1. The document provides examples of unit conversions between various systems of measurement including centimeters, meters, yards, miles, inches, and picas. 2. Size comparisons are made between everyday objects like grapes and stars, as well as microscopic objects like atoms, cell nuclei, and protons. 3. Calculations are shown for time intervals, rates, and conversions between seconds, minutes, hours, days, weeks, and years.

1. 1
CHAPTER 1 SPACE, TIME, AND MASS
Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions
Manual, available for purchase. Answers to all solutions below are underscored.
1-1. Assume a height of 5 ft 10 in. Then, 5′10″ = 70 in = 70 in × 2.54 cm/in = 178 cm.
1-2. There are approximately 300 actual pages in this text (vol. 1). The thickness is 2.5 cm. Therefore
each page = 2.5 cm/300 pg 3
8.3 × 10 cm.−
=
1-3. 100 yd × 0.914 m/yd = 91.4 m; 53 1/3 yd × 0.914 m/yd 48.7 m=
1-4. 31 step 1000 m steps
× 1.7 × 10
0.60 m 1 km km
N = =
1-5.
1 pica
11 in × 66 picas.
1
in
6
L = =
17 1 pica
in × 51 picas
12 in
6
W = =
1-6. Virus: 2 × 10−8
m × 10/0.3048 ft/m × 12 in/ft 7
8 × 10 in−
=
Similarly: Atom: 1 × 10−10
m × 39.4 in/m 9
4 × 10 in−
=
Fe Nucleus: 8 × 10−15
m × 39.4 in/m 13
3 × 10 in−
=
Proton: 2 × 10−15
m × 39.4 in/m 14
8 × 10 in−
=
†1-7. Let’s convert 1/2 inch to mm using conversion factors, then use proportional reasoning to do the
others until the number of significant figures becomes large.
1
in × 25.4 mm/in 12.7 mm
2
=
1 12.7 mm
in 6.35 mm
4 2
= =
1 6.35 mm
in 3.175 mm 3.18 mm
8 2
= = = (to three significant figures)
1 3.175 mm
in 1.5875 mm 1.59 mm
16 2
= = = (to three significant figures)
The number of digits is becoming large, so let’s do direct conversions for the rest of the problems.
1
in × 25.4 mm/in 0.794 mm
32
=
1
in × 25.4 mm/in 0.397 mm
64
=
1-8.
3 2 6
10 in 2.54 cm 10 m 10 µm
1 mil × × × × 25.4 µm.
mil in cm m
− − −
=
1000 µm 1 mil
1 mm × × 39.4 mil
mm 25.4 µm
=

2. CHAPTER 1
2
1-9. (a) Grapefruit diameter ≈ 0.1 m
Ratio of grapefruit/sun = 0.1 m/(1.4 × 109
m) = 7 × 10−11
Earth diameter ≈ 13 × 106
m
Comparative size of Earth = 13 × 106
m × (7 × 10−11
)
= 9 × 10−4
m 1 mm.≈
Nearest star distance = 4 × 1016
m
Comparative distance = 4 × 1016
m × (7 × 10−11
) 6
2.8 × 10 m=
(b) Head diameter ≈ 0.2 m
Earth diameter ≈ 13 × 106
m
Earth/head ratio ≈ 7 × 107
Size of atom = 10−10
m
Comparative size of atom = 10−10
m × (7 × 107
) = 7 × 10−3
m = 7 mm
Size of red blood cell ≈ 7.5 × 10−6
m
Comparative size of cell = 7.5 × 10−6
m × (7 × 107
) 500 m 1/ 2 km≈ =
1-10. Distance to Q1208 + 1011 = 12.4 × 109
× 9.47 × 1015
= 1.17 × 1026
m
Distance on the diagram (PRELUDE, p. 6)
=
26
5
20
1.17 × 10
= 7.8 × 10 m
1.5 × 10
1-11. Size (diameter) of the sun = 2 × 6.46 × 108
= 1.4 × 109
m distance on the diagram (PRELUDE, p.
6) =
9
3
12
1.4 × 10
10 m 1 mm
1.5 × 10
−
=
1-12.
9
6 1310 m
10 × 633 nm × 6.33 × 10 m.
nm
l
−
− −
∆ = = According to Table 1-1, the diameter of an
atom is about 1 × 10−10
m, so this is 6.33 × 10−3
times the diameter of an atom, or roughly 1/100
the diameter of an atom.
†1-13. 1 turn = 360°, so 5° × 1 turn/360° = 0.0139 turn. For an English thread,
6
1in 0.0254m 10 µm
0.0139 turn × × × = 4.41 µm.
80 turns in m
For a metric thread,
3 6
0.5mm 10 m 10 µm
0.0139 turn × × × = 6.94 µm.
turn mm m
−
1-14. 1 nmi = 1852 m; Circumference of Earth = 4.00 × 107
m
Circumference = (4.00 × 107
m)/1852 m/nmi 21, 600 nmi=
Also 360° × 60 min/deg 21, 600 min, so 1nmi 1min= ⇒

3. CHAPTER 1
3
†1-15. For one of the triangles, (R + 1.75 m)2
= R2
+ (4700 m)2
. Expand
this to get 2 2 2 2
2(1.75 m) (1.75 m) (4700 m) .R R R+ + = +
We expect R to be much larger than 1.75 m, so we can ignore
(1.75 m)2
relative to all the other terms. The R2
terms cancel,
leaving
(3.50 m)R = (4700 m)2
, which gives R = 6.3 × 106
m.
1-16. 22 yr, 5 mo, 23 days = (8035 + 153 + 23) 8211 days=
(This excludes leap years and assumes average 30.5-day month.)
1 day = 1 day × 24 h/day × 60 min/h × 60 s/min = 86,400 s
8211 days = 8211 days × 86,400 s/day 8
7.1 × 10 s=
1-17. 1 yr = 365.25 days. Therefore, 4.5 × 109
yr
= 4.5 × 109
yr × 365.25 day/yr × 86,400 s/day 17
1.4 × 10 s=
1-18. 12
9
60min 60s 1 calculation
1 h × × × 3.6 × 10 calculations/h.
h min 10 s
N −
= =
1-19.
3600s 60s
2h 9min 21s 2h × 9min × 21s 7761 s
h min
= + + =
1-20.
3600s 60s
2h 24min 51s 2h × 24min × 51s 8692 s
h min
= + + =
†1-21.
365.25 solar days/year 24 h
1 sidereal day × ×
366.25 sidereal days/year solar day
= 23.934 h/sidereal day. Using 1 h = 60
min to convert the 0.934 h to minutes gives 1 sidereal day = 23 h 56 min.
1-22. 7 9s
4 ticks × 3.2 × 10 × 10 years 1.2 × 10 ticks
year
N = =
1-23. 6 1h 1day
10 s × × 11.6 days
3600s 24h
=
1-24.
7 days 24 h
1week × ×
week day
= 168 h. 53600 s
168h × 6.048 × 10 s
hr
=
1-25.
7
7beats 1 min 3.2 × 10 s
71 ticks × × 3.8 × 10 beats/year
min 60 s yr
N = =
1-26. (a) June 24−25: (20 − 4) s/day = 16 s/24 h 0.67 s/h=
June 25−26: (34 − 20) s/day = 14 s/24 h 0.58 s/h=
June 26−27: (51 − 34) s/day = 17 s/24 h 0.71 s/h=
(b) Average rate = (51 − 4)/3 day = 47 s/(24 × 3)h 0.65 s/h=
R
R + 1.75 m
9400

4. CHAPTER 1
4
(c) 10h
30m
on June 30 is 70.5 h after noon June 27. By average loss, watch should have lost 70.5
h × 0.65 s/h = 46 s. Combined with loss of 51 s on June 27 gives total loss of 97 s. Therefore the
correct WWV time is h m s
10 31 37 . With the largest rate, loss is 70.5 h × 0.71 s/h = 50 s. Combined
loss is 51 + 50 = 101 s. Then estimated WWV time is h m s
10 31 41 . The wristwatch can be trusted to
about ±4 s on June 30.
†1-27. 1 day = 24 hr = 86,400 s. The earth rotates 360º per day, which corresponds to a rotation rate of
360 60 min
× 0.250 min/s.
86,400 s degree
°
= A timing error of 1 s will result in an angular error of 0.250
min. According to Problem 1-14, 1 min = 1852 m, so the corresponding error in position is 0.250
min/s × 1852 m = 463 m = 0.463 km.
1-28. 140 lb-mass = 140 lb-mass × 0.454 kg/lb-mass = 64 kg
140 lb-mass = 140 lb-mass × 0.454 kg/lb-mass × 1/14.6 slug/kg 4.35slug=
140 lb-mass = 140 lb-mass × 0.454 kg/lb-mass × 1 amu/(1.67 × 10−27
kg) 28
3.8 × 10 amu=
†1-29. 24 27
(0.33 4.9 5.98 0.64 1900 553 87.3 10.8 0.66) × 10 kg 2.56 × 10 kgplanetsm = + + + + + + + + =
(to three significant figures). msun = 1.99 × 1030
kg, so the total mass is
30 27 30
1.99 × 10 kg 2.56 × 10 kg 1.99 × 10 kgtotal sun planetsm m m= + = + = (to three significant
figures). The fraction of the total mass included in the planets is
27
30
2.56 × 10
× 100% × 100% 0.134%.
1.99 × 10
planets
total
m
m
= = The fraction of the mass in the sun is 100% −
0.134% = 99.9%.
1-30.
26
15
largest length 1 × 10 m
smallest length 2 × 10 m−
= = 5 × 1040
.
17
24
longest time 4 × 10 s
shortest time 1 × 10 s−
= = 4 × 1041
.
53
31
largest mass 1 × 10 kg
smallest mass 9 × 10 kg−
= = 1 × 1083
. The first two ratios are within an order of magnitude of
each other, and the third is roughly equal to the square of the other two. (Note that the first two
ratios will keep increasing because the universe is expanding and aging.)
†1-31. From the periodic table in the Appendix, we see that the uranium nucleus contains about 238
nucleons, each with the mass of a proton. Table 1.7 gives 1.7 × 10−27
kg for the mass of a proton.
Then the total mass of the electrons is (92)( 9.1 × 10−31
kg) = 8.4 × 10−29
kg, and the total mass of
the nucleus is (238)( 1.7 × 10−27
kg) = 4.0 × 10−25
kg. To two significant figures, the total mass of
the atom is 4.0 × 10−25
kg. The fraction of the total mass in the electrons is 8.4 × 10−29
/4.0 × 10−25
= 2.1 × 10−4
= 0.021%. The fraction of mass in the nucleus is 99.98%.
1-32. Let m represent the mass of material and M represent the mass of one mole. Then
6
23 14atoms 0.1 × 10 g
6.02204 × 10 × 3 × 10 atoms.
mol 197 g/mol
A
m
N N
M
−
⎛ ⎞⎛ ⎞
= = =⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
1-33. 1 lb avoirdupois = 0.453 59 kg = 435.59 g (given in the text)
0.453 59 kg
1 lb troy 0.822 86 lb avoidupois ×
lb avoidupois
= = 0.373 24 kg = 373.24 g
1-34.
21
23 4atoms 10 kg
6.02204 × 10 × 1.1 × 10 atoms
mol 0.055 85 kg/mol
A
m
N N
M
−
⎛ ⎞⎛ ⎞
= = =⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠

5. CHAPTER 1
5
1-35. (a) Since the density of water is 1 g/cm3
, 250 cm3
of water has a mass of 250 g, which is 250
g/(18 g/mol) = 14 mol. Therefore, the number of molecules is 14 mol × 6.02 × 1023
molecules/mol 24
8.4 × 10 molecules.=
(b) Mass of sea-water (density = 1030 kg/m3
)
is 1.3 × 1018
m3
× 1030 kg/m3
= 1.3 × 1021
kg = 1.3 × 1024
g
Number of molecules of sea-water is
1.3 × 1024
g × 1 mol/18 g × 6.02 × 1023
molecules/mol 46
4.3 × 10 molecules.=
(c) Ratio of molecules of water from cup to molecules from sea is
8.4 × 1024
/4.3 × 1046 22
2.0 × 10 .−
=
The probability of a single molecule drawn from the ocean being originally from the cup is
therefore 2.0 × 10−22
. Drawing a cup full then should result in 2.0 × 10−22
× 8.4 × 1024
1680 molecules.=
1-36. The mass of H2 and He, respectively, is 0.70 × 1.99 × 1030
kg = 1.39 × 1030
kg and 0.30 × 1.99 ×
1030
kg = 0.60 × 1030
kg. The atomic mass of hydrogen and helium, respectively, is 1 g/mol and 4
g/mol. Therefore, the number of hydrogen atoms is
NH = 1.39 × 1033
g × 1 mol/g × 6.02 × 1023
molecules/mol 56
8.38 × 10 molecules.=
The number of helium atoms is NHe = 0.60 × 1033
g × 1 mol/4 g × 6.02 × 1023
molecules/mol
55
9.03 × 10 atoms.=
Total number 56
9.28 × 10=
†1-37. Molecular mass of N2 = 28 g/mol
Molecular mass of O2 = 32 g/mol
Molecular mass of Ar = 40 g/mol
Therefore, 1000 g of air will contain:
755 g N2 = 755 g/(28 g/mol) = 27.0 mol
232 g O2 = (232/32) mol = 7.25 mol
13 g Ar = (13/40) mol = 0.325 mol
The percentage by number of molecules of these substances is:
N2: 27.0/(27.0 + 7.25 + 0.325) × 100% = 27.0/34.575 × 100% = 78.1%
O2: (7.25/34.575) × 100% = 21%
Ar: (0.325/34.575) × 100% = 0.9%
Therefore, the “molecular mass” of air is
(0.781 × 28) + (0.21 × 32) + (0.009 × 40) 28.95g/mol.=
1-38. The mass of elements is:
Oxygen: 0.65 × 73 kg = 47.4 kg; atomic mass = 16
Carbon: 0.185 × 73 kg = 13.5 kg; atomic mass = 12
Hydrogen: 0.095 × 73 kg = 6.94 kg; atomic mass = 1.008
Nitrogen: 0.033 × 73 kg = 2.41 kg; atomic mass = 14
Calcium: 0.015 × 73 kg = 1.09 kg; atomic mass = 40.08
Phosphorous: 0.01 × 73 kg = 0.73 kg; atomic mass = 31

6. CHAPTER 1
6
The number of atoms of these substances is:
O: 47.4 × 103
g × 1/16 mol/g × 6.02 × 1023
/mol = 1.78 × 1027
C: 13.5 × 103
g × 1/12 mol/g × 6.02 × 1023
/mol = 6.77 × 1026
H: 6.94 × 103
g × 1 mol/g × 6.02 × 1023
/mol = 4.17 × 1027
N: 2.41 × 103
g × 1/14 mol/g × 6.02 × 1023
/mol = 1.04 × 1026
Ca: 1.09 × 103
g × 1/40 mol/g × 6.02 × 1023
/mol = 1.64 × 1025
P: 0.73 × 103
g × 1/31 mol/g × 6.02 × 1023
/mol = 1.42 × 1025
Total number of atoms: = 6.76 × 1027
†1-39. 0.53° = 0.53/360 × 2π rad = 9.25 × 10−3
rad
d = 9.25 × 10−3
rad × 1.5 × 1011
m
= 1.4 × 109
m
8
6.9 × 10 mr =
1-40. 8 15m days h s
1 ly 3.00 × 10 × 365.25 × 24 × 3600 9.47 × 10 m
s year day h
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
1-41. 1 ly = 9.47 × 1015
m
2.2 × 106
ly = 2.2 × 106
ly × 9.47 × 1015
m/ly 22
2.1 × 10 m=
1-42. 1 light-s (ls) = 3.00 × 108
m/s × 1 s = 3.00 × 108
m
1 light-min (lm) = 3.00 × 108
m/s × 60 s = 1.80 × 1010
m
E-S distance = 1.50 × 1011
m × 1/(1.80 × 1010
) lm/m 8.3 lm=
E-M distance = 3.84 × 108
m × 1/(3.00 × 108
) ls/m 1.28 ls=
1-43. 1 second of arc =
1
3600
degree =
2π 1
×
360 3600
rad = 4.85 × 10−6
rad
(a) 61 AU
4.85 × 10
pc
−
=
1 pc = 1 × 1/(4.85 × 10−6
)AU 5
2.06 × 10 AU=
(b) 1 pc = 2.06 × 105
AU × 1.496 × 1011
m/AU = 3.08 × 1016
m
1 ly = 9.47 × 1015
m
1 pc = 3.08 × 1016
m × 1 ly/(9.47 × 1015
m) 3.25 ly=
(c) 1 ly = 3.00 × 108
m/s × 1 yr = 3.00 × 108
m/s × 3.156 × 107
s 15
9.47 × 10 m=
1-44.
2
2 1 ft
1m ×
0.3048 m
⎛ ⎞
⎜ ⎟
⎝ ⎠
= 10.76 ft2
. Note that the conversion factor must be squared.
†1-45.
3
3 1 ft
1m ×
0.3048 m
⎛ ⎞
⎜ ⎟
⎝ ⎠
= 35.31 ft3
. Note that the conversion factor must be cubed.
1-46. Use the result from 1−44:
2
2
1 m
(78 ft)(27 ft) ×
10.76 ft
A z= = 196 m2
.
†1-47. His height is measured to a precision of 0.1 inch, so we want to see how many significant digits
this implies. 8 feet = 96 inches (exactly), so to the nearest 0.1 inch his height can be expressed as
107.1 inches, which contains four significant figures. Converting 11.1 inches to feet gives his
height in feet: 8 ft + 11.1 in = 8.925 ft, which also contains four significant figures. Thus his
height in meters should be specified to four figures:
m
8.925 ft × 0.3048
ft
= 2.720 m.

7. CHAPTER 1
7
1-48.
2 2
2
3 ft 1 m
100 yd × 53.33 yd × ×
yd 10.76 ft
A
⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
= 4.46 × 103
m2
, using the result from 1−44.
†1-49.
3
3 3 31 kg 100 cm
8.9 g/cm × × 8.9 × 10 kg/m .
1000 g m
⎛ ⎞ ⎛ ⎞
=⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
Note that the cm to m conversion factor
must be cubed! 1 ft = 0.3048 m, 1 lb = 0.454 kg.
3
3 3 31 lb 0.3048 m
8.9 × 10 kg/m × × 555 lb/ft ,
0.454 kg ft
⎛ ⎞ ⎛ ⎞
=⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
or 5.6 × 102
lb/ft3
to two significant
figures. Again, note that the m to ft conversion factor must be cubed. 1 ft = 12 in, so
3
3 31 ft
555 lb/ft × 0.32 lb/in .
12 in
⎛ ⎞
=⎜ ⎟
⎝ ⎠
1-50. Assume a mass of 73 kg and a density of 1000 kg/m3
.
V =
mass
density
= 73 kg × 1/1000 m3
/kg 3
0.073 m=
1-51.
3 6 3
3
cm 3600 s 10 m
92 × 24h × ×
s h cm
−
= 7.9 m3
/day.
1-52. 1 liter = 103
cm3
, so 5.2 liter = 5.2 × 103
cm3
.
3 3
3
5.2 × 10 cm
57 s.
92 cm /s
t = =
†1-53.
4 2
2
2
2
12
10 m
1 cm ×
cm
m
10
transistor
N
−
−
= = 108
transistors. If they’re stacked, N = the number of transistors per
layer (108
) × number of layers. If the cube is 1 cm high and each layer is 10−7
m, or 10−5
cm,
thick, then the cube holds 105
layers and the cube can hold 108
× 105
= 1013
transistors!
1-54. 1 gal = 3.785 liter.
3
3 3
3
g kg cm 1 lb liter
Density 1.00 × 10 × 10 × × 3.875
cm g liter 0.4536 kg gal
−
= = 8.34
lb/gal
†1-55. (a) (3.6 × 104
) × (2.049 × 10−2
) = (3.6)(2.049) × 104−2
= 7.4 × 102
(b) (2.581 × 102
) − (7.264 × 101
) = (2.581 − 0.7264) × 102
= 1.855 × 102
(c) 0.079832 ÷ 9.43 =
2
0
7.9832 × 10
9.43 × 10
−
= 0.847 × 10−2−0
= 8.47 × 10−3
1-56.
30 3
3 6 3
3 8 3 3
3
3 3(2.0 × 10 kg) g m
Density × 10 × 10 1.5 g/cm
4 4 4 (7.0 × 10 m) kg cm
3
m m m
V RR π ππ
−
= = = = =
†1-57.
30 3
6
3 3 3 3 3
3
3 3(2.0 × 10 kg) 1 metric ton m
Density × × 10
4 4 4 (20 × 10 m) 10 kg cm
3
m m m
V RR π ππ
−
= = = =
= 6.0 × 107
metric tons/cm3
1-58. Oceans of the earth have 1.3 × 1018
m3
of water. The mass of the oceans is 1.3 × 1018
m3
× 1030
kg/m3
= 1.3 × 1021
kg. Mass of the earth is 5.98 × 1024
kg, so that the percentage of the mass of
the earth that is water is:
(1.3 × 1021
kg/5.98 × 1024
kg) × 100% 0.02%.=

8. CHAPTER 1
8
†1-59. From Table 1.10, 1 liter = 10−3
m3
. According to data given in the “Conversion of Units” section
of the chapter, the density of water is 1000 kg/m3
, so
3 3
3 310 m 1 min
300 liters/min × × 5.00 × 10 m /s,
liter 60 s
−
−
= and
3 3 3
5.00 × 10 m /s × 1000 kg/m 5.00 kg/s.−
=
1-60. 1 in = 1 in × 2.54 cm/in × 1/100 m/cm = 0.0254 m. Therefore volume on 1 m2
is V = 0.0254 m ×
1 m2 3 2
0.0254m /m .=
Mass of this much water is 0.0254 m3
/m2
× 1000 kg/m3
= 25.4 kg/m2
.
†1-61. This can be solved using proportional reasoning. 3
3
3
Density ,
4 4
3
m m m
V RR ππ
= = = which means
3 3
,
copperlead
lead copper
mm
R R
= from which we get
1/ 3 1/ 3
15 3.5
(4.8 × 10 m)
1.06
lead
lead copper
copper
m
R R
m
−
⎛ ⎞ ⎛ ⎞
= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
= 7.1 × 10−15
m. Likewise we get
1/ 3 1/ 3
15 0.27
(4.8 × 10 m)
1.06
oxygen
oxygen copper
copper
m
R R
m
−
⎛ ⎞ ⎛ ⎞
= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
= 3.0 × 10−15
m. Note that it was not necessary to include the factor (× 10−25
) when expressing the
masses because it cancels in the ratio.
1-62. Density is calculated as
mass
volume
where V =
4
3
πr3
.
Planet Mass (kg) Vol (m3
) Density (kg/m3
)
Pluto 6.6 × 1023
? 1.13 × 1020
5800 ?
Mercury 3.3 × 1023
5.94 × 1019
5600
Earth 5.98 × 1024
1.09 × 1021
5500
Venus 4.9 × 1024
9.51 × 1020
5100
Mars 6.40 × 1023
1.62 × 1020
4000
Neptune 1.03 × 1026
4.58 × 1022
2250
Uranus 8.73 × 1025
5.65 × 1022
1550
Jupiter 1.90 × 1027
1.52 × 1024
1250
Saturn 5.53 × 1026
9.23 × 1023
600
1-63. One dimension of each roof segment must be divided by cos 45°, or multiplied by 2. Then total
roof area = floor area × 2 = 250 m2
× 2 = 354 m2
.
1-64. Use the radius of the earth from Table 1-1.
6
15m
6.4 × 10 m
s
R
θ
∆
∆ = = = 2.3 × 10−6
radian. In
degrees, this is (2.3 × 10−6
radian) × (180°/π
radian) = 1.3 × 10−4
degree.
†1-65. The slope is the tangent of the required angle. For a slope of 1:5, 1 1
tan 11 .
5
θ −
= = ° For a slope of
1:10, 1 1
tan 5.7 .
10
θ −
= = ° For a slope angle of 0.1°, tan θ = 1.7 × 10−3
, so the slope is
1:( 1.7 × 10−3
)−1
= 1: 5.7 × 102
, so the rise is about 1 atom per 570 atoms.
∆θ
R
∆s = 15 m

9. CHAPTER 1
9
1-66. tan ,
h
D
θ = where D = 75 m. Both D and θ are
specified to two significant figures, so the value
calculated for h must be specified to two figures.
Thus tan (75 m)( tan 78 )h D θ= = ° = 3.5 × 102
m.
†1-67. In this calculation, assume that the calendar year is exactly 365 days long and a circle contains
exactly 360°, so these are not interpreted as numbers with only three significant figures. The
angle the earth moves through in one calendar year is
365
× 360
365.24
θ = ° = 359.76°. In four
years, including one leap year with exactly 366 days, the total number of days is 3 × 365 + 366 =
1461 days. The angle the earth moves through in four years is
1461
× 360
365.24
θ = ° = 1440.0°,
which is exactly four complete circles to five significant figures. (The angle is actually a little
larger than 1440.0°, which is why there are some four-year intervals that do not include an extra
day.)
1-68. The sun will set at Marchena n-minutes after 8 PM where
n =
minutes in a day × distance between islands
circumference of the earth
=
3
7
24 × 60 × 60 × 10
4 × 10
= 2.15 min
Therefore, the sun sets at Marchena at 2.15 min after 8 PM.
1-69. The diameter, d of the tree trunk is related to its length, L by
d = AL3/2
, where A is a constant
7.6 = A(81)3/2
if L = 90 m, then
d = A(90)3/2
=
3 / 2
90
7.6 = 8.9m.
81
⎛ ⎞
⎜ ⎟
⎝ ⎠
The mass, m of the tree trunk is related to d and L by
m = ρπ
2
2
d⎛ ⎞
⎜ ⎟
⎝ ⎠
L, where ρ is the mass density.
6100 = ρ
π
(7.6)
4
2
81
for L = 90 m,
m = ρ 2π
(8.9)
4
90 = 6100
2
8.9
7.6
⎛ ⎞
⎜ ⎟
⎝ ⎠
90
= 9295 tons
81
1-70. Distance from pole to equator = 1/4 circumference. Therefore
d =
1
(2π
4
r) =
π
2
r =
π
2
× 6.37 × 106
m 7
1.00 × 10 m.=
Straight line distance is, using the Pythagorean theorem:
d = 2 2 2 6 6
2 2 × 6.37 × 10 m 9.0 × 10 m.r r r+ = = =
θ
h
D

10. CHAPTER 1
10
†1-71.
23
235 g/mol
atoms
6.02204 × 10
mol
atom
A
M
m
N
= = = 3.902 × 10−22
g = 3.902 × 10−25
kg. In atomic mass units,
this is
25
27
3.902 × 10 kg
kg
1.66054 × 10
u
atomm
−
−
= = 235.0 u.
1-72. The molar mass M of water (H2O) is 18.0 g/mol. The mass m of 1 liter (1000 cm3
) of water is
1000 g. 231000 g molecules
= × = × 6.02204 × 10
18.0 g/mol mol
A
m
N N
M
= 3.35 × 1025
molecules. Each
molecule contains one oxygen atom and two hydrogen atoms, so there are 3.35 × 1025
oxygen
atoms and 6.70 × 1025
hydrogen atoms.
1-73. Molecular mass of N2 is 2 × 14 g/mol = 28 g/mol. The density of air is 1.3 kg/m3
. The number of
molecules in 1 cm3
of air is then number of molecules = 10−6
m3
× 1.3 kg/m3
× 1000 g/kg × 1
mol/28 g × 6.02 × 1023
molecules/mol 19
2.8 × 10 molecules.=
1-74.
3 3
6 3
3
cells mm m
5.1 × 10 × 1000 × 10 × 5.2 liters
mm m liter
N −⎛ ⎞
= ⎜ ⎟
⎝ ⎠
= 2.7 × 1013
cells.
1-75. Volume of paint = area × thickness.
3
3 2volume m
Thickness (1 liter) × 10 (8 m )
area liter
−⎛ ⎞
= = ÷⎜ ⎟
⎝ ⎠
= 1.25 × 10−4
m (0.125 mm)
1-76. The area is
2
6 2 3 12 2m
9.4 × 10 km × 10 9.4 × 10 m .
km
⎛ ⎞
=⎜ ⎟
⎝ ⎠
6 3 12 2 4 2mass kg
Area density 8 × 10 metric tons × 10 (9.4 × 10 m ) 9 × 10 kg/m
area metric ton
−⎛ ⎞
= = ÷ =⎜ ⎟
⎝ ⎠
1-77. 55 mi/h = 55 mi/h × 1.609 km/mi 88.5 km/h=
= 55 mi/h × 5280 ft/mi × 1/3600 h/s 80.7 ft/s=
= 88.5 km/h × 1000 m/km × 1/3600 h/s 24.6 m/s=
1-78.
26
3 15 3
3
3 3(9.5 × 10 kg)
Density
4 4 4 (4.6 × 10 m)
3
m m m
V RR π ππ
−
−
= = = = = 2.33 × 1017
kg/m3
. In metric tons per
cm3
this is
2
17
3 3 2
kg 1 metric ton 1 m
2.33 × 10 × ×
m 10 kg 10 cm
⎛ ⎞
⎜ ⎟
⎝ ⎠
= 2.3 × 108
metric tons/cm3
.
1-79. 120 yr × 365 days/yr + 237 days = 44067 days. 9h 3600 s
44067 days × 24 × 3.81 × 10 s
day h
=
1-80.
1
slope × 300 m × 300 m = 33.3 m.
9
h = =
2 2
(33.3 m) (300 m) 302 m.d = + =
h
d
300 m

11. CHAPTER 1
11
1-81. The distance traveled by the plane gliding:
x = 5 × 103
/tan 15°
= 18.7 × 103
m 18.7 km.=
Therefore the pilot can reach San Francisco.
1-82. θ = cos−1
R
Rth
⎛ ⎞
⎜ ⎟
⎝ ⎠
= cos−1
6
6 3
6.4 × 10
6.4 × 10 2.3 × 10
⎛ ⎞
⎜ ⎟
+⎝ ⎠
= 1° = 0.027 rad
τ = Rθ = 6.4 × 106
× 0.027
= 1.71 × 105
m
= 171 km
†1-83. The arc AB has a length of 3900 km, so the half angle
θ is 1950 km/RE, where RE is the radius of the earth.
Using the value from Table 1-1 gives
θ = (1950 km)/(6.4 × 103
km) = 0.305 radian = 17.5°.
(a)The linear distance d from A to B is
3
2 sin 2(6.4 × 10 km)(sin17.5 )Ed R θ= = ° = 3840
km.
(b)The depth h at the midpoint is
cos (1 cos ),E E Eh R R Rθ θ= − = − which gives
3
(6.4 × 10 km)(1 cos17.5 )h = − ° = 296 km.
(c) “Horizontal” means parallel to the surface of the earth, or tangent to the surface of the earth at
that location. The tangent line is shown in the diagram, and the slope angle φ between the tangent
and the line AB is shown. Since the radius OA is perpendicular to the tangent line and the radius
OC is perpendicular to the line AB, the angle φ must be the same as θ. Thus φ = 17.5° and the
slope is the tangent of that angle: slope = tan 17.5° = 0.315. Using the ratio form of slope gives
slope = 1:(0.315)−1
= 1:3.2. (This is a quite steep slope!) Note that to observers standing at each
end, the tunnel appears to be sloping down into the earth.
RE
RE
d/2
θ
A
B
h
φ
O
C
θ