Instructions Manual to Serway and Jewett's Physics for Scientists

1307 Pages·2005·24.69 MB·English

INSTRUCTOR'S SOLUTIONS MANUAL
FOR
SERWAY AND JEWETT'S
PHYSICS
FOR SCIENTISTS AND ENGINEERS
SIXTH EDITION
Ralph V. McGrew
Broome Community College
James A. Currie
Weston High School
Australia • Canada • Mexico • Singapore • Spain • United Kingdom • United States
1
Physics and Measurement
CHAPTER OUTLINE ANSWERS TO QUESTIONS
1.1 Standards of Length, Mass,
and Time Q1.1 Atomic clocks are based on electromagnetic waves which atoms
1.2 Matter and Model-Building
emit. Also, pulsars are highly regular astronomical clocks.
1.3 Density and Atomic Mass
1.4 Dimensional Analysis
1.5 Conversion of Units Q1.2 Density varies with temperature and pressure. It would be
1.6 Estimates and Order-of- necessary to measure both mass and volume very accurately in
Magnitude Calculations
order to use the density of water as a standard.
1.7 Significant Figures

Q1.3 People have different size hands. Defining the unit precisely
would be cumbersome.

Q1.4 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms

Q1.5 (b) and (d). You cannot add or subtract quantities of different
dimension.

Q1.6 A dimensionally correct equation need not be true. Example:
1 chimpanzee = 2 chimpanzee is dimensionally correct. If an
equation is not dimensionally correct, it cannot be correct.
Q1.7 If I were a runner, I might walk or run 101 miles per day. Since I am a college professor, I walk about
100 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on
vacation.
Q1.8 On February 7, 2001, I am 55 years and 39 days old.
F I F I
G365.25 dJ G86400 sJ
55 yrH K+39 d=20128 dH K =1.74×109 s~109 s.
1 yr 1 d
Many college students are just approaching 1 Gs.
Q1.9 Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10.
Q1.10 The mass of the forty-six chapter textbook is on the order of 100 kg.
Q1.11 With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.
1
2 Physics and Measurement
SOLUTIONS TO PROBLEMS
Section 1.1 Standards of Length, Mass, and Time
No problems in this section
Section 1.2 Matter and Model-Building
P1.1 From the figure, we may see that the spacing between diagonal planes is half the distance between
diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the
Pythagorean theorem, L = L2 +L2 . Thus, since the atoms are separated by a distance
diag
1
L=0.200 nm, the diagonal planes are separated by L2 +L2 = 0.141 nm .
2
Section 1.3 Density and Atomic Mass
4 4 e j3
*P1.2 Modeling the Earth as a sphere, we find its volume as πr3 = π6.37×106 m =1.08×1021 m3. Its
3 3
m 5.98×1024 kg
density is then ρ= = = 5.52×103 kg m3 . This value is intermediate between the
V 1.08×1021 m3
tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to
3000 kg m3. The average density of the Earth is significantly higher, so higher-density material
must be down below the surface.
a fb g e j m
P1.3 With V= base area height V= πr2 h and ρ= , we have
V
F I
m 1 kg G109 mm3J
ρ= = a f a fH K
πr2h π19.5 mm 2 39.0 mm 1 m3
ρ= 2.15×104 kg m3 .
m
*P1.4 Let V represent the volume of the model, the same in ρ= for both. Then ρ =9.35 kg V and
iron
V
F I
mgold ρgold mgold G19.3×103 kg/m3 J
ρ = . Next, = and m =9.35 kgH K = 23.0 kg .
gold V ρ 9.35 kg gold 7.86×103 kg /m3
iron
4 e j
P1.5 V=V −V = πr3 −r3
o i 2 1
3
ρ= m, so m=ρV=ρFHG4πIKJer3 −r3j= 4πρer23 −r13j .
2 1
V 3 3
Chapter 1 3
4 4
P1.6 For either sphere the volume is V= πr3 and the mass is m=ρV=ρ πr3. We divide this equation
3 3
for the larger sphere by the same equation for the smaller:
m ρ4πr33 r3
(cid:65) = (cid:65) = (cid:65) =5.
m ρ4πr33 r3
s s s
a f
Then r =r 3 5 =4.50 cm1.71 = 7.69 cm .
(cid:65) s
P1.7 Use 1 u=1.66×10−24 g.
F I
(a) For He, m0 =4.00 uHG1.66×10-24 gKJ = 6.64×10−24 g .
1 u
F I
(b) For Fe, m0 =55.9 uHG1.66×10-24 gKJ = 9.29×10−23 g .
1 u
F I
(c) For Pb, m0 =207 uHG1.66×10−24 gKJ = 3.44×10−22 g .
1 u
*P1.8 (a) The mass of any sample is the number of atoms in the sample times the mass m of one
0
atom: m=Nm . The first assertion is that the mass of one aluminum atom is
0
m =27.0 u=27.0 u×1.66×10−27 kg 1 u=4.48×10−26 kg.
0
Then the mass of 6.02×1023 atoms is
m=Nm =6.02×1023 ×4.48×10−26 kg=0.0270 kg=27.0 g.
0
Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be
written m=Nm .
0
0.027 kg
0.0270 kg=6.02×1023m , so m = =4.48×10−26 kg,
0 0 6.02×1023
in agreement with the first assertion.
(b) The general equation m=Nm applied to one mole of any substance gives M g=NM u,
0
where M is the numerical value of the atomic mass. It divides out exactly for all substances,
giving 1.0000000×10−3 kg=N1.660540 2×10−27 kg. With eight-digit data, we can be quite
sure of the result to seven digits. For one mole the number of atoms is
F I
G 1 J
N=H K10−3+27 = 6.022137×1023 .
1.6605402
(c) The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one
b g
molecule of H O is 2 1.0080 +15.999 u=18.0 u. Then the molar mass is 18.0 g .
2
b g
(d) For CO we have 12.011 g+2 15.999 g = 44.0 g as the mass of one mole.
2
4 Physics and Measurement
F I
P1.9 Mass of gold abraded: ∆m =3.80 g−3.35 g=0.45 g=b0.45 ggHG 1 kg KJ =4.5×10−4 kg.
103 g
F I
Each atom has mass m =197 u=197 uHG1.66×10−27 kgKJ =3.27×10−25 kg.
0
1 u
Now, ∆m = ∆N m , and the number of atoms missing is
0
∆m 4.5×10−4 kg
∆N = = =1.38×1021 atoms.
m 3.27×10−25 kg
0
The rate of loss is
F IF IF IF I
∆N 1.38×1021 atomsG 1 yr JG 1 d JG 1 h JG1 minJ
= H KH KH KH K
∆t 50 yr 365.25 d 24 h 60 min 60 s
∆N
= 8.72×1011 atoms s .
∆t
P1.10 (a) m=ρL3 =e7.86 g cm3je5.00×10−6 cmj3 = 9.83×10−16 g =9.83×10−19 kg
(b) N= m = e9.83×10−19 kg j= 1.06×107 atoms
m0 55.9 u1.66×10−27 kg 1 u
P1.11 (a) The cross-sectional area is
a fa f a fa f
A=2 0.150 m 0.010 m + 0.340 m 0.010 m
.
=6.40×10−3 m2.
The volume of the beam is
e ja f
V=AL= 6.40×10−3 m2 1.50 m =9.60×10−3 m3.
Thus, its mass is
FIG. P1.11
e je j
m=ρV= 7.56×103 kg/m3 9.60×10−3 m3 = 72.6 kg .
F I
(b) The mass of one typical atom is m =a55.9 ufHG1.66×10−27 kgKJ =9.28×10−26 kg. Now
0
1 u
m 72.6 kg
m=Nm and the number of atoms is N= = = 7.82×1026 atoms .
0 m 9.28×10−26 kg
0
Chapter 1 5
F I
P1.12 (a) The mass of one molecule is m =18.0 uHG1.66×10−27 kgKJ =2.99×10−26 kg. The number of
0
1 u
molecules in the pail is
m 1.20 kg
Npail = m = 2.99×10−26 kg = 4.02×1025 molecules .
0
(b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere.
F I F I
N =N HG mpail KJ =(4.02×1025 molecules)HG 1.20 kg KJ,
both pail M 1.32×1021 kg
total
or
Nboth = 3.65×104 molecules .
Section 1.4 Dimensional Analysis
P1.13 The term x has dimensions of L, a has dimensions of LT−2, and t has dimensions of T. Therefore, the
equation x=kamtn has dimensions of
L=eLT−2jmaTfn or L1T0 =LmTn−2m.
The powers of L and T must be the same on each side of the equation. Therefore,
L1 =Lm and m=1 .
Likewise, equating terms in T, we see that n−2m must equal 0. Thus, n=2 . The value of k, a
dimensionless constant, cannot be obtained by dimensional analysis .
*P1.14 (a) Circumference has dimensions of L.
(b) Volume has dimensions of L3.
(c) Area has dimensions of L2.
e j1/2
Expression (i) has dimension L L2 =L2, so this must be area (c).
Expression (ii) has dimension L, so it is (a).
e j
Expression (iii) has dimension L L2 =L3, so it is (b). Thus, (a)=ii; (b)=iii, (c)=i .
6 Physics and Measurement
P1.15 (a) This is incorrect since the units of ax are m2 s2, while the units of v are m s.
a f
(b) This is correct since the units of y are m, and cos kx is dimensionless if k is in m−1.
∑F ∑F
*P1.16 (a) a∝ or a=k represents the proportionality of acceleration to resultant force and
m m
the inverse proportionality of acceleration to mass. If k has no dimensions, we have
F L F M⋅L
a = k , =1 , F = .
m T2 M T2
M⋅L kg⋅m
(b) In units, = , so 1 newton=1 kg⋅m s2 .
T2 s2
P1.17 Inserting the proper units for everything except G,
L O 2
Mkg mP G kg
N Q= .
s2 m 2
2 2 m3
Multiply both sides by m and divide by kg ; the units of G are .
kg⋅s2
Section 1.5 Conversion of Units
a fa f
*P1.18 Each of the four walls has area 8.00 ft 12.0 ft =96.0 ft2. Together, they have area
F I
e jG 1 m J2
4 96.0 ft2 H K = 35.7 m2 .
3.28 ft
P1.19 Apply the following conversion factors:
1 in=2.54 cm, 1 d=86400 s, 100 cm=1 m, and 109 nm=1 m
FG 1 IJb2.54 cm inge10−2 m cmje109nm mj
H in dayK = 9.19 nm s .
32 86400 s day
This means the proteins are assembled at a rate of many layers of atoms each second!
F I
G0.0254 mJ3
*P1.20 8.50 in3 =8.50 in3H K = 1.39×10−4 m3
1 in
Chapter 1 7
P1.21 Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should
a fa f
expect the area to be about A≈ 30 m 50 m =1500 m2.
Categorize: We model the lot as a perfect rectangle to use Area = Length × Width. Use the
conversion: 1 m=3.281 ft.
F I F I
a f G 1 m Ja fG 1 m J
Analyze: A=LW = 100 ft H K 150 ft H K= 1390 m2 = 1.39×103 m2 .
3.281 ft 3.281 ft
Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper
units of m2. Unit conversion is a common technique that is applied to many problems.
a fa fa f
P1.22 (a) V= 40.0 m 20.0 m 12.0 m =9.60×103 m3
b g
V=9.60×103 m3 3.28 ft 1 m 3 = 3.39×105 ft3
(b) The mass of the air is
e je j
m=ρ V= 1.20 kg m3 9.60×103 m3 =1.15×104 kg.
air
The student must look up weight in the index to find
e je j
F =mg= 1.15×104 kg 9.80 m s2 =1.13×105 N.
g
Converting to pounds,
e jb g
Fg = 1.13×105 N 1 lb 4.45 N = 2.54×104 lb .
P1.23 (a) Seven minutes is 420 seconds, so the rate is
30.0 gal
r= = 7.14×10−2 gal s .
420 s
(b) Converting gallons first to liters, then to m3,
F IF I
r=e7.14×10−2 gal sjHG3.786 LKJHG10−3 m3KJ
1 gal 1 L
r= 2.70×10−4 m3 s .
(c) At that rate, to fill a 1-m3 tank would take
F IF I
G 1 m3 JG 1 h J
t=H KH K = 1.03 h .
2.70×10−4 m3 s 3600
8 Physics and Measurement
F I
G1.609 kmJ
*P1.24 (a) Length of Mammoth Cave =348 miH K = 560 km=5.60×105 m=5.60×107 cm .
1 mi
F I
G0.3048 mJ
(b) Height of Ribbon Falls =1612 ftH K = 491 m=0.491 km=4.91×104 cm .
1 ft
F I
G0.3048 mJ
(c) Height of Denali =20320 ftH K = 6.19 km=6.19×103 m=6.19×105 cm .
1 ft
F I
G0.3048 mJ
(d) Depth of King’s Canyon =8200 ftH K = 2.50 km=2.50×103 m=2.50×105 cm .
1 ft
P1.25 From Table 1.5, the density of lead is 1.13×104 kg m3, so we should expect our calculated value to
be close to this number. This density value tells us that lead is about 11 times denser than water,
which agrees with our experience that lead sinks.
m
Density is defined as mass per volume, in ρ= . We must convert to SI units in the calculation.
V
F IF I
23.94 g G 1 kg JG100 cmJ3
ρ= H KH K = 1.14×104 kg m3
2.10 cm3 1000 g 1 m
At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our
result is indeed close to the expected value. Since the last reported significant digit is not certain, the
difference in the two values is probably due to measurement uncertainty and should not be a
concern. One important common-sense check on density values is that objects which sink in water
must have a density greater than 1 g cm3, and objects that float must be less dense than water.
P1.26 It is often useful to remember that the 1 600-m race at track and field events is approximately 1 mile
in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to
F IF I
a fG 1 mi2 JG1609 mJ2
1 acre H KH K = 4.05×103 m2 .
640 acres mi
F IF IF I
tonG2000 lbJG 1 h JG1 minJ
*P1.27 The weight flow rate is 1200 H KH KH K = 667 lb s .
h ton 60 min 60 s
P1.28 1 mi=1609 m=1.609 km; thus, to go from mph to km h, multiply by 1.609.
(a) 1 mi h= 1.609 km h
(b) 55 mi h= 88.5 km h
(c) 65 mi h=104.6 km h. Thus, ∆v= 16.1 km h .