1
Physics and Measurement
CHAPTER OUTLINE
1.1 Standards of Length, Mass, and Time
1.2 Modeling and Alternative Representations
1.3 Dimensional Analysis
1.4 Conversion of Units
1.5 Estimates and Order-of-Magnitude Calculations
1.6 Significant Figures
SOLUTIONS TO THINK–PAIR–SHARE AND ACTIVITIES
TP1.1 (a) The fourth experimental point from the top is a circle: This point
lies just above the best-fit curve that passes through the point (400 cm2,
0.20 g). The interval between horizontal grid lines is 1 space = 0.05 g.
An estimate from the graph shows that the circle has a vertical
separation of 0.3 spaces = 0.015 g above the best-fit curve.
© 2026
, (b) The best-fit curve passes through 0.20 g, so the percentage
difference is
(c) The best-fit curve passes through the origin and the point (600
cm2, 0.32 g). Therefore, the slope of the best-fit curve is
(d) For shapes cut from this copy paper, the mass of the cutout is
proportional to its area: m = aA. The proportionality constant a is 5.3
g/m2.
(e) This result is to be expected if the paper has thickness and
density that are uniform within the experimental uncertainty.
(f) The slope is the areal density of the paper, its mass per unit area.]
TP1.2 All results should be close to 2.54, representing the conversion factor
2.54 cm/in.
TP1.3 Solution: The difference is due to the average density, which is related
to the composition of the penny. Before 1982, U.S. pennies were 95%
copper and 5% zinc. After that date, they are 97.5% zinc, with a coating
of 2.5% copper. Both copper and zinc pennies were produced in 1982.
Perhaps a measurement of the mass of a sample of 1982 pennies would
be interesting.
TP1.4 Answers may vary. 200 pages of a typical textbook are usually about 70
to 80 mm thick.
© 2026
, 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
Its density is then
(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.
P1.2 (a) where d is the
diameter.
Then
(b)
P1.3 For either sphere, the volume is and the mass is
We divide this equation for the larger sphere by the
same equation for the smaller:
© 2026
, Then
P1.4 The volume of a spherical shell can be calculated from
From the definition of density, , so
P1.5 Let us find the angle subtended by the width of the Great Wall at the
height of the spacecraft orbit. From the description of a subtended
angle in the problem statement, we obtain
The angle subtended by the width of the Great Wall at a height of 200
km is 3.5 × 10–5 rad, which is smaller than the normal visual acuity of
the eye by about a factor of ten. Therefore, despite its great length, its
width cannot be seen. In the same way, a single human hair cannot be
seen from several meters away, despite its length. Your argument
should be based on this calculation.]
Answer: The angle subtended by the Great Wall is less than the visual
acuity of the eye.
© 2026
Physics and Measurement
CHAPTER OUTLINE
1.1 Standards of Length, Mass, and Time
1.2 Modeling and Alternative Representations
1.3 Dimensional Analysis
1.4 Conversion of Units
1.5 Estimates and Order-of-Magnitude Calculations
1.6 Significant Figures
SOLUTIONS TO THINK–PAIR–SHARE AND ACTIVITIES
TP1.1 (a) The fourth experimental point from the top is a circle: This point
lies just above the best-fit curve that passes through the point (400 cm2,
0.20 g). The interval between horizontal grid lines is 1 space = 0.05 g.
An estimate from the graph shows that the circle has a vertical
separation of 0.3 spaces = 0.015 g above the best-fit curve.
© 2026
, (b) The best-fit curve passes through 0.20 g, so the percentage
difference is
(c) The best-fit curve passes through the origin and the point (600
cm2, 0.32 g). Therefore, the slope of the best-fit curve is
(d) For shapes cut from this copy paper, the mass of the cutout is
proportional to its area: m = aA. The proportionality constant a is 5.3
g/m2.
(e) This result is to be expected if the paper has thickness and
density that are uniform within the experimental uncertainty.
(f) The slope is the areal density of the paper, its mass per unit area.]
TP1.2 All results should be close to 2.54, representing the conversion factor
2.54 cm/in.
TP1.3 Solution: The difference is due to the average density, which is related
to the composition of the penny. Before 1982, U.S. pennies were 95%
copper and 5% zinc. After that date, they are 97.5% zinc, with a coating
of 2.5% copper. Both copper and zinc pennies were produced in 1982.
Perhaps a measurement of the mass of a sample of 1982 pennies would
be interesting.
TP1.4 Answers may vary. 200 pages of a typical textbook are usually about 70
to 80 mm thick.
© 2026
, 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
Its density is then
(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.
P1.2 (a) where d is the
diameter.
Then
(b)
P1.3 For either sphere, the volume is and the mass is
We divide this equation for the larger sphere by the
same equation for the smaller:
© 2026
, Then
P1.4 The volume of a spherical shell can be calculated from
From the definition of density, , so
P1.5 Let us find the angle subtended by the width of the Great Wall at the
height of the spacecraft orbit. From the description of a subtended
angle in the problem statement, we obtain
The angle subtended by the width of the Great Wall at a height of 200
km is 3.5 × 10–5 rad, which is smaller than the normal visual acuity of
the eye by about a factor of ten. Therefore, despite its great length, its
width cannot be seen. In the same way, a single human hair cannot be
seen from several meters away, despite its length. Your argument
should be based on this calculation.]
Answer: The angle subtended by the Great Wall is less than the visual
acuity of the eye.
© 2026
