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
=−
ln1 + .
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
ln1 + = 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
1a
≈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
− 21 − 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 = ω1 − 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
= w1 −
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
= x1 +
∂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!
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= *
×
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/
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= $ $" ×
.
n=
(
−$
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pV
=
RT *
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(* / ×
)
V! = V *T! T ( = * "
( nRT V = / ! ×
! =
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(* % × $ (
⋅ ( *!0
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(
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$
)
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,- *
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! =
( 8 nV
0
/×
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.
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<
p ! = p *V V! ( = *
8
V
( V! =
.
p T!
p! T
= *$
(* !0"
!// ( = $ /%
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(
>
p !V
=
nR
*
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( T! =
(1
'
>
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(*$
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(
⋅ (
= $%$
= / $°4
)
* (1
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nRT *$
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( 6
F = pA = *nRT V ( A = *nRT ( L
A V = L 6 T = ! ° 4 = !0$
'
F=
(* $ % #
!
⋅ ( *!0$
(
= $ "" ×
%
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⋅ (*$/$
(
=%" ×
%
?
°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
$
$"% )
)(
*0
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$ % #
⋅
A m = nM = *$
= ! %×
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=
,
(p=*
0 × −0 . (*
×
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⋅ (*$
%*
= %% ×
)(!/$
(* % ×
$
=
A
−"
ρ
'
!$
0 ")
0× $ )
= 0
!%
!=
R
M=
ρ = ρ e−
!%
@
p = ρRT M =
=!! × % .
%
RT
y=
6
,
$
(
(
=!! ×
pV
RT
−
$
.
)
− "
=! %
−$
(=
)
$
A
!$
*!
×
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) 5
( = %0 )
)
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*" !$ ×
"
−!
%
(
0 ")
(
(
nRT
RT
=
=
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V
p=
×
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*" !$ ×
$
− !
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⋅ (*/
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1
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(B
9
) '
C
V = *!! % ×
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= ! !$ ×
(
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=
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=
n A
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=
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− !"
=% 0 ×
⋅ (*$
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(
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$
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
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)
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×
5
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)
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v
v
=
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= %0
v
v
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v
v
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= $
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(
n
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×
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−
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∑ 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 = *$ !(* $
−!$
×
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m
=
(
v =
$RT
$* $ % #
=
M
*$! ×
−$
)
−$
#
(
⋅ (*$
(
= % %×
)
(
-
= ! $% ×
!
s!
!
* (
M
*$! × −$ )
v =
( mv =
*" !$ × !$
A
= ! / × −!$ ) ⋅
1
−!
(="! ×
(*$
!*" ! × −! #(
( *" !$ × !$
(
*$! ×
#
(
(
*% % ×
!
(
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!*" ! × −! #(*$! ×
*" !$ × !$
=
!mK
−$
)
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(
=
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mv
=
!L v
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p
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P P =
(
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1
?
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*
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8
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(
pV
RT
! " ⋅
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− /
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(
− 0
= !% ×
?
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)(
!% ×
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. =
)
×
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)
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⋅
)
)($
!
8
)(
5
!!
* (
#(
'
'
1
×
p=$
,
− $
1
= % "×
$
"0
λ = "×
> T? ! = T; ! * M ? ! M ; ! ( = (!0$
= $ /0 ×
$
(
⋅
! !(
°4
$kT
$* $ × −!$ # (*$
(
=
m
$ × −")
-
(1
"( ! )R = $R = !% 0 #5
)*!
$R
M
(
)
$( $ % #
=(
×
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⋅
)
= " %% ×
−$
(
'
)
) = $0 ×
'
$
#5) ⋅
(
'
(
−
(6
) ( $$ #5) ⋅ °4)(
) 5
)=
( Cv = (C ) (
#
⋅ °4
°4 ( "% #5) ⋅ °4)(
) 5
) = !0 #5) ⋅ °4 − " °4 (! " #5) ⋅ °4) ×
) 5
) = $/ #5 ⋅ °4 − °4 ( B
$R
; !+
( CV
(&
,- *
,- *
! ( $
!"( Q = (!
)(!
/0 #
* ( 0$" #
⋅
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)=
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c=
(
/%
=
% 0
CV
! /" #5
⋅
=
$
M ! ×
) 5
'
//
m?C ? T? = m C
(
= /% #5) ⋅
m? =
T
m C
C?
* (
m? =
[(
pV = nRT
" ) 1
)5 (
" )
6
)](
! ) 5
1
*
⋅
! "
!(
5
V =
⋅
nRT
=
p
)(!0$ ) = %
* " (v
$kT $RT
v!
v !=
=
=
m
M
* " (!
*
,
% (
T=
Mv !
*! × −$ )
=
$* " ( ! R $* " (! * $ % #
(
v ! = *% $
⋅ (
( *% $
×
−%
⋅
!
!
(*
(! = 0 /
( *% $
×
−%
⋅
!
!
(*
( ! = %$
( *% $
×
−%
⋅
!
!
(*
×
−%
⋅
!
(! =
ε=
D )
m
f * v ( = %π
!πkT
$!
!
mv !
$!
!ε − ε kT
π m
− ε kT
=
e
εe
m !πkT
m
!
(v !
,
,- *
f = Aε e − ε kT
$$(
A
1
df
ε − ε kT
ε
e
= Ae − ε kT −
= Ae − ε kT −
de
kT
kT
1
f
,- *
ε = mv ! = kT
!
9
$%(
k
R
=
m M
?
)
8
8
=
R
M
−$
( !* $ % #
⋅ (*$
( *%% ×
(
⋅ (*$
( *π *%% ×
* $ % #
⋅ (*$
( $* $ % #
( *%% ×
T=
'
−$
−$
( = $ $/ ×
)
(( = $
)
( = % !×
)
!
×
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!
°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%
(
'
*?
$×
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)(
$ (=
. (*$
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
x1 − 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 Q1
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