Instructor's Solutions Manual for Physics for Scientists and Engineers with Modern Physics

2437 Pages·2016·43.67 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
2
2 ⎛ 0.304 8 m⎞ 2
1 420 ft ⎜ ⎟ = 132 m
⎝ 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
3
cube, (2 m) . 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).

9
OQ1.7 The population is about 6 billion = 6 × 10 . Assuming about 100 lb per
person = about 50 kg per person (1 kg has the weight of about 2.2 lb),
9 11
the total mass is about (6 × 10 )(50 kg) = 3 × 10 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.
2
OQ1.9 Mass is measured in kg; acceleration is measured in m/s . Force =
2
mass × acceleration, so the units of force are answer (a) kg⋅m/s .
7
OQ1.10 0.02(1.365) = 0.03. The result is (1.37 ± 0.03) × 10 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.
3 –3 –6 –9
Examples: 1 km = 10 m, 1 mg = 10 g = 10 kg, 1 ns = 10 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 3 4 6 3 21 3
πr = π (6.37 × 10 m) = 1.08 × 10 m
3 3
Its density is then
24
m 5.98 × 10 kg 3 3
ρ = = = 5.52 × 10 kg/m
21 3
V 1.08 × 10 m
(b) This v alue is intermediate between the tabulated densities of
aluminum and iron. Typical rocks have densities around 2000 to
3
3000 kg/m . The average density of the Earth is significantly
higher, so higher-density material must be down below the
surface.
2 m
P1.2 With V = (base area)(height), (πr )h and ρ = , we have

V = V
9 3
m 1 kg ⎛ 10 mm ⎞
ρ = 2 = 2
⎜ 3 ⎟
π r h π (19.5 mm) (39.0 mm)⎝ 1 m ⎠
4 3
ρ = 2.15 × 10 kg/m

m
P1.3 Let V represent the volume of the model, the same in ρ = , for both.

V
mgold
Then
ρiron = 9.35 kg/V and
ρgold = V .
ρgold mgold
Next, =

ρiron 9.35 kg
3 3
m ⎛ 19.3 × 10 kg/m ⎞
and gold = (9.35 kg)
⎜ 3 3 ⎟ = 22.9 kg
⎝ 7.87 × 10 kg/m ⎠

3 3 3
P1.4 (a) ρ = m/V and V = (4/3)πr = (4/3)π (d/2) = πd /6, where d is the

diameter.
−27
3 6(1.67 × 10 kg) 17 3
Then ρ = 6m/πd =
−15 3 = 2.3 × 10 kg/m
π (2.4 × 10 m)

17 3
2.3 × 10 kg/m 13
(b)
3 3 = 1.0 × 10 times the density of osmium
22.6 × 10 kg/m

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4 Physics and Measurement

P1.5 For either sphere the volume is 4πr3 and the mass is
V =
4 3 3
ρV = ρ πr . We divide this equation for the larger sphere by the
m = 3
s ame equation for the smaller:
3 3
 = ρ(4/3)πr
3 = r3 = 5
ms ρ(4/3)πrs rs
Then
 = rs 3 5 = (4.50 cm) 3 5 = 7.69 cm

*P1.6 The volumre of a spherical shell can be calculated from

o − Vi = 4π (r23 − r13 )
3
V = V m
From the d efinition of density, ρ = , so
V
3 3
ρV = ρ
( 34π )(r23 − r13 ) = 4π ρ(r32 − r1 )

m =

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
2 2
theorem,
diag = L + L . Thus, since the atoms are separated by a
distance LL = 0.200 nm, the diagonal planes are separated by

L2 + L2 = 0.141 nm .
1
P1.8 ( 2a) Treat this as a conversion of units using
–25 –2
1 Cu-atom = 1.06 × 10 kg, and 1 cm = 10 m:

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

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Chapter 1 5

(b) Thinking in terms of units, invert answer (a):
3
−1 ⎛ 1 cm ⎞
(density) = ⎜ 22 ⎟
⎝ 8.42 × 10 Cu-atoms⎠
−23 3
= 1.19 × 10 cm /Cu-atom

(c) For a cube of side L,
3 −23 3 −8
= 1.19 × 10 cm → L = 2.28 × 10 cm

L

Section 1.3 Dimensional Analysis
P1.9 (a) Write out dimensions for each quantity in the equation
vf = vi + ax
The variables vf and vi are expressed in units of m/s, so
[vf] = [vi] = LT –1
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,

(b) Write out diemquenatsiioonn s(a f)oirs e naocth dqiumaenntsitiyo ninal tlyh ec oerqruecat.ion

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] = [2 m][cos (kx)] = L

These are the same, so

equation (b) is dimensionally correct.

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6 Physics and Measurement

2
P1.10 Circumference has dimensions L, area has dimensions L , and volume
3 2 1/2 2
has dimensions L . Expression (a) has dimensions L(L ) = L ,
expression (b) has dimensions L, and expression (c) has dimensions
2 3
L(L ) = L . 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 2 2
⎡⎛ p ⎞ ⎤ [p] kg ⋅m
[ ] = ⎢⎜ ⎟ ⎥ = = 2
K ⎣⎝ 2m⎠ ⎦ kg s
2
Solvin g for [p ] and [p] then gives

p 2 kg2 ⋅ m2 kg ⋅m
[ ] = 2 → [p] =
s 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,

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

Therefore, the units of momentum are N ⋅s .

⎡ kg ⋅m⎤ [M][L]
P1.12 We substitute [ ] = [M], [m]= [L], and [F] =
⎢ 2 ⎥ = 2 into
⎣ s ⎦ [T]
kg
Newton’s law of universal gravitation to obtain

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

Solving for [G] then gives
3
3
[L] m
[ ]= 2 = 2
[M][T] kg ⋅s
G
−2
*P1.13 The term x has dimensions of L, a has dimensions of LT , and t has

m n
dimensions of T. Therefore, the equation x = ka t has dimensions of
−2 m n 1 0 m n−2m
( ) (T) or L T = L T

L = LT
The power s 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.
3
(a) [X] = [At ] + [Bt]
3 3
L = [A]T + [B]T→ [A] = L/T , and [B] = L/T .

2
(b)
[dx/dt] = ⎣⎡3At ⎦⎤ + [B]= L/T .

Section 1.4 Conversion of Units
4 3
P1.15 From Table 14.1, the density of lead is 1.13 × 10 kg/m , so we should
expect our calculated value to be close to this value. The density of
3 3
water is 1.00 × 10 kg/m , 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.

ρ = ⎛⎜
⎝ 22.130.9 c4m g3 ⎞⎟⎠ ⎜⎛⎝ 1 10 0k0g g⎞⎟⎠ (1010 m cm)3
= ⎛⎜
⎝ 22.130.9 c4m g3 ⎟⎠⎞ ⎛⎜⎝ 1 10 0k0g g⎞⎟⎠ (1 0001  0m003 cm3 )
4 3
= 1.14  × 10 kg/m

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 ⎞ 2
( )⎜ ⎟ (125 ft)⎜ ⎟ = 871 m
⎝ 3.281 ft ⎠ ⎝ 3.281 ft ⎠
A = LW = 75.0 ft
P1.18 Apply the following conversion factors: 1 in = 2.54 cm, 1 d = 86 400 s,
9
100 cm = 1m, and 10 nm = 1 m. Then, the rate of hair growth per
second is
−2 9
⎛ 1 ⎞ (2.54 cm/in)(10 m/cm)(10 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!
2
P1.19 The area of the four walls is (3.6 + 3.8 + 3.6 + 3.8) m × (2.5 m) = 37 m .
2
Each sheet in the book has area (0.21 m)(0.28 m) = 0.059 m . The
2 2
number of sheets required for wallpaper is 37 m /0.059 m = 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
2
given in the problem and the conversion 43 560 ft
= 1 acre. Then,

V = Bh
2
1 ⎡⎣(13.0 acres)(43 560 ft /acre)⎦⎤
=
3 × (481 ft)
7 3
= 9.08 × 10 ft
ANS FIG. P1.20
or
−2 3
7 3 ⎛ 2.83 × 10 m ⎞
( ft )⎜ 3 ⎟
⎝ 1 ft ⎠
V = 9.08 × 10
6 3
   = 2.57 × 10 m

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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:
6
g = (2.50 tons/block)(2.00 × 10 blocks)(2 000 lb/ton)
F
10
= 1.00 × 10 lbs

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

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

3
(c) To find the time to fill a 1.00-m tank, find the rate time/volume:
3 −4 3
−4 m ⎛ 2.70 × 10 m ⎞
2.70 × 10 =
⎜ ⎟
s ⎝ 1 s ⎠

−4 3 −1
⎛ 2.70 × 10 m ⎞ ⎛ 1 s ⎞ 3 s
or
⎝⎜ 1 s ⎟⎠ = ⎜⎝ 2.70 × 10−4 m3 ⎟⎠ = 3.70 × 10 m3

3 ⎛ 1 h ⎞
and so: 3.70 × 10 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
2 2
⎛ 1 mi ⎞ ⎛ 1 609 m⎞ 3 2
(1 acre)
⎜
⎝ 640 acres⎟⎠ ⎝ mi ⎠ = 4.05 × 10 m

*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⎞
( )⎜ ⎟ × (26 ft)⎜ ⎟
⎝ 1 ft ⎠ ⎝ 1 ft ⎠
= 50.0 ft
⎛ 0.304 8 m⎞
× (8.0 ft)⎜ ⎟
⎝ 1 ft ⎠
3 3
= 294.5 m = 290 m
3
3 ⎛ 100 cm⎞ 8 3
= (294.5 m )⎜ ⎟ = 2.9 × 10 cm
⎝ 1 m ⎠

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