TEST BANK FOR Fundamentals of Physics 7th Edition David Halliday, Robert Resnick, Jearl Walker
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1. Using the given conversion factors, we find
(a) the distance d in rods to be
(4.0 furlongs)(201.168 m furlong)
4.0 furlongs = 160 rods,
5.0292 m rod
d= =
(b) and that distance in chains to be
(4.0 furlongs)(201.168 m furlong)
40 chains.
20.117 m chain
d= =
2. The conversion factors 1 gry =1/10 line , 1 line=1/12 inch and 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 = 103 m = (103 m)(106 μ m 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 = 10−2 m = (10−2m)(106 μ m 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,
1.0 yd = (0.91m)(106 μ m m) = 9.1 × 105 μm.
4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we
obtain
( ) 1 inch 6 picas
0.80 cm = 0.80 cm 1.9 picas.
2.54 cm 1 inch
≈
(b) With 12 points = 1 pica, we have
( ) 1 inch 6 picas 12 points
0.80 cm = 0.80 cm 23 points.
2.54 cm 1 inch 1 pica
≈
5. Various geometric formulas are given in Appendix E.
(a) Substituting
R = c6.37 × 106 mhc10−3 km mh = 6.37 × 103 km
into circumference = 2πR, we obtain 4.00 × 104 km.
(b) The surface area of Earth is
( )2 3 2 8 2 A = 4πR = 4π 6.37 × 10 km = 5.10 × 10 km .
(c) The volume of Earth is
( )4 3 4 3 3 12 3
6.37 10 km 1.08 10 km .
3 3
[Solved] TEST BANK FOR Fundamentals of Physics 7th Edition David Halliday, Robert Resnick, Jearl Walker
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- Submitted On 14 Nov, 2021 06:54:02
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