Fundamentals of Physics, Seventh Edition Solution Manual

4345 Pages·2007·31.27 MB·english

by  David Halliday

1. Using the given conversion factors, we find

(a) the distance d in rods to be

( )( )
4.0 furlongs 201.168 m furlong
d = 4.0 furlongs = =160 rods,
5.0292 m rod

(b) and that distance in chains to be

( )( )
4.0 furlongs 201.168 m furlong
d = = 40 chains.
20.117 m chain
2. The conversion factors 1 gry=1/10 line, 1 line=1/12 inchand 1 point = 1/72 inch
imply that 1 gry = (1/10)(1/12)(72 points) = 0.60 point. Thus, 1 gry2 = (0.60 point)2 =
0.36 point2, which means that 0.50 gry2= 0.18 point2.
3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside
front cover of the textbook (see also Table 1–2).
(a) Since 1 km = 1 × 103 m and 1 m = 1 × 106 µm,
( )( )
1km =103m = 103m 106 µm m =109 µm.
The given measurement is 1.0 km (two significant figures), which implies our result
should be written as 1.0 × 109 µm.
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−2 m,
( )( )
1cm =10−2m = 10−2m 106µm m =104 µm.
We conclude that the fraction of one centimeter equal to 1.0 µm is 1.0 × 10−4.
(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
1.0yd = (0.91m)(106µm m) = 9.1×105 µm.
4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we
obtain
( ) (cid:167) 1inch (cid:183)(cid:167)6 picas(cid:183)
0.80 cm = 0.80 cm ≈1.9 picas.
(cid:168) (cid:184)(cid:168) (cid:184)
2.54 cm 1inch
(cid:169) (cid:185)(cid:169) (cid:185)
(b) With 12 points = 1 pica, we have
( ) (cid:167) 1inch (cid:183)(cid:167)6 picas(cid:183)(cid:167)12 points(cid:183)
0.80 cm = 0.80 cm ≈ 23 points.
(cid:168) (cid:184)(cid:168) (cid:184)(cid:168) (cid:184)
2.54 cm 1inch 1pica
(cid:169) (cid:185)(cid:169) (cid:185)(cid:169) (cid:185)
5. Various geometric formulas are given in Appendix E.
(a) Substituting
c hc h
R = 6.37 × 106m 10−3km m = 6.37 ×103 km
into circumference = 2πR, we obtain 4.00 × 104 km.
(b) The surface area of Earth is
( )
A=4πR2 = 4π 6.37 ×103 km 2 = 5.10 ×108 km2.
(c) The volume of Earth is
V = 4π R3 = 4π (6.37 ×103 km)3 =1.08×1012 km3.
3 3
6. We make use of Table 1-6.
(a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz?
We note from the already completed part of the table that 1 cahiz equals a dozen fanega.
Thus, 1 fanega = 1 cahiz, or 8.33 × 10−2cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the
12
already completed part) implies that 1 cuartilla = 1 cahiz, or 2.08 × 10−2 cahiz.
48
Continuing in this way, the remaining entries in the first column are 6.94 × 10−3 and
3.47×10−3.
(b) In the second (“fanega”) column, we similarly find 0.250, 8.33 × 10−2, and 4.17 × 10−2
for the last three entries.
(c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries.
1
(d) Finally, in the fourth (“almude”) column, we get = 0.500 for the last entry.
2
(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our
amount of 7.00 almudes must be equal to 14.0 medios.
(f) Using the value (1 almude = 6.94 × 10−3cahiz) found in part (a), we conclude that
7.00 almudes is equivalent to 4.86 × 10−2 cahiz.
(g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501
m3 or 55501 cm3. Thus, 7.00 almudes = 7.00 fanega = 7.00(55501 cm3) = 3.24 × 104 cm3.
12 12
7. The volume of ice is given by the product of the semicircular surface area and the
thickness. The are of the semicircle is A = πr2/2, where r is the radius. Therefore, the
volume is
π
V = r2 z
2
wherez is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have
( ) (cid:167)103m(cid:183) (cid:167)102cm(cid:183)
r = 2000km = 2000 ×105 cm.
(cid:168) (cid:184) (cid:168) (cid:184)
1km 1m
(cid:169) (cid:185) (cid:169) (cid:185)
In these units, the thickness becomes
( ) (cid:167)102cm(cid:183)
z=3000m= 3000m = 3000×102 cm
(cid:168) (cid:184)
1m
(cid:169) (cid:185)
which yields,
π ( ) ( )
V = 2000 ×105 cm 2 3000 ×102cm =1.9 ×1022 cm3.
2
8. From Figure 1.6, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is
equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z.
(a) In units of W,
( ) (cid:167)258 W(cid:183)
50.0 S = 50.0 S = 60.8 W
(cid:168) (cid:184)
212 S
(cid:169) (cid:185)
(b) In units of Z,
( ) (cid:167)156 Z(cid:183)
50.0 S = 50.0 S = 43.3 Z
(cid:168) (cid:184)
180 S
(cid:169) (cid:185)
9. We use the conversion factors found in Appendix D.
1 acre⋅ft = (43,560 ft2)⋅ft = 43,560 ft3
Since 2 in. = (1/6) ft, the volume of water that fell during the storm is
V =(26 km2)(1/6 ft)=(26 km2)(3281ft/km)2(1/6 ft) = 4.66×107 ft3.
Thus,
4.66 × 107 ft3
V = = 1.1× 103 acre⋅ ft.
4.3560 × 104 ft3 acre⋅ft
10. The metric prefixes (micro (µ), pico, nano, …) are given for ready reference on the
inside front cover of the textbook (also, Table 1–2).
(a)
( ) (cid:167) 100 y (cid:183) (cid:167)365 day(cid:183) (cid:167) 24 h (cid:183) (cid:167)60 min(cid:183)
1µcentury = 10−6 century = 52.6 min.
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
1century 1 y 1day 1h
(cid:169) (cid:185) (cid:169) (cid:185) (cid:169) (cid:185) (cid:169) (cid:185)
(b) The percent difference is therefore
52.6 min − 50 min
= 4.9%.
52.6 min