Ohanian 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 = 8.3 × 10−3 cm.

1-3.

100 yd × 0.914 m/yd = 91.4 m; 53 1/3 yd × 0.914 m/yd = 48.7 m
1 step
1000 m
steps
N=
×
= 1.7 × 103
0.60 m
1 km
km
1 pica
17
1 pica
L = 11 in ×
= 66 picas. W =
in ×
= 51 picas
1
1
2
in
in
6
6
−8
Virus: 2 × 10 m × 10/0.3048 ft/m × 12 in/ft = 8 × 10−7 in

1-4.
1-5.
1-6.

Similarly: Atom: 1 × 10−10 m × 39.4in/m = 4 × 10−9 in
Fe Nucleus: 8 × 10−15 m × 39.4 in/m = 3 × 10−13 in
Proton: 2 × 10−15 m × 39.4 in/m = 8 × 10−14 in
†1-7.

1-8.

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 (to three significant figures)
8
2
1
3.175 mm
in =
= 1.5875 mm = 1.59 mm (to three significant figures)
16
2
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
10−3 in 2.54 cm 10−2 m 10−6 µm
1 mil ×
×
×
×
= 25.4 µm.
mil
in
cm
m
1000 µm
1 mil
1 mm ×
×= 39.4 mil
mm
25.4 µm

1

CHAPTER
1-9.

1

(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) = 2.8 × 106 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)
1.17 × 1026
=
= 7.8 × 105 m
20
1.5 ×10

1-11.

Size (diameter) of the sun = 2 × 6.46 × 108 = 1.4 × 109 m distance on the diagram (PRELUDE, p.
1.4 × 109
6) =
10−3 m = 1 mm
1.5 × 1012

1-12.

†1-13.

10−9 m
= 6.33 × 10−13 m. According to Table 1-1, the diameter of an
nm
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 turn = 360°, so 5° × 1turn/360° = 0.0139 turn. For an English thread,
1in
0.0254 m 106 µm
0.0139 turn ×
×
×
= 4.41 µm. For a metric thread,
80 turns
in
m
∆l = 10−6 × 633 nm ×

0.5 mm 10−3 m 106 µm
×
×
= 6.94 µm.
turn
mm
m
1 nmi = 1852 m; Circumference of Earth = 4.00 × 107 m
Circumference = (4.00 × 107 m)/1852 m/nmi = 21, 600 nmi
0.0139 turn ×

1-14.

Also 360° × 60 min/deg = 21, 600 min, so1nmi ⇒ 1min

2

CHAPTER
†1-15.

For one of the triangles, (R + 1.75 m)2 = R2 + (4700 m)2. Expand
this to get R 2 + 2(1.75 m) R + (1.75 m) 2 = R 2 + (4700 m) 2 .
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.

9400

R
R + 1.75 m

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 = 7.1 × 108 s

1-17.
1-18.
1-19.

1-20.
†1-21.

1-22.
1-23.
1-24.
1-25.
1-26.

1 yr = 365.25 days. Therefore, 4.5 × 109 yr
= 4.5 × 109 yr × 365.25 day/yr × 86,400 s/day = 1.4...