16. Since a change of longitude equal to 360 corresponds to a 24 hour change, then one expects to change longitude by 360 /24 = 15 before resetting one's watch by 1.0 h.
13. None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important is that the clock advance by the same amount in each 24-h period. T
12. The number of seconds in a year is 3.156107 . This is listed in Appendix D and results from the product (365.25 day/y)(24 h/day)(60 min/h)(60 s/min) . (a) The number of shakes in a second is 108 ; therefore, there are indeed more shakes per secon
19. If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then ME = N m or N = ME /m. We convert mass m to kilograms using Appendix D (1 u = 1.661 10-27 kg). Thus, ME 5.98 1024 kg N= = = 9.0 1049 . m
11. We use the conversion factors given in Appendix D and the definitions of the SI prefixes given in Table 12 (also listed on the inside front cover of the textbook). Here, "ns" represents the nanosecond unit, "ps" represents the picosecond unit, an
10. The metric prefixes (micro (), pico, nano, . . .) are given for ready reference on the inside front cover of the textbook (also, Table 1-2). 1 century = = 10-6 century 52.6 min . 100 y 1 century 365 day 1y 24 h 1 day 60 min 1h
The percent differ
15. We convert meters to astronomical units, and seconds to minutes, using 1000 m = 1 km 1 AU = 1.50 108 km 60 s = 1 min . Thus, 3.0 108 m/s becomes 3.0 108 m s 1 km 1000 m AU 1.50 108 km 60 s min = 0.12 AU/min .
17. The last day of the 20 centuries is longer than the first day by (20 century)(0.001 s/century) = 0.02 s . The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the cumulat
18. We denote the pulsar rotation rate f (for frequency). f= 1 rotation 1.55780644887275 10-3 s
(a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if we ignore significant figure considerations for a moment), we
14. The time on any of these clocks is a straight-line function of that on another, with slopes = 1 and y-intercepts = 0. From the data in the figure we deduce tC tB = = 2 594 tB + 7 7 33 662 tA - . 40 5
These are used in obtaining the following res
20. To organize the calculation, we introduce the notion of density (which the students have probably seen in other courses): m = . V (a) We take the volume of the leaf to be its area A multiplied by its thickness z. With density = 19.32 g/cm3 and m
21. We introduce the notion of density (which the students have probably seen in other courses): = and convert to SI units: 1 g = 1 10-3 kg. (a) For volume conversion, we find 1 cm3 = (1 10-2 m)3 = 1 10-6 m3 . Thus, the density in kg/m3 is 1g cm3
22. The volume of the water that fell is V = = = = (26 km2 )(2.0 in.) (26 km2 ) 1000 m 1 km
2
(2.0 in.)
0.0254 m 1 in.
(26 106 m2 )(0.0508 m) 1.3 106 m3 .
We write the mass-per-unit-volume (density) of the water as: = m 3 = 1 103 kg/m . V
The
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 )(3281 ft/km)2 (1/6 ft) = 4.66 107 ft3 .
23. We introduce the notion of density (which the students have probably seen in other courses): = m V
and convert to SI units: 1000 g = 1 kg, and 100 cm = 1 m. (a) The density of a sample of iron is therefore = 7.87 g/cm
3
1 kg 1000 g
100 cm 1m
8. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area A = 20 12 = 240 m2 ) in addition to a rectangular box (height h = 6.0 m and same base). Therefore, V = 1 hA + h A = 2 h +h 2 A = 1800 m3 .
(a)
7. The volume of ice is given by the product of the semicircular surface area and the thickness. The semicircle area is A = r2 /2, where r is the radius. Therefore, the volume is V = 2 r z 2
where z is the ice thickness. Since there are 103 m in 1
6. (a) Using the fact that the area A of a rectangle is widthlength, we find Atotal = = = (3.00 acre) + (25.0 perch)(4.00 perch) (40 perch)(4 perch) (3.00 acre) + 100 perch2 1 acre 580 perch2 .
We multiply this by the perch2 rood conversion factor
40. (a) When is measured in radians, it is equal to the arclength s divided by the radius R. For a very large radius circle and small value of , such as we deal with in Fig. 1-9, the arc may be approximated as the straight line-segment of length 1 A
39. Using the (exact) conversion 2.54 cm = 1 in. we find that 1 ft = (12)(2.54)/100 = 0.3048 m (which also can be found in Appendix D). The volume of a cord of wood is 8 4 4 = 128 ft3 , which we convert (multiplying by 0.30483 ) to 3.6 m3 . Therefo
38. Although we can look up the distance from Cleveland to Los Angeles, we can just as well (for an order of magnitude calculation) assume it's some relatively small fraction of the circumference of Earth which suggests that (again, for an order of
37. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain 1 ken2 1.972 m2 = = 3.88 . 2 1m 1 m2 (b) Similarly, we find 1.973 m3 1 ken3 = = 7.65 . 1 m3 1 m3 (c) The volume of a cylinder is the circular area of its base multiplie
36. (a) For the minimum (43 cm) case, 9 cubit converts as follows: (9 cubit) 0.43 m 1 cubit = 3.9 m .
And for the maximum (43 cm) case we obtain (9 cubit) 0.53 m 1 cubit = 4.8 m .
(b) Similarly, with 0.43 m 430 mm and 0.53 m 530 mm, we find 3.9
35. (a) When is measured in radians, it is equal to the arclength divided by the radius. For very large radius circles and small values of , such as we deal with in this problem, the arcs may be . . approximated as . R Sun .. . . . . . . . . . . . .
34. (a) We find the volume in cubic centimeters (193 gal) 231 in3 1 gal 2.54 cm 1 in
3
= 7.31 105 cm3
and subtract this from 1 106 cm3 to obtain 2.69 105 cm3 . The conversion gal in3 is given in Appendix D (immediately below the table of Volume
33. (a) In atomic mass units, the mass of one molecule is 16 + 1 + 1 = 18 u. Using Eq. 1-9, we find (18 u) 1.6605402 10-27 kg 1u = 3.0 10-26 kg .
(b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of wate
32. The mass in kilograms is (28.9 piculs) 100 gin 1 picul 16 tahil 1 gin 10 chee 1 tahil 10 hoon 1 chee 0.3779 g 1 hoon
which yields 1.747 106 g or roughly 1750 kg.