Chapter 16
ANSWERS TO EVEN-NUMBERED PROBLEMS
P16.2
(a) 37.0°C, 310 K; (b) –20.6°C, 253 K
P16.4
(a) −270°C; (b) 1.27 atm, 1.74 atm
P16.6
(a) 810°F; (b) 450 K
P16.8
Required T = −376° C is below absolute zero.
P16.10
(a)
P16.12
55.0°C
P16.14
(a) 437°C; (b) 2.1 × 10–3°C; (c) No. Aluminum melts at 660°C. Also, brass melts at
about 900°C.
P16.16
r  0.663 mm to the right at 78.2° below the horizontal
P16.18
0.548 gal
P16.20
6.64 × 10–27 kg
P16.22
877
P16.24
1.89 MPa
P16.26
In each pump-up-and-discharge cycle, the volume of air in the tank doubles. Thus
1.00 L of water is driven out by the air injected at the first pumping, 2.00 L by the
second, and only the remaining 1.00 L by the third. Each person could more
efficiently use his device by starting with the tank half full of water, instead of 80%
full.
0
1  T  ; (b) m
m1  m 2 
P16.28
P0VM  1 1 

R  T1 T2 
P16.30
(a) 1.17 × 10−3 kg; (b) 11.5 mN; (c) 1.01 kN; (d) molecules must be moving very fast
P16.32
~102 kg
P16.34
7.13 m
P16.36
4.39 kg
P16.38
(a) 2.28 kJ; 6.21 × 10–21 J
P16.40
(a) 8.76 × 10−21 J; (b) For helium, vrms = 1.62 km/s and for argon,
vrms = 514 m/s
P16.42
3 PV
2 KN A
P16.44
(a) 3.54 × 1023 atoms; (b) 6.07 × 10−21 J; (c) 1.35 km/s
P16.46
See P16.46 for the full explanation.
P16.48
132 m/s
P16.50
(a) –9.73°C/km; (b) Air contains water vapor. Air does not behave as an ideal gas.
As a parcel of air rises in the atmosphere and its temperature drops, its ability to
contain water vapor decreases, so water will likely condense out as liquid drops or
as ice crystals. (The condensate may or may not be visible as clouds.) The
condensate releases its heat of vaporization, raising the air temperature above the
value that would be expected according to part (a); (c) –4.60°C/km; (d) 4.34 km;
P16.52
0.523 kg
P16.56
(a) Particle in equilibrium model; (b) On the piston,
F  F
gas
P16.58
P16.60
;
P16.54
h    3 
V T
A
 Fg  Fair  0 :  F  PA  mg  P0 A  0 ; (c) h 
nRT
mg  P0 A
 2   1 Li T
; (b) In the expression from part (a),  is directly proportional to
r
∆T and also to  2   1 . Therefore,  is zero when either of these quantities
becomes zero; (c) the bimetallic strip bends the other way

(a) A  2 AT ; (b) T  1
P16.62
(a) See P16.62(a) for the full explanation; (b) as the temperature increases, the
density decreases; (c) 5 × 10–5 (°C)–1; (d) –2.5 × 10–5 (°C)–1
P16.64
y  L T/ 2
P16.66
(a)  
P16.68
(a)
P16.70
95.0°; T falls below 100°C, so steam condenses, and the expensive apparatus falls
(assuming that the boiling point does not change significantly with the change in
pressure).
P16.72
(a) See ANS. FIG. P16.72(a); (b) v mp  510 m/ s ; (c) 575 m/s, 624 m/s; (d) 44%
P16.74
(a) 2.38 mm; (b) 294.4 K = 21.4°C
1
3
1
; (b) 3.66 × 10−3 K−1; (c) He  3.665  10 K , this agrees within 0.06% of
T
3
1
the tabulated value; (d)  He  3.67  10 K , this agrees within 0.2% of the
tabulated value
PM
; (b) 1.33 kg/m3
RT
Chapter 17
ANSWERS TO EVEN-NUMBERED PROBLEMS
P17.2
0.105° C
P17.4
0.234 kJ/kg  °C
P17.6
(a) 16.1°C; (b) 16.1°C; (c) It makes no difference whether the drill bit is dull or
sharp, or how far into the block it cuts. The answers to (a) and (b) are the same
because all of the work done by the bit on the block constitutes energy being
transferred into the internal energy of the steel.
P17.8
(a) 1.82 × 103 J/kg . °C; (b) We cannot make definite identification. It might be
beryllium; (c) The material might be an unknown alloy or a material not listed in
the table.
mAl cAl  mccw Tc  mhcwTh
P17.10
mAl cAl  mccw  mhcw
P17.12
88.2 W
P17.14
12.9 g steam
P17.16
(a) Isolated system (momentum). The collision is a perfectly inelastic collision,
where momentum is conserved, but kinetic energy is not. (It is transformed to
internal energy); (b) 20.0 m/s to the right; (c) 1.18 × 103 J; (d) No; (e) 327.3°C; (f) 3.1 g
of solid lead and 16.9 g of liquid lead
P17.18
(a) 7; (b) As the car stops, it transforms part of its kinetic energy into internal
energy due to air resistance. As soon as the brakes rise above the air temperature,
they transfer energy by heat into the air and transfer it very fast if they attain a
high temperature.
P17.20
11.1 W
P17.22
(a) –4 PiVi; (b) It is proportional to the square of the volume, according to
T  (Pi / nRV i )V 2 .
P17.24
–nR(T2 – T1)
P17.26
4.29 × 104 J
P17.28
(a) 12.0 kJ; (b) −12.0 kJ
P17.30
From the first law of thermodynamics, ∆Eint = Q + W = 10.0 J + 12.0 J = +22.0 J. The
change in internal energy is a positive number, which would be consistent with an
increase in temperature of the gas, but the problem statement indicates a decrease in
temperature.
P17.32
(a) 1 300 J; (b) 100 J; (c) −900 J; (d) −1 400 J
P17.34
(a) −3.10 kJ; (b) 37.6 kJ
P17.36
(a) −4 PiVi; (b) 4 PiVi
P17.38
(a) 0.007 65 m3; (b) 305 K
P17.40
13.5PV
P17.42
Between 10–3 °C and 10–2 °C
P17.44
(a) 5.66 × 107 J; (b) 1.12 kg
P17.46
(a) 1.39 atm; (b) 366 K and 253 K; (c) Q = 0; (d) −4.66 kJ; (e) −4.66 kJ
P17.48
1/ 
(a) See ANS FIG P17.48(a); (b) 3 Vi ; (c) 3Ti; (d) Ti;
 1 

1  31/   1  31/  
(e) PiVi 


   1 

P17.50
941 K
P17.52
(a) 28.0 kJ; (b) 46.0 kJ; (c) 10.0 atm; (d) 25.1 atm
P17.54
25.0 kW
P17.56
1  dEint  1
 kBT 
 nRT 
(a) Eint  N f 
  f 
 ; (b) CV  
  fR ; (c)
2
2
n dT  2
f 2
CP
1

CP  CV  R   f  2 R ; (d)  
CV
f
2
3
R , so this gas cannot exist.
2
P17.58
The lowest possible value of CV, is CV 
P17.60
30.3 kcal/h
P17.62
2.22 × 10–2 W/m . °C
P17.64
(a) Intensity is defined as power per area perpendicular to the direction of energy
flow. The direction of sunlight is along the line from the Sun to an object. The
perpendicular area is the projected flat circular area enclosed by the terminator. The
object radiates infrared light outward in all directions. The area perpendicular to
this energy flow is its spherical surface area; (b) 279 K, it is chilly, well below room
temperatures we find comfortable.
P17.66
(a) See P17.66(a) for a full explanation; (b) See P17.66(b) for a full explanation; (c)
See P17.66(c) for a full explanation
P17.68
(a) 0.203 mol; (b) 900 K; (c) 900 K; (d) 15.0 L; (e) A  B: lock piston in place and put
the cylinder into an oven at 900 K, gradually heating the gas. B  C: keep the
sample in the oven while gradually letting the gas expand to lift a load on the
piston as far as it can. C  A: carry the cylinder back into the room at 300 K and let
the gas gradually cool and contract without touching the piston; (f) A  B: W = 0,
∆Eint = 1.52 kJ, Q = 1.52 kJ. B  C: W = –1.67 kJ, ∆Eint = 0, Q = 1.67 kJ. C  A: W =
1.01 kJ, ∆Eint = –1.52 kJ, Q = –2.53 kJ; (g) QABCA = 0.656 kJ, WABCA = –0.656 kJ, (∆Eint)ABCA
=0
P17.70
P17.72
(a) 0.964 kg or more; (b) The test samples and the inner surface of the insulation can
be pre-warmed to 37.0° as the box is assembled. Then, nothing changes in
temperature during the test period and the masses of the test samples and
insulation make no difference.
P
RT
P17.74
Most of the morning is used in making coffee, and there is no time left for a
morning hike.
P17.76
(a) 
PiV i
; Put the cylinder into a refrigerator at absolute temperature T/2. Let the
2
piston move freely as the gas cools; (b) +1.39PiVi; With the gas in a constant-
temperature bath at Tf , slowly push the piston in; (c) W = 0; Lock the piston in
place and hold the cylinder over at hotplate at 3Tf .
P17.78
(a) 800 J/kg ∙ °C; (b) This differs from the tabulated value by 11%, so the values
agree within 15%.
P17.80
(a) 15.0 mg; (b) −5.00 J; (c) 5.00 J; (d) 15.0 mg; (e) 5.00 J; (f) ∆Eint = 0;
(g) 4.04 × 10−3 °C; (h) Q = 0; ∆Emech = −5.00 J; (i) Q = 0; ∆Eint = 2.50 J
P17.82
10.2 h
P17.84
(a) See P17.84(a) for full explanation; (b) See P17.84(b) for full explanation; (c) The
expressions are equal because PV = nRT and   CV  R / CV  1  R/ CV give
R    1CV , so PV  n   1CV T and PV /   1  nCV T .
P17.86

2W
2 W 
(a) Ti 
; (b) Pi  1 
3 nR
3 nRTi 

P17.88
(a) 0.645 kg/h; (b) 0.068 9
5/ 2
Chapter 18
ANSWERS TO EVEN-NUMBERED PROBLEMS
P18.2
(a) 29.4 L/h; (b) 185 hp; (c) 527 N · m; (d) 1.91 × 105 W
P18.4
13.7°C
P18.6
 0.5Th  383 
Qc
 1.40 
where Qc / t
t
Th  383 

(a)
is in megawatts and T is in kelvins; (b)
the exhaust power decreases as the firebox temperature increases; (c) 1.87 MW; (d)
3.84 × 103 K; (e) No answer exists. The energy exhaust cannot be that small.
P18.8
(a) e1 + e2 – e1e2; (b) 1 
Tc
; (c) The combination of reversible engines is itself a
Th
reversible engine so it has the Carnot efficiency. No improvement in net efficiency
1
has resulted; (d) Ti  Th  Tc  ;
2
1/ 2
(e) Ti  ThTc 
P18.10
 Th 
 Tc 
(a) Pt 
; (b) Pt 

 Th  Tc 
 Th  Tc 
P18.12
The efficiency of a Carnot engine operating between these temperatures is 6.83%.
Therefore, there is no way that the inventor’s engine can have an efficiency of 0.110
= 11.0%.
P18.14
(a) 5.12%; (b) 5.27 TJ/h; (c) As fossil-fuel prices rise, this way to use solar energy
will become a good buy.
P18.16
(a) See Table P18.16(a); (b) See Table P18.16(b); (c) 23.7%; (d) 23.7%
P18.18
77.8 W
P18.20
(a) 2.93; (b) (COP)refrigerator; (c) with EER 5, $510, with EER 10, $255; Thus, the cost for
air conditioning is half as much for an air conditioner with EER 10 compared with
an air conditioner with EER 5.
P18.22
(a) See P18.22(a) for the full solution; (b) See P18.22(b) for the full solution.
P18.24
(a) 7.69 × 108 J; (b) 5.67 × 108 J
P18.26
72.2 J
P18.28
−610 J/K
P18.30
244 J/K
P18.32
(a) See Table P18.32; (b) 2 heads and 2 tails; (c) either all heads or all tails; (d) 2
heads and 2 tails
P18.34
(a) S h  
P18.36
3.28 J/K
P18.38
717 J/K
P18.40
1 W/K
P18.42
9 × 1014 W
P18.44
(a) 214 J and 64.3 J; (b) 35.7 J and 35.7 J. The total energy the firebox puts out equals
to the total energy transferred to the environment; (c) The net flow of energy by
heat from the cold to the hot reservoir without work input is possible; (d) Weng S =
Q
Q
 1 1
; (b) Sc   ; (c) Q   
Th
Tc
 Tc Th 
233 J, Q h, S = 333 J; (e) 83.3 J; (f) 83.3 J; (g) 0; (h) The output of 83.8 J of energy from
the heat engine by work in a cyclic process without any exhaust by heat is
impossible; (i) −0.111 J/K; (j) A decrease in total entropy is impossible.
P18.46
The computed change in entropy is 4.88 J/K, which is not zero. While the
statements about energy transfer by heat are true, the mixing process is
irreversible. After the water has come to equilibrium, it will not spontaneously
separate again into warm and cool water. Therefore, there is an entropy increase of
the mixture during the irreversible process.
P18.48
See P18.48 for the full derivation
P18.50
(a) 0.476 J/K; (b) 417 J; (c) See P18.50(c) for the full derivation
P18.52
See P18.52 for the full derivation
P18.54
PTc
Th  Tc cT
P18.56
(a) 13.4 J/K; (b) 310 K; (c) 11.1 J/K; (d) smaller by less than 1%
P18.58
(a) 10.5 nRTi; (b) 8.50 nRTi; (c) 0.190; (d) 0.833; The Carnot efficiency is much higher.
P18.60
nCp ln 3
P18.62
(a) S  nCp 1n3 ; (b) The pair of processes considered here carries the gas from the
initial state in P18.60 to the final state here. Entropy is a function of state. Entropy
change does not depend on path. Therefore, the entropy change in P18.60 equals
∆Sisothermal + ∆Sadiabatic in this problem. Since ∆Sadiabatic = 0, the answers to P18.60 and
P18.62(a) must be the same.
P18.64
See P18.64 for the full derivation
Chapter 19
ANSWERS TO EVEN-NUMBERED PROBLEMS
P19.2
(a) 2.62 × 1024; (b) 2.38 electrons for every 109 already present
P19.4
57.5 N
P19.6
The unknown charge on each dust particle is about half of the smallest possible
free charge, the charge of the electron. No such free charge exists. Therefore, the
forces cannot balance.
P19.8
(a)
P19.10
(a) 0; (b) 30.0 N; (c) 21.6 N; (d) 17.3 N; (e) –13.0 N; (f) 17.3 N; (g) 17.0 N;
(h) 24.3 N at 44.5° above the +x direction
P19.12
1.57 μN to the left
P19.14
The field at the origin can be to the right, if the unknown charge is
–9Q, or the field can be to the left, if and only if the unknown charge is +27Q.
P19.16
(a)
q1
q1  q2
d ; (b) Yes, if the third bead has a positive charge.
 5.58  1011 N C ĵ
2ke sin  0
y
1.02  107
N C ĵ
2k e 
y
P19.18
(a)
P19.20
4akeq
x3
P19.22
(a) 1.29  104 ĵ N C ; (b) 3.86  102 ĵ N
P19.24
(a)
P19.26
2keQ
Q

2
3 3a
6 3 0 a2
P19.28
(a) See ANS. FIG. P19.28; (b) at the center; (c) 1.73ke
P19.30
1
(a)  ; q1 is negative, q2 is positive
3
P19.32
; (b)
; (b)
k e0
; (b) to the left
x0
q
ĵ
a2
(a) Particle under constant velocity; (b) Particle under constant acceleration; (c) the
proton moves in a parabolic path just like a projectile in a gravitational field; (d)
m p v i2 sin 2
; (e) 36.9° or 53.1°;
eE
(f) 166 ns or 221 ns
K
; (b) in the direction of motion
ed
P19.34
(a)
P19.36
355 kN  m 2 / C
P19.38
(a)
P19.40
2
2
(a) 1.36 MN  m / C ; (b) 678 kN  m / C ; (c) no
P19.42
P19.44
P19.46
Q
Q
; (b)
2 0
2 0
Q6 q
6 0
6
2
7
2
(a) 3.20  10 N  m / C ; (b) 1.92  10 N  m / C ; (c) The answer to part (a) would
change because the charge could now be at different distances from each face of the
cube. The answer to part (b) would be unchanged because the flux through the
entire closed surface depends only on the total charge inside the surface.
(a) 4.86 × 109 N/C away from the wall; (b) So long as the distance from the wall is
small compared to the width and height of the wall, the distance does not affect the
field.
P19.48
r
rad ially aw ay from the cylind er axis
2 0
P19.50
Q
Qr 3
(a)
;
(b)
; (c) See ANS. FIG. P19.50(c)
0
0 a3
P19.52
(a) +913 nC; (b) 0
P19.54
(a) 51.4 kN/C, radially outward; (b) 646 N . m2/C
P19.56
The electric field just outside the surface occurs at 16.0 kN/C. The peak in the figure
occurs at about 6.5 kN/C. Therefore, it is not possible that this figure represents the
electric field for the given situation.
P19.58
The two fields are
Q
, the same in magnitude, and both are perpendicular to
2A 0
the plates, vertically upward if Q is positive.
P19.60
P19.62
(a)  ; (b) 3 ; (c) 6ke

r
, rad ially outw ard
(a) 1.06 nC/m2, negative; (b) –542 kC; (c) 4.88 × 103 N away from Earth; (d) The
gravitational force is in the opposite direction and 4.08 × 1016 times larger. Electrical
forces are negligible in accounting for planetary motion.
2 R 2  d 2
P19.64
(a) 0; (b)
P19.66
See P19.66 for complete solution
P19.68
(a) It is possible in just one way; (b) x = –16.0 cm; (c) 51.3 C
P19.70
L
P19.72
(a)  2   1 ; (b) See P19.72(b) for complete solution
P19.74
(a) 2.18
P19.76
1
2
0
k L  Li 
ke
keq
s2
; (b) the electric field is in the k̂ direction
keQq
ma3
Ex  0 and Ey 
P19.78
a
3 0
Chapter 20
ANSWERS TO EVEN-NUMBERED PROBLEMS
P20.2
1.35 MJ
P20.4
(a) –6.00 × 10–4 J; (b) –50.0 V
P20.6
(a) isolated; (b) electric potential energy and elastic potential energy; (c)
Particle in equilibrium; (e)
2QE
; (d)
k
QE
d2 x 
kx 

; (f)
;
2
k
dt
m
m
; (h) The period does not depend on the electric field. The electric field
k
just shifts the equilibrium point for the spring, just like a gravitational field does
for an object hanging from a vertical spring.
(g) 2
keQ 2
s
P20.8
5.41
P20.10
(a) no point; (b)
P20.12
27.4 fm
5keq2

9d
P20.14
 Eint
P20.16
(a) v 1 
and v 2 
P20.18
2k eq
a
2m2 keq1q2  1
1
 

m1 m1  m2  r1  r2 d 
2m1keq1q2  1
1
  ; (b) faster than calculated in (a)

m 2 m1  m 2  r1  r2 d 
(a) a positive 7.00-nC charge is 7.00 cm from the origin with position vector 7.00 cm
at 70.0°, and a negative 8.00-nC charge is 3.00 from the origin with position vector
3.00 cm at 90.0°;


(b) 7.03î  109 ĵ  105 N ; (c) 2.40 × 10–5 J
P20.20
(a) 6.00 m; (b) –2.00µC
P20.22
(a) 0; (b)
P20.24
(a) At x = 0, V = 10.0 V, at x = 3.00 m, V = –11.0 V, at x = 6.00 m,
V = –32.0 V; (b) 7.00 N/C in the +x direction
k eQ
r2
L 
C


ke  L  d ln  1   
2 ; (b)

d 
m

P20.26
(a)
P20.28
ke   2 ln 3
P20.30
1.56 × 1012 electrons
P20.32
(a) 0, 1.67 MV; (b) 5.84 MN/C away, 1.17 MV; (c) 11.9 MN/C away, 1.67 MV
P20.34
(a) 1.00 µF; (b) 100 V
P20.36
(a) See P20.36(a) for full solution; (b) See P20.36(b) for full solution.
P20.38
2N  1    R
2
0
d
P20.40
mgd tan 
q
P20.42
(a) 17.0 μF; (b) 9.00 V; (c) Q5 = 45.0 μC, Q12 = 108 μC
P20.44
(a) 6.05 F ; (b) 83.7  C
P20.46
None of the possible combinations of the extra capacitors is
4
C , so the desired
3
capacitance cannot be achieved.
1
C 
2 p
1 2
1
Cp  CpCs and C2  Cp 
4
2
1 2
C  C p Cs
4 p
P20.48
C1 
P26.50
(a) 2C; (b) Q1  Q 3  Q 2 ; (c) V1  V2  V3 ; (d) Q3 and Q1 increase; Q2 decreases
P20.52
120  C ; (b) 40.0  C and 80.0  C
P20.54
Q2
2 0 A
P20.56
(a) C V  ; (b) V  
P20.58
R 1Q
R 2Q
keq12 ke Q  q1 
keQ 2
; (d)
;

; (c)
;
(a)
(b)
R1  R2
R1  R2
2R1
2R 2
2R
4V
V  ;
; (c) 4C
(d) Positive work is done by the agent
3
3
pulling the plates apart.
2
2
2
(e) V1 
keQ
R1  R 2
and V 2 
k eQ
; (f) 0
R1  R2
P20.60
4.47 × 103 V
P20.62
(a) Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of
plastic between them; (b) 10–6 F; (c) 102 V
P20.64
1.04 m
P20.66
(a) 369 pC; 1.20 × 10–10 F, 3.10 V; (c) –45.5 nJ
P20.68
9.79 kg
P20.70
(a)
P20.72
Because n is not an integer, this is not possible. Therefore, the energy given cannot
be possible for an allowed state of the atom.
P20.74
2keq1q2 m1  m 2 
 m  m2 
 2m1 
m1 v
; (b)
; (c)  1
v
î
;
(d)
2
 m  m  v î
m1  m 2
m 1m 2 v
 m1  m2 
1
2
 ra 
V  1 
(a) V  2ke ln   ; (b) E 
 
ln ra rb  r 
 rb 
k e q2
3am
P20.76
v
P20.78
k eQ 2
2R
P20.80
Even if the charge were to accelerate to infinity, it would only achieve a maximum
speed of 30.0 m/s, so it cannot strike the wall of your laboratory at 40.0 m/s.
P20.82
(a) See P20.82(a) for full solution; (b) Er 
2ke p cos 
k p sin 
, E  e 3
;
3
r
r
(c) Yes, the results are reasonable; (d) No, because the magnitude of the electric
field between the charges of the dipole is not infinite;
ke p 2y 2  x 2
ke py
3ke pxy
V

;
E

,
E

(e)
(f) x
32
y
52
52
x2  y2
x2  y2
x2  y2

P20.84


(a) On the lower plate the charge is 

P20.86





Q
, and on the upper plate the charge is
3
2Qd
2Q
; (b)
3 0 A
3
(a)
Q 02 d l  x 
Q 02 d
Q 02
Q 02
;
to
the
right;
;
; (e) They are precisely
(b)
(c)
(d)
2 0 l 3
2 0 l 3
2 0 l 4
2 0 l 4
the same.
Chapter 21
ANSWERS TO EVEN-NUMBERED PROBLEMS
0.632 I0 ;
(b) 0.999 95I0 ; (c) I0
P21.2
(a)
P21.4
q
2
P21.6
(a) 99.5 kA/m2; (b) The current is the same; (c) The current density is smaller; (d)
0.800 cm; (e) I = 5.00 A; (f) 2.49 × 104 A/m2
P21.8
6.43 A
P21.10
500 mA
mR
; (b)
4  m 
  m R 
1/ 4
P21.12
(a)
P21.14
1.71 
P21.16
(a) 5.58 × 10–2 kg/mol; (b) 1.41 × 105 mol/m3; (c) 8.49 × 1028
m
atoms
;
m3
(d) 1.70 × 1029 electrons/m3; (e) 2.21 × 10–4 m/s
P21.18
15.0 µW
P21.20
(a) 2.1 W; (b) 3.42 W; (c) It would not be as safe. If surrounded by thermal
insulation, it would get much hotter than a copper wire.
P21.22
See P21.22 for full explanation.
P21.24
~ $1
P21.26
(a) $1.48; (b) $0.005 34; (c) $0.381
P21.28
(a) Any diameter d and length
2
8
related by d  4.77  10 l , where d and
are
in meters; (b) Yes; for V = 0.500 cm3 of Nichrome, l  3.65 m and d = 0.418 mm.
P21.30
(a) 0.530; (b) 221 J; (c) 15.1°C
P21.32
6.53 
P21.34
(a) 4.59 ; (b) 8.16%
P21.36
(a) See P21.36(a) for the full solution; (b) The current never exceeds
50 µA.
P21.38
None of these is
4
R, so the desired resistance cannot be achieved.
3
,
2
4
2
, V 3 
, V 4 
; (c)
9
9
3
P21.40
(a) V4  V3  V1  V2 ; (b) V1 
P21.42
(a) 470 W; (b) 1.60 mm or more; (c) 2.93 mm or more
P21.44
(a) The 120-V potential difference is applied across the series combination of the
two conductors in the extension cord and the lightbulb. The potential difference
across the lightbulb is less than 120 V, and its power is less than 75 W; (b) See ANS.
FIG. P21.44; (c) 73.8 W
P21.46
(a) See ANS. FIG. P21.46; (b) 11.0 mA in the 220- resistor and out of the positive
pole of the 5.80-V battery; The current is 1.87 mA in the 150- resistor and out of
the negative pole of the 3.10-V battery; 9.13 mA in the 370- resistor
P21.48
(a) See ANS. FIG. P21.48
P21.50
(a) 13.0I1  18.0I2  30.0 ; (b) 18.0I2  5.00I3   24.0 ; (c) I1  I2  I3  0 ; (d)
I3  I1  I2 ; (e) 5.00I1  23.0I12  24.0 ; (f) I 2  0.416 A and I1 = 2.88 A; (g) I3 = 3.30 A;
V 2 
3
I
2I
; (e) I4 increases and I1, I2, and I3
I1  I4  I2  I3 ; (d) I1  I, I 2  I 3  , I 4 
3
3
3I
3I
decrease; (f) I1  , I 2  I3  0, I 4 
4
4
(h) The negative sign in the answer for I2 means that this current flows in the
opposite direction to that shown in the circuit diagram and assumed during the
solution. That is, the actual current in the middle branch of the circuit flows from
right to left and has a magnitude of 0.416 A.
P21.52
(a) 172 A downward; (b) 1.70 A downward; (c) No, the current in the dead battery
is upward in Figure P21.52, so it is not being charged. The dead battery is
providing a small amount of power to operate the starter, so it is not really "dead."
P21.54
(a) 3.91 s; (b) 782 µs
P21.56
587 k
P21.58
 1
1 t R2 C

e
(a) (R1 + R2)C; (b) R2C; (c)  

 R1 R 2
P21.60
(a) 3.00 × 108 W; (b) 1.75 × 1017 W
P21.62
(a) Lightbulb A = 576  and Lightbulb B = 144 ; (b) 4.80 s; (c) The charge is the
same. It is at a location that is lower in potential;
(d) 0.040 0 s; (e) The energy is the same. Energy enters the lightbulb by electric
transmission and leaves by heat and electromagnetic radiation; (f) $1.98
P21.64
(a) V/L in the positive x direction; (b) 4 L/ d2; (c) V  d / 4L;
V  V
(d) V/L in the positive x direction; (e)  J       E
 L  L
P21.66
(a)
P21.68
See P21.68 for full explanation.
P21.70
See P21.70 for full explanation.
P21.72
  
(a) ln 
  0.011 8 t  0.088 2 ; (b) The time constant is 84.7 s and the capacitance
V 
2
r 
2 LV

ln  b  ; (b)  
I ln rb ra 
2 L  ra 
is 8.47 µF.
P21.74
P21.76
R1  2R 2 C ln 2
7
(a) See the table in P21.76(a); (b) 9.93  10  m; (c) The average value is within
6
1% of the tabulated value of 1.00  10   m given in Table 21.1.
Chapter 22
ANSWERS TO EVEN-NUMBERED PROBLEMS
P22.2
(a) up; (b) out of the page, since the charge is negative; (c) no deflection; (d) into the
page
P22.4
48.9° or 131°
P22.6
Gravitational force: 8.93  10–30 N down, electric force: 1.60  10–17 N up, and
magnetic force: 4.80  10–17 N down.
P22.8
2.34  10–18 N
P22.10
7.88  10–12 T
P22.12
e2 B2 2
r1  r22 

2m e
P22.14
(a) 4.31  107 rad/s; (b) 5.17  107 m/s
P22.16
(a) 7.66  107 rad/s; (b) 2.68  107 m/s; (c) 3.76  106 eV; (d) 3.13  103 revolutions; (e)
2.57  10–4 s
P22.18
(a) 2 rIB sin  ; (b) up, away from magnet
P22.20
(a) 0; (b) 40.0î mN ; (c) 40.0k̂ mN ; (d) 40.0î  40.0k̂ mN ; (e) The forces on the


four segments must add to zero, so the force on the fourth segment must be the
negative of the resultant of the forces on the other three.
P22.22
P22.24
P22.26
See P22.22 for full explanation.
r r
U    B
(a) +x direction; (b) torque is in the –z direction; (c) –x direction;
(d) torque is in the +z direction; (e) No; (f) Both the forces and the torques are equal
in magnitude and opposite in direction, so they sum to zero and cannot affect the
motion of the loop; (g) in the yz plane at 130° counterclockwise from the +y axis; (h)
2
the +x direction; (i) zero; (j) counterclockwise; (k) 0.135 A  m ; (l) 130°; (m)
0.155 N  m
P22.28
P22.30
4
(a) 6.40  10 N  m; (b) 0.241 W; (c) 2.56  10–3 J; (d) 0.154 W
0 I
into the paper
4 x
P22.32

1  I
B  1   0 d irected into the page 
   2R
P22.34
(a) 2I1 out of the paper; (b) 6I1 into the paper
P22.36
See ANS. FIG. P22.36
P22.38
12.5 T
P22.40
20.0 μT
P22.42
(a) The situation is possible in just one way; (b) 12.0 cm to the left of wire 1; (c) 2.40
A down
P22.44
 0 I1 I 2  a 

 to the left
2   c c  a
P22.46
This is the required center-to-center separation distance of the wires, but the wires
cannot be this close together. Their minimum possible center-to-center separation
distance occurs if the wires are touching, but this value is 2r = 2(25.0 μm) = 50.0 μm,
which is much larger than the required value above. We could try to obtain this
force between wires of smaller diameter, but these wires would have higher
resistance and less surface area for radiating energy. It is likely that the wires
would melt very shortly after the current begins.
P22.48
(a) 6.34  10–3 N/m; (b) inward toward the center of the bundle;
(c) greatest at the outer surface
P22.50
500 A
P22.52
(a) zero; (b) B 
P22.54
20.0 T toward the bottom of the page
0 IN 

2l 

0 I
0 I 2
tangent to the w all; (c)
inward
2 R
2 R 2
xl


 ; (b) See P22.56(b) for full explanation.
x 2  a2 
x
P22.56
(a)
P22.58
207 W
P22.60
(a) 9.27  10–24 A . m2; (b) downward
P22.62
(a) 750 J; (b) 6.40 g
P22.64
 0 Js
2
P22.66
(a) See P22.66(a) for full explanation; (b)
x  l 2  a2
IB
qV H t
P22.68
 k mg
Id
P22.70
(a) See P22.70(a) for full explanation; (b) See P22.70(b) for full explanation
P22.72
(a) The electric current experiences a magnetic force; (b) JLB; (c) Charge moves
within the fluid inside the length L, but charge does not accumulate: the fluid is not
charged after it leaves the pump; (d) It is not current-carrying; (e) It is not
magnetized.
P22.74
This current would instantly vaporize any wire of reasonable size. For example, if
we imagine a 1.00-m segment of copper wire 10 cm in diameter, a huge wire, this
current delivers over a terawatt of power to this short segment! Furthermore, the
power delivered to such a wire wrapped around the Earth is on the order of 1020
W, which is larger than all of the solar power delivered to the Earth by the Sun.
P22.76

4
(a) 2.74  10–4 T; (b) 2.74  10 T  ĵ ; (c) Under the assumption that the rails are
infinitely long, the length of rail to the left of the bar does not depend on the
location of the bar; (d) 1.15  10–3 N; (e) +x direction; (f) Yes, length of the bar,
current, and field are constant, so force is constant; (g) 0.999 m s  î
P22.78
P22.80
0 q
2.5 5 R
The particle will not exit the field at the top but rather will complete a semicircle in
the magnetic field region and will exit at the bottom, traveling in the opposite
direction with the same speed.
Chapter 23
ANSWERS TO EVEN-NUMBERED PROBLEMS
P23.2
(a) Each coil has a pulse of voltage tending to produce counterclockwise current as
the projectile approaches, and then a pulse of clockwise voltage as the projectile
recedes; (b) 625 m/s
P23.4
6.03 × 10–12 V
0 n r22 I
; (b)
02 n r22 I
P23.6
(a)
P23.8
(a) See P23.8(a) for full explanation; (b) The emf induced in the coil is proportional
to the line integral of the magnetic field around the circular axis of the toroid.
Ampère’s law says that this line integral depends only on the amount of current
the coil encloses.
P23.10
2R
t
  68.2e1.60t ,
4r1R
t
; (c) left
where t is in seconds and
 is in mV
P23.12
(a) 0.013 3 A; (b) The current is counterclockwise in the lower loop and clockwise in
the upper loop.
P23.14
2.83 mV
P23.16
1.00 m/s
P23.18
Rmv
B2 l 2
P23.20
P23.22
The speed of the car is equivalent to about 640 km/h or 400 mi/h, much faster than
the car could drive on the curvy road and much faster than any standard
automobile could drive in general.
N 2 B2 w 2 v
N 2 B2 w 2 v
to
the
left;
to the left again
(a)
(b) 0; (c)
R
R
P23.24
3.92 m/s is the highest speed the loop can have while the upper edge is above the
field, so it cannot possibly be moving at 4.00 m/s.
P23.26
(a) to the right; (b) out of the page; (c) to the right
P23.28
(a) 2.81 × 10–3 N/C; (b) tangent to the electric field line passing through at point P2
and counterclockwise
P23.30
1.36 μH
P23.32
See P23.32 for full explanation.
P23.34
19.2 T  m 2
P23.36
9.77 mm
P23.38
See P23.38 for full explanation.
P23.40
(a) 0.800; (b) 0
P23.42
7.67 mH
P23.44
(a)
P23.46
(a) 1.00 A; (b) V12  12.0 V, V1 200  1.20 kV, VL  1.21 kV
P23.48
(a) 8.06 MJ; (b) 6.32 kJ
P23.50
(a) 66.0 W; (b) 45.0 W; (c) 21.0 W;
P23.52
(a) 101 μV tending to produce clockwise current; (b) twice as large in magnitude
and in the opposite sense.
P23.54
(a) 1.43 x 105 A; (b) 82.4 T/s; (c) 86.3 cm
P23.56
(a)
P23.58
(a) amplitude doubles and period is unchanged; (b) doubles the amplitude and
cuts the period in half; (c) amplitude unchanged and period is cut in half
P23.60

1  e
5R
 5Rt 2L

; (b) 10R
6  e
 5Rt 2 L

Bl v
; (b) particle in equilibrium; (c) 281 m/s; (d) 1.88 A; (e) 169 W; (f) 169 W; (g)
R
yes; (h) increase; (i) yes; (j) larger
(a) I1 = I2 + I; (b)   I1R1  I2 R 2  0; (c)   I1R 1  L
(d)    IR   L
dI
 0;
dt
dI
 0 ; (e) See P23.60(e) for full explanation.
dt
P23.62
(a) 2.54 × 105 m/s; (b) 215 V
P23.64
(a) See P23.64(a) for full description; (b) See P23.64(b) for full description.
P23.66
(a) 0.125 V clockwise; (b) 0.020 0 A clockwise
P23.68
0 N 2 h  b 
ln  
 a
2
P23.70
3.97  1025 
P23.72
v
P23.74
L
2R

1  e
Bd
 B2 d 2 t mR

Chapter 25
ANSWERS TO EVEN-NUMBERED PROBLEMS
P25.2
(a) See P25.2(a) for full explanation; (b) See P25.2(b) for full explanation.
P25.4
(a) 78.3°; (b) 2.56 m; (c) 9.72°; (d) 442 nm; (e) The light wave slows down as it
moves from air to water, but the sound wave speeds up by a larger factor. The light
wave bends toward the normal and its wavelength shortens, but the sound wave
bends away from the normal and its wavelength increases.
P25.6
(a) See ANS. FIG. P25.6; (b) 42.0°; (c) 63.1°; (d) 26.9°; (e) 107 m
P25.8
19.5° above the horizon
P25.10
~10–11 s, ~10–3 wavelengths
P25.12
(a) 1.52; (b) 417 nm; (c) 4.74 × 1014 Hz; (d) 198 Mm/s
P25.14
158 Mm/s
P25.16
 2  19.5 ; 3  19.5 ;  4  30.0
P25.18
  180  2
P25.20
(a)  = 30.4°; (b)  = 22.3°
P25.22
(a) 0.387 cm; (b) 106 ps
P25.24





 sin    


1 
1  sin   
sin 1 nV sin    sin 1 

sin
n
sin


sin




R
 n   
 nV   
R









P25.26
(a) See ANS. FIG. P25.26(a); (b) As the waves move to shallower water, the wave
fronts slow down, and those closer to shore slow down more. The rays tend to
bend toward the normal of the contour lines; or equivalently, the wave fronts bend
to become more nearly parallel to the contour lines; (c) See ANS. FIG. P25.26(c); (d)
We suppose that the headlands are steep underwater, as they are above water. The
rays are everywhere perpendicular to the wave fronts of the incoming refracting
waves. As shown, because the rays tend to bend toward the normal of the contour
lines, the rays bend toward the headlands and deliver more energy per length at
the headlands.
P25.28
(a) 10.7°; (b) air; (c) Sound in air falling on the wall from directions is 100%
reflected.
P25.30




 1.00  
  sin 1
 n   ; 1
1  sin 1  n sin    sin 1 
n
2
 1 sin   cos 

P25.32
(a) See P25.32(a) for full explanation; (b) n  1.41 and n  2.12
P25.34
The beam will exit after making 81 reflections, so it does not make 85 reflections.
P25.36
(a)
P25.38
See P25.38 for full explanation.
P25.40
(a) angle of incidence: 30.0°, angle of refraction: 18.8°; (b) angle of incidence: 30.0°,
angle of refraction: 50.8°; (c) and (d) See Table 25.40.
P25.42
(a)
P25.44
(a) 45.0°; (b) Yes. The ray will exit if it enters at an angle  that satisfies the
1
condition tan   , where n = 1, 2, 3, …
n
P25.46
(a) 0.172 mm/s; (b) 0.345 mm/s; (c) and (d) northward and downward at 50.0°
P25.48
h
n2  1

; (b) 4.73 cm; (c) For n = 1, h = 0. For n = 2, h = ∞. For
(a)
d
4  n2
n > 2, h has no real solution.
P25.50
(a)
P25.52
The person and lightbulb are separated by a distance d, and the light travels at a
distance 3d. This gives a maximum ratio of 3.00, and we see that a ratio of 3.10 is
impossible.
P25.54
(a) Total internal reflection occurs for all values of , or the maximum angle is 90°;
(b) 30.3°; (c) Total internal reflection never occurs as the light moves from lowerindex polystyrene to higher-index carbon disulfide.
P25.56
(a) See ANS. FIG. P25.56; (b) The straightness of the graph line demonstrates Snell’s
proportionality of the sine of the angle of refraction to the sine of the angle of
incidence; (c) 1.328  0.8%
P25.60
No light from above the water will approach the scuba diver’s eyes from 48.8°
found in Example 25.4.
P25.62
(a) The optical day is longer; (b) 164 s
nd
; (b) Yes; for very small d, the light strikes the interface at very large angles
n1
of incidence; (c) Yes: as n increases, the critical angle becomes smaller; (d) 350 μm
h  n  1.00 
 n  1.00 
 times larger

 ; (b) 
2 
c
2
4n
; (b) 68.5%
 n  12
Chapter 26
ANSWERS TO EVEN-NUMBERED PROBLEMS
P26.2
(a) Younger; (b) ~10–9 s
P26.4
4.58 m
P26.6
See ANS. FIG. P26.6 for the locations of the five images.
P26.8
(a) +2.22 cm; (b) +10.0
P26.10
A convex mirror diverges light rays incident upon it, so the mirror in this problem
cannot focus the Sun’s rays to a point.
P26.12
(a) See P26.12(a) for full explanation; (b) real image at 0.639 s and virtual image at
0.782 s
P26.14
(a)
P26.16
(a) q = 45.0 cm, M = –0.500; (b) q = –60.0 cm, M = 3.00; (c) The image (a) is real,
inverted, and diminished. That of (b) is virtual, upright, and enlarged. See ANS.
FIG. P26.16 for the ray diagrams.
P26.18
(a) –26.7 cm; (b) upright; (c) 0.0267
P26.20
(a) 0.708 m in front of the sphere; (b) upright
P26.22
(a) 25.6 m; (b) 0.058 7 rad; (c) 2.51 m; (d) 0.023 9 rad; (e) 62.8 m
P26.24
38.2 cm below the top surface
P26.26
The image is inverted, real, and diminished.
P26.28
1.50 cm/s
P26.30
(a) 12.3 cm to the left of the lens; (b) 0.615; (c) See ANS. FIG. P26.30.
P26.32
(a) q = 40.0 cm, M = –1.00, real, inverted; (b) q = infinity, no image;
(c) q = –20.0 cm; M = 2.00, upright, virtual
P26.34
dq  
P26.36
(a) 5.36 cm; (b) –18.8 cm; (c) See ANS. FIG. P26.36(c). The image is enlarged,
upright, and virtual; (d) A magnifying glass with focal length 7.50 cm is used to
form an image of a stamp, enlarged 3.50 times. Find the object distance. Locate and
describe the image.
P26.38
(a) qa = 26.3 cm, qd = 46.7 cm, –8.75 cm, –23.3 cm; (b) See ANS. FIG. P26.38(b); (c) See
ad
ad
; (b) 2
a 1
a 1
2
q2
dp
p2
P26.38(c) for full explanation; (d) The integral stated adds up the areas of ribbons
covering the whole image, each with vertical dimension |h| and horizontal width
dq; (e) 328 cm2.
P26.40
P26.42
See P26.40 for full explanation.
d
d2
 fd ; (b) The smaller solution for p gives a larger value for q, with a
(a) 
2
4
real, enlarged, inverted image. The larger solution for p describes a real,
diminished, inverted image.
P26.44
(a) –4.00 diopters; (b) diverging lens
P26.46
(a) Yes, a single lens can correct the patient’s vision; (b) f = 56.3 cm,
P = +1.78 diopters; (c) –1.18 diopters
P26.48
(a) –25.0 cm; (b) nearsighted; (c) –3.70 diopters
P26.50
(a) 50.8 diopters  P  60.0 diopters; (b) –0.800 diopters, diverging
P26.52
(a) bottom half; (b) 1.50 mm; (c) 48.2°; (d) 96.4°;
P26.54
(a)
1 1
1
1
1
1



; (b) 
; (c) 0.300 m;
f p1 1.50  p1
f p1  0.900 0.600  p1
(d) 0.240 m
P26.56
f 
Md
Md
when the lens is converging
2 when the lens is diverging; f 
1  M 
M  12
P26.58
See P26.58 for full explanation
P26.60
See P26.60 for full explanation
P26.62
p  f1  f 2  f1 f 2
p  f1
P26.64
(a) See P26.64(a) for full explanation; (b) 1.75 times
P26.66
(a) 67.5 cm; (b) The lenses can be displaced in two ways. The first lens can be
moved 1.28 cm farther from the object and the second lens 17.7 cm toward the
object. Alternatively, the first lens can be moved 0.927 cm toward the object and the
second lens 4.44 cm toward the object.
P26.68
(a) The image is real, inverted, and actual size; (b) Light travels the same path
regardless of direction, so light shined on the image is directed to the actual object
inside, and the light then reflects and is directed back to the outside. Light directed
into the hole in the upper mirror reflects as shown in the lower figure, to behave as
if it were reflecting from the image.
P26.70
(a) 20.0 cm; (b) –6.00; (c) inverted; (d) q2 = 6.67 cm and Moverall = –2.00, inverted
P26.72
(a) 1.99; (b) 10.0 cm to the left of the lens; (c) –2.50; (d) inverted
Chapter 30
ANSWERS TO EVEN-NUMBERED PROBLEMS
P30.2
~1028 protons; (b) 1028 neutrons; (c) ~1028 electrons
P30.4
r2 
P30.6
(a) 5.18 fm; (b)  is much less than the distance of closest approach
P30.8
6.1 × 1015 N toward each other
P30.10
~200 MeV
P30.12
greater for
P30.14
0.200 mCi
P30.16
P30.18
P30.20
m2
r
m1 1
2
ln 2
R0T1 2
15
7
t1 T1 2
N by 3.54 MeV
2
t 2 T1 2

5 1
13
(a) 1.55  10 s ; (b) 12.4 h; (c) 2.39  10 atoms; (d) 1.88 mCi
(a) See P30.20(a) for full explanation; (b) See P30.20(b) for full explanation; (c) See
ANS. FIG. P30.20(c); (d) 10.9 min;
ln 1 2 
(e) t m 
; (f) 10.9 min
1  2
TY ln 2
P30.22
P30.24
ln  2 r1 r2  Y

T
t


21
9
12 1
(a) 1.05  10 ; (b) 1.37  10 ; (c) 3.83  10 s ;
3
3
(d) 3.17  10 d ecays w eek ; (e) 951 decays/week; (f) 9.95  10 yr
P30.26
65

211
(a) 28 Ni ; (b) 82 Pb; (c)
P30.28
(a) See P30.28(a) for full explanation; (b) 18.6 keV
P30.30
(a) 173 MeV; (b) 0.078 8%
P30.32
(a)
P30.34
31
8
(a) 2.53  10 J; (b) 5.34  10 yr
P30.36
3.60 × 1038 protons/s
P30.38
57
12
(a) 4.28  10 J; (b) 1.19  10 atoms; (c) 105 billion years; (d) The time interval in
197
79
Au  01 n 
198
79
55
27
Co
Au * 
198
80
Hg 
0
1
e   ; (b) 7.89 MeV
(c) is an order of magnitude larger than the expected remaining lifetime of the Sun.
Only the hydrogen in a relatively small core is available as a nuclear fuel. Only in
the core are temperatures and densities high enough for the fusion reaction to be
self-sustaining.
P30.40
(a) See P30.40(a) for full explanation; (b) 26.7 MeV
P30.42
(a) See P30.42(a) for full explanation; (b) 1.16 u
P30.44
(a) See P30.44(a) for full explanation; (b) 1.53 MeV
P30.46
(a) The process cannot occur because energy input would be required; (b) Required
energy can come from the electrostatic repulsion; (c) 1.20 MeV
P30.48
3
1
(a) See ANS. FIG. P30.48; (b) 4.17  10 min , 2.77 h; (c) 4.02  103 counts min ; (d)
9.65  106 atoms
P30.50
(a) See P30.50(a) for full explanation; (b) 1.95 × 10–3 eV
P30.52
(a) 3.91  109 yr ; (b) no older
P30.54
(a)
P30.56
The nucleus 57Co cannot decay by e+ emission because the Q value is
–0.187 MeV.
P30.58
(a) 2.24 × 107 kWh; (b) 17.6 MeV for each D-T fusion; (c) 2.34 × 108 kWh; (d) 9.36
kWh; (e) Coal is cheap at this moment in human history. We hope that safety and
waste disposal problems can be solved so that nuclear energy can be affordable
before scarcity drives up the price of fossil fuels. Burning coal in the open puts
carbon dioxide into the atmosphere, worsening global warming. Plutonium is a
very dangerous material to have sitting around.
P30.60
(a) See P30.60(a) for full explanation; (b)
P30.62
(a) 2.65 × 109 J; (b) 78.0 times larger
P30.64
ln 2 

; (b) 
ln f

R

mN A E
M U-235  cw 100  Tc   Lv  cs Th – 100
P30.66
(a) 0.155 m; (b) 51.7 MeV; (c) P = QR; (d) 2.27 × 105 J/yr; (e) 3.18 J/yr
P30.68
While electric charge is conserved (5 + 2 = 1 + 6), the number of nucleons is not (10 +
4  1 + 12). Therefore, this reaction cannot occur.
P30.70
(a) 103 Pa; (b) 6 × 109 m3; (c) 6 × 1012 J; (d) ~1014 J; (e) ~ 104 ton TNT