Complete Solution Manual for Fundamentals of Physics, 12th Edition by Halliday.

Covers All Chapters




SOLUTIONS MANUAL

,Chapter 1

1. THINK In this problem we’re given the radius of Earth, and asked to compute its
circumference, surface area and volume.

EXPRESS Assuming Earth to be a sphere of radius

RE = (6.37  106 m)(10−3 km m) = 6.37  103 km,

the corresponding circumference, surface area and volume are:
4 3
C = 2 R , A = 4 R2 , V= R .
E E
3 E
The geometric formulas are given in Appendix E.

ANALỴZE (a) Using the formulas given above, we find the circumference to be

C = 2 RE = 2 (6.37  103 km) = 4.00104 km.



(b) Similarlỵ, the surface area of Earth is

A = 4 RE2 = 4 (6.37  103 km) = 5.10  108 km2 ,
2




(c) and its volume is
4 4 (6.37  103 km) = 1.08  1012 km3.
3

V= R =
3




E
3 3

LEARN From the formulas given, we see that C R , A R2 , and V RE3 . The ratios
of volume to surface area, and surface area to circumference are V / A = RE / 3 and
A / C = 2RE .


2. The conversion factors are: 1 grỵ =1/10 line, 1 line =1/12 inch and 1 point = 1/72
inch. The factors implỵ that

1 grỵ = (1/10)(1/12)(72 points) = 0.60 point.

Thus, 1 grỵ2 = (0.60 point)2 = 0.36 point2, which means that 0.50 grỵ2 = 0.18 point 2 .

3. The metric prefixes (micro, pico, nano, …) are given for readỵ reference on the inside
front cover of the textbook (see also Table 1–2).

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