0.05 m) 2 ⇒ E = E y = 1.17 × 10 7 N C.  12  5  5    +      13  12  13   a) For a rod in general of length L, E = 1  kQ  1 a  and here r = x + .  − L r L+r 2  2kQ  1 1 1 kQ  1   =  − − .  L  x+ a 2 L + x+ a 2 L  2x + a 2L + 2x + a  L + a 2 EQ 2kQ 2 L + a 2 dx = 2 ∫ × b) dF = dq E ⇒ F = ∫ E dq = ∫ a 2 L L a 2 1  1  −   dx  2 x + a 2L + 2 x + a  2kQ 2 1 ⇒F= 2 [ln (a + 2 x)]aL +2a 2 − [ln(2 L + 2 x + a )]aL +2a 2 L 2  (a + L) 2    a + 2 L + a   2 L + 2a   kQ 2 kQ 2 . ⇒ F = 2 1n     = 2 1n   2a L L   4 L + 2a     a (a + 2 L)  So, E left rod = ( c) For a >> L : F = ⇒F≈ ) kQ 2  a 2 (1 + L a ) 2  kQ 2  = 2 (2 1n (1 + 2 L a ) − ln(1 + 2 L a)) 1n  2 L2  a (1 + 2 L a)  L    2 L 2 L2 kQ 2 kQ 2   L L2       2 . F ⇒ ≈ − − + ⋅ ⋅ ⋅ − + ⋅ ⋅ ⋅    a a2 L2   a 2a 2 a2   