Chapter 1
1. Using the given conversion factors, we find
(a) the distance d in rods to be
(b) and that distance in chains to be
2. The conversion factors
,
and 1 point = 1/72 inch
imply that 1 gry = (1/10)(1/12)(72 points) = 0.60 point. Thus, 1 gry2 = (0.
CHAPTER 2 36
28. We choose the positive direction to be that of the initial velocity of the car
(implying that a < 0 since it is slowing down). We assume the acceleration is constant
and use Table 2-1.
(a) Substituting v0 = 137 km/h = 38.1 m/s, v = 90 km/
CHAPTER 2 28
14. We use the functional notation x(t), v(t) and a(t) and find the latter two quantities
by differentiating:
with SI units understood. These expressions are used in the parts that follow.
(a) From
, we see that the only positive value of t f
25
8. The values used in the problem statement make it easy to see that the first part of
the trip (at 100 km/h) takes 1 hour, and the second part (at 40 km/h) also takes 1 hour.
Expressed in decimal form, the time left is 1.25 hour, and the distance that
1. The speed (assumed constant) is (90 km/h)(1000 m/km) (3600 s/h) = 25 m/s.
Thus, during 0.50 s, the car travels (0.50)(25) 13 m.
2. Hubers speed is
v0=(200 m)/(6.509 s)=30.72 m/s = 110.6 km/h,
where we have used the conversion factor 1 m/s = 3.6 km/h. S
CHAPTER 120
50. The volume of one unit is 1 cm3 = 1 106 m3, so the volume of a mole of them is
6.02 1023 cm3 = 6.02 1017 m3. The cube root of this number gives the edge length:
. This is equivalent to roughly 8 102 kilometers.
51. A million milligrams com
CHAPTER 118
41. (a) The difference between the total amounts in freight and displacement tons,
(8 7)(73) = 73 barrels bulk, represents the extra M&Ms that are shipped. Using the
conversions in the problem, this is equivalent to (73)(0.1415)(28.378) = 293
CHAPTER 114
33. The metric prefixes (micro (), pico, nano, ) are given for ready reference on the
inside front cover of the textbook (see also Table 12). The surface area A of each grain
of sand of radius r = 50 m = 50 106 m is given by A = 4(50 106)2 = 3
CHAPTER 18
19. We introduce the notion of density:
and convert to SI units: 1 g = 1 103 kg.
(a) For volume conversion, we find 1 cm3 = (1 102m)3 = 1 106m3. Thus, the density
in kg/m3 is
Thus, the mass of a cubic meter of water is 1000 kg.
(b) We divide th
5
11. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600
seconds. Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix,
this is roughly 1.21 1012 s.
12. A day is equivalent to 86400 seconds and a meter