Physics for Scientists and Engineers with Modern Physics Instructor Solution Manual

2437 Pages·2014·43.666 MB·English

by  Raymond A. Serway, John W. Jewett

1
Physics and Measurement

CHAPTER OUTLINE

1.1 Standards of Length, Mass, and Time
1.2 Matter and Model Building
1.3 Dimensional Analysis
1.4 Conversion of Units
1.5 Estimates and Order-of-Magnitude Calculations
1.6 Significant Figures

* An asterisk indicates a question or problem new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS

OQ1.1 The meterstick measurement, (a), and (b) can all be 4.31 cm. The
meterstick measurement and (c) can both be 4.24 cm. Only (d) does not
overlap. Thus (a), (b), and (c) all agree with the meterstick
measurement.
OQ1.2 Answer (d). Using the relation

= ⎛2.54 cm⎞⎛ 1 m ⎞ =
1 ft 12 in⎜ ⎟⎜ ⎟ 0.304 8 m
⎝ 1 in ⎠⎝100 cm⎠

we find that

⎛0.304 8 m⎞2
1420 ft2⎜ ⎟ =132 m2
⎝ 1 ft ⎠

OQ1.3 The answer is yes for (a), (c), and (e). You cannot add or subtract a
number of apples and a number of jokes. The answer is no for (b) and
(d). Consider the gauge of a sausage, 4 kg/2 m, or the volume of a
cube, (2 m)3. Thus we have (a) yes; (b) no; (c) yes; (d) no; and (e) yes.
1
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2 Physics and Measurement

OQ1.4 41 € ≈ 41 € (1 L/1.3 €)(1 qt/1 L)(1 gal/4 qt) ≈ (10/1.3) gal ≈ 8 gallons,
answer (c).
OQ1.6 The number of decimal places in a sum of numbers should be the same
as the smallest number of decimal places in the numbers summed.
21.4 s
15 s
17.17 s
4.003 s
57.573 s = 58 s, answer (d).

OQ1.7 The population is about 6 billion = 6 × 109. Assuming about 100 lb per
person = about 50 kg per person (1 kg has the weight of about 2.2 lb),
the total mass is about (6 × 109)(50 kg) = 3 × 1011 kg, answer (d).
OQ1.8 No: A dimensionally correct equation need not be true. Example: 1
chimpanzee = 2 chimpanzee is dimensionally correct.
Yes: If an equation is not dimensionally correct, it cannot be correct.
OQ1.9 Mass is measured in kg; acceleration is measured in m/s2. Force =
mass × acceleration, so the units of force are answer (a) kg⋅m/s2.
OQ1.10 0.02(1.365) = 0.03. The result is (1.37 ± 0.03) × 107 kg. So (d) 3 digits are
significant.

ANSWERS TO CONCEPTUAL QUESTIONS

CQ1.1 Density varies with temperature and pressure. It would be necessary
to measure both mass and volume very accurately in order to use the
density of water as a standard.
CQ1.2 The metric system is considered superior because units larger and
smaller than the basic units are simply related by multiples of 10.
Examples: 1 km = 103 m, 1 mg = 10–3 g = 10–6 kg, 1 ns = 10–9 s.
CQ1.3 A unit of time should be based on a reproducible standard so it can be
used everywhere. The more accuracy required of the standard, the less
the standard should change with time. The current, very accurate
standard is the period of vibration of light emitted by a cesium atom.
Depending on the accuracy required, other standards could be: the
period of light emitted by a different atom, the period of the swing of a
pendulum at a certain place on Earth, the period of vibration of a
sound wave produced by a string of a specific length, density, and
tension, and the time interval from full Moon to full Moon.
CQ1.4 (a) 0.3 millimeters; (b) 50 microseconds; (c) 7.2 kilograms

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Chapter 1 3

SOLUTIONS TO END-OF-CHAPTER PROBLEMS

Section 1.1 Standards of Length, Mass, and Time

P1.1 (a) Modeling the Earth as a sphere, we find its volume as

4 4 ( )
πr3 = π 6.37×106 m 3 =1.08×1021 m3
3 3
Its density is then

m 5.98×1024 kg
ρ= = = 5.52×103 kg/m3
V 1.08×1021 m3
(b) This value is intermediate between the tabulated densities of
aluminum and iron. Typical rocks have densities around 2000 to
3000 kg/m3. The average density of the Earth is significantly
higher, so higher-density material must be down below the
surface.
m
P1.2 With V = (base area)(height), V =(πr2)h and ρ= , we have
V

m 1 kg ⎛109 mm3⎞
ρ= =
πr2h π(19.5 mm)2(39.0 mm)⎝⎜ 1 m3 ⎠⎟
ρ= 2.15×104 kg/m3

m
P1.3 Let V represent the volume of the model, the same in ρ= , for both.
V
m
Then ρ = 9.35 kg/V and ρ = gold.
iron gold V
ρ m
Next, gold = gold
ρ 9.35 kg
iron
⎛19.3×103 kg/m3⎞
and m =(9.35 kg) = 22.9 kg
gold ⎝⎜7.87×103 kg/m3⎠⎟

P1.4 (a) ρ=m/Vand V =(4/3)πr3 =(4/3)π(d/2)3 =πd3/6, where d is the

diameter.
6(1.67×10−27kg)
Then ρ=6m/πd3 = = 2.3×1017kg/m3
π(2.4×10−15m)3

2.3×1017 kg/m3
(b) = 1.0×1013 times the density of osmium
22.6×103 kg/m3

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4 Physics and Measurement

4
P1.5 For either sphere the volume is V = πr3 and the mass is
3
4
m=ρV =ρ πr3. We divide this equation for the larger sphere by the
3
same equation for the smaller:

m ρ(4/3)πr3 r3
 =  =  =5
m ρ(4/3)πr3 r3
s s s
Then r =r 3 5 =(4.50 cm)3 5 = 7.69 cm
 s
*P1.6 The volume of a spherical shell can be calculated from
4
V =V −V = π(r3 −r3)
o i 3 2 1

m
From the definition of density, ρ= , so
V

(4 ) 4πρ(r3 −r3)
m=ρV =ρ π (r3 −r3)= 2 1
3 2 1 3

Section 1.2 Matter and Model Building

P1.7 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
diag
distance L = 0.200 nm, the diagonal planes are separated by
1
L2 +L2 = 0.141 nm.
2
P1.8 (a) Treat this as a conversion of units using
1 Cu-atom = 1.06 × 10–25 kg, and 1 cm = 10–2 m:

⎛ kg⎞⎛10−2 m⎞3⎛ Cu-atom ⎞
density =⎝⎜8 920m3⎠⎟⎝⎜ 1 cm ⎠⎟ ⎝⎜1.06×10−25 kg⎠⎟
Cu-atom
= 8.42×1022
cm3

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Chapter 1 5

(b) Thinking in terms of units, invert answer (a):

⎛ 1 cm3 ⎞
(density)−1 =
⎜ ⎟
⎝8.42×1022 Cu-atoms⎠
= 1.19×10−23 cm3/Cu-atom

(c) For a cube of side L,

L3 =1.19×10−23 cm3 →L= 2.28×10−8cm

Section 1.3 Dimensional Analysis

P1.9 (a) Write out dimensions for each quantity in the equation
v = v + ax
f i
The variables v and v are expressed in units of m/s, so
f i
[v] = [v] = LT –1
f i
The variable a is expressed in units of m/s2; [a] = LT –2
The variable x is expressed in meters. Therefore, [ax] = L2 T –2
Consider the right-hand member (RHM) of equation (a):
[RHM] = LT –1+L2 T –2
Quantities to be added must have the same dimensions.
Therefore, equation (a)is not dimensionally correct.

(b) Write out dimensions for each quantity in the equation
y = (2 m) cos (kx)
For y, [y] = L
for 2 m, [2 m] = L
and for (kx), [kx]= ⎡(2 m–1)x⎤= L–1L
⎣ ⎦
Therefore we can think of the quantity kx as an angle in radians,
and we can take its cosine. The cosine itself will be a pure number
with no dimensions. For the left-hand member (LHM) and the
right-hand member (RHM) of the equation we have
[LHM]=[y]= L [RHM]=[2m][cos (kx)]= L

These are the same, so equation (b)is dimensionally correct.

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6 Physics and Measurement

P1.10 Circumference has dimensions L, area has dimensions L2, and volume
has dimensions L3. Expression (a) has dimensions L(L2)1/2 = L2,
expression (b) has dimensions L, and expression (c) has dimensions
L(L2) = L3. The matches are: (a) and (f), (b) and (d), and (c) and (e).

P1.11 (a) Consider dimensions in terms of their mks units. For kinetic
energy K:
[ ]
2
[ ]= ⎡⎛ p2 ⎞⎤= p = kg⋅m2
K
⎢⎜ ⎟⎥
⎝2m⎠ kg s2
⎣ ⎦

Solving for [p2] and [p] then gives

kg2⋅m2 kg⋅m
[p]2 = → [p]=
s2 s

The units of momentum are kg⋅m/s.

(b) Momentum is to be expressed as the product of force (in N) and
some other quantity X. Considering dimensions in terms of their
mks units,

[N]⋅[X]=[p]
kg⋅m kg⋅m
⋅[X]=
s2 s
[X]=s

Therefore, the units of momentum are N⋅s.

⎡kg⋅m⎤ [M][L]
P1.12 We substitute [kg]=[M], [m]=[L], and [F]= = into
⎣⎢ s2 ⎦⎥ [T]2

Newton’s law of universal gravitation to obtain

[M][L] [G][M]2
=
[T]2 [L]2

Solving for [G] then gives

[L]3 m3
[G]= =
[M][T]2 kg⋅s2

*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=(LT−2)m(T)n or L1T0 =LmTn−2m

The powers of L and T must be the same on each side of the equation.
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Chapter 1 7

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 Summed terms must have the same dimensions.
(a) [X] = [At3] + [Bt]

L=[A]T3 +[B]T→ [A]=L/T3, and [B]=L/T .

(b) [dx/dt]= ⎡3At2⎤+[B]= L/T .
⎣ ⎦

Section 1.4 Conversion of Units

P1.15 From Table 14.1, the density of lead is 1.13 × 104 kg/m3, so we should
expect our calculated value to be close to this value. The density of
water is 1.00 × 103 kg/m3, so we see that lead is about 11 times denser
than water, which agrees with our experience that lead sinks.
Density is defined as ρ=m/V. We must convert to SI units in the

calculation.
⎛ 23.94 g ⎞⎛ 1 kg ⎞(100 cm)3
ρ=⎜ ⎟⎜ ⎟
⎝2.10 cm3⎠⎝1 000 g⎠ 1 m
⎛ 23.94 g ⎞⎛ 1 kg ⎞(1 000 000 cm3)
= ⎜ ⎟⎜ ⎟
⎝2.10 cm3⎠⎝1 000 g⎠ 1 m3
= 1.14 × 104 kg/m3

Observe how we set up the unit conversion fractions to divide out the
units of grams and cubic centimeters, and to make the answer come
out in kilograms per cubic meter. 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 from the tabulated values
is possibly due to measurement uncertainty and does not indicate a
discrepancy.
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8 Physics and Measurement

P1.16 The weight flow rate is

⎛ ton⎞⎛2000 lb⎞⎛ 1 h ⎞⎛1 min⎞
⎜1 200 ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 667 lb/s
⎝ h ⎠⎝ ton ⎠⎝60 min⎠⎝ 60 s ⎠

P1.17 For a rectangle, Area = Length × Width. We use the conversion
1 m = 3.281 ft. The area of the lot is then

⎛ 1 m ⎞ ⎛ 1 m ⎞
A=LW =(75.0 ft)⎜ ⎟(125 ft)⎜ ⎟ = 871 m2
⎝3.281 ft⎠ ⎝3.281 ft⎠

P1.18 Apply the following conversion factors: 1 in = 2.54 cm, 1 d = 86 400 s,
100 cm = 1m, and 109 nm = 1 m. Then, the rate of hair growth per
second is

⎛ 1 ⎞(2.54 cm/in)(10−2 m/cm)(109 nm/m)
rate=⎜ in/day⎟
⎝32 ⎠ 86 400 s/day
= 9.19 nm/s

This means the proteins are assembled at a rate of many layers of
atoms each second!
P1.19 The area of the four walls is (3.6 + 3.8 + 3.6 + 3.8) m × (2.5 m) = 37 m2.
Each sheet in the book has area (0.21 m)(0.28 m) = 0.059 m2. The
number of sheets required for wallpaper is 37 m2/0.059 m2 = 629 sheets
= 629 sheets(2 pages/1 sheet) = 1260 pages.

The number of pages in Volume 1 are insufficient.

P1.20 We use the formula for the volume of a pyramid
given in the problem and the conversion 43 560 ft2
= 1 acre. Then,

V =Bh

1
= ⎡(13.0 acres)(43 560 ft2/acre)⎤
⎣ ⎦
3
×(481 ft)
=9.08×107 ft3
ANS FIG. P1.20
or

⎛2.83×10−2 m3⎞
V =(9.08×107 ft3)
⎜ ⎟
⎝ 1 ft3 ⎠
   = 2.57×106 m3

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Chapter 1 9

P1.21 To find the weight of the pyramid, we use the conversion
1 ton = 2 000 lbs:

F =(2.50 tons/block)(2.00×106 blocks)(2 000 lb/ton)
g
= 1.00×1010 lbs

⎛ 30.0 gal ⎞⎛1 mi⎞ gal
P1.22 (a) rate=⎜ ⎟⎜ ⎟ = 7.14×10−2
⎝7.00 min⎠⎝ 60 s⎠ s

gal⎛231 in3⎞⎛2.54 cm⎞3⎛ 1 m ⎞3
(b) rate=7.14×10−2 ⎜ ⎟⎜ ⎟ ⎜ ⎟
s ⎝ 1 gal ⎠⎝ 1 in ⎠ ⎝100 cm⎠
m3
= 2.70×10−4
s

(c) To find the time to fill a 1.00-m3 tank, find the rate time/volume:
m3 ⎛2.70×10−4 m3⎞
2.70×10−4 =
⎜ ⎟
s ⎝ 1 s ⎠

⎛2.70×10−4 m3⎞−1 ⎛ 1 s ⎞ s
or ⎝⎜ 1 s ⎠⎟ =⎝⎜2.70×10−4 m3⎠⎟ = 3.70×103 m3

⎛ 1 h ⎞
and so: 3.70×103 s⎜ ⎟ = 1.03 h
⎝3 600 s⎠

*P1.23 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
⎛ 1 mi2 ⎞⎛1609 m⎞2
(1 acre) = 4.05×103 m2
⎜ ⎟
⎝640 acres⎠⎝ mi ⎠

*P1.24 The volume of the interior of the house is the product of its length,
width, and height. We use the conversion 1 ft = 0.304 8 m and
100 cm = 1 m.

V =LWH
⎛0.304 8 m⎞ ⎛0.304 8 m⎞
=(50.0 ft)⎜ ⎟ ×(26 ft)⎜ ⎟
⎝ 1 ft ⎠ ⎝ 1 ft ⎠
⎛0.304 8 m⎞
×(8.0 ft)⎜ ⎟
⎝ 1 ft ⎠
=294.5 m3 = 290 m3
=(294.5 m3)⎛⎜100 cm⎞⎟3 = 2.9×108 cm3
⎝ 1 m ⎠

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