, 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
⎛ 0.304 8 m ⎞
1 420 ft 2 ⎜ ⎟⎠ = 132 m
2
⎝ 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
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 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.
0.02(1.365) = 0.03. The result is (1.37 ± 0.03) × 10 kg. So (d) 3 digits are
7
OQ1.10
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
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 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
π r = π ( 6.37 × 106 m ) = 1.08 × 1021 m 3
4 3 4 3
3 3
Its density is then
m 5.98 × 1024 kg
ρ= = = 5.52 × 103 kg/m 3
V 1.08 × 10 m
21 3
(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 = (π r 2 ) h and ρ = , we have
V
m 1 kg ⎛ 109 mm 3 ⎞
ρ= =
π r 2 h π ( 19.5 mm )2 ( 39.0 mm ) ⎜⎝ 1 m 3 ⎟⎠
ρ = 2.15 × 10 4 kg/m 3
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
⎛ 19.3 × 103 kg/m 3 ⎞
and mgold = ( 9.35 kg ) ⎜ = 22.9 kg
⎝ 7.87 × 10 kg/m ⎟⎠
3 3
ρ = m/V and V = ( 4/3 ) π r 3 = ( 4/3 ) π ( d/2 ) = π d 3 /6, where d is the
3
P1.4 (a)
diameter.
6 ( 1.67 × 10−27 kg )
Then ρ = 6m/ π d = 3
= 2.3 × 1017 kg/m 3
π ( 2.4 × 10 m)
−15 3
2.3 × 1017 kg/m 3
(b) = 1.0 × 1013 times the density of osmium
22.6 × 10 kg/m
3 3
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
⎛ 0.304 8 m ⎞
1 420 ft 2 ⎜ ⎟⎠ = 132 m
2
⎝ 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
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 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.
0.02(1.365) = 0.03. The result is (1.37 ± 0.03) × 10 kg. So (d) 3 digits are
7
OQ1.10
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
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 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
π r = π ( 6.37 × 106 m ) = 1.08 × 1021 m 3
4 3 4 3
3 3
Its density is then
m 5.98 × 1024 kg
ρ= = = 5.52 × 103 kg/m 3
V 1.08 × 10 m
21 3
(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 = (π r 2 ) h and ρ = , we have
V
m 1 kg ⎛ 109 mm 3 ⎞
ρ= =
π r 2 h π ( 19.5 mm )2 ( 39.0 mm ) ⎜⎝ 1 m 3 ⎟⎠
ρ = 2.15 × 10 4 kg/m 3
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
⎛ 19.3 × 103 kg/m 3 ⎞
and mgold = ( 9.35 kg ) ⎜ = 22.9 kg
⎝ 7.87 × 10 kg/m ⎟⎠
3 3
ρ = m/V and V = ( 4/3 ) π r 3 = ( 4/3 ) π ( d/2 ) = π d 3 /6, where d is the
3
P1.4 (a)
diameter.
6 ( 1.67 × 10−27 kg )
Then ρ = 6m/ π d = 3
= 2.3 × 1017 kg/m 3
π ( 2.4 × 10 m)
−15 3
2.3 × 1017 kg/m 3
(b) = 1.0 × 1013 times the density of osmium
22.6 × 10 kg/m
3 3
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
