44. According to Appendix D, a nautical mile is 1.852 km, so 24.5 nautical miles would be 45.374 km. Also, according to Appendix D, a mile is 1.609 km, so 24.5 miles is 39.4205 km. The difference is 5
25. The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally: (a) 11 tuffets = (1
26. If we estimate the typical large domestic cat mass as 10 kg, and the typical atom (in the cat) as 10 u 2 1026 kg, then there are roughly (10 kg)/( 2 1026 kg) 5 1026 atoms. This is close to be
23. We introduce the notion of density, = 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 c
21. 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 = Nm or N = ME/m. We convert mass m to kilograms using Appendix D (1 u = 1.661 1027 kg
20. To organize the calculation, we introduce the notion of density:
=
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 mass
19. We introduce the notion of density:
=
and convert to SI units: 1 g = 1 103 kg.
m V
(a) For volume conversion, we find 1 cm3 = (1 102m)3 = 1 106m3. Thus, the density in kg/m3 is 1g 1 g cm = c
18. 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 fi
17. 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
2 594 tB + 7 7 33 662 tB = tA . 40 5 tC =
29. (a) Dividing 750 miles by the expected 40 miles per gallon leads the tourist to believe that the car should need 18.8 gallons (in the U.S.) for the trip. (b) Dividing the two numbers given (to hig
37. 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 conv
40. (a) In atomic mass units, the mass of one molecule is 16 + 1 + 1 = 18 u. Using Eq. 1 9, we find 18u = (18u ) 1.6605402 1027 kg 1u = 3.0 1026 kg.
(b) We divide the total mass by the mass of each
45. (a) The receptacle is a volume of (40)(40)(30) = 48000 cm3 = 48 L = (48)(16)/11.356 = 67.63 standard bottles, which is a little more than 3 nebuchadnezzars (the largest bottle indicated). The rema
48. The volume of the water that fell is
V = ( 26 km
) ( 2.0 in.) = ( 26 km ) = ( 26 10 m ) ( 0.0508 m )
2 2 6 2
1000 m 1 km
2
( 2.0 in.)
0.0254 m 1 in.
= 1.3 106 m 3 .
We write the mass-per
46. The volume of the filled container is 24000 cm3 = 24 liters, which (using the conversion given in the problem) is equivalent to 50.7 pints (U.S). The expected number is therefore in the range from
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: 8.4 105 m3 . This is equivalent
47. 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 = 0
49. Equation 1-9 gives (to very high precision!) the conversion from atomic mass units to kilograms. Since this problem deals with the ratio of total mass (1.0 kg) divided by the mass of one atom (1.0
51. A million milligrams comprise a kilogram, so 2.3 kg/week is 2.3 106 mg/week. Figuring 7 days a week, 24 hours per day, 3600 second per hour, we find 604800 seconds are equivalent to one week. Thu
39. (a) For the minimum (43 cm) case, 9 cubit converts as follows: 9cubit = ( 9cubit ) And for the maximum (43 cm) case we obtain 9cubit = ( 9cubit ) 0.53m = 4.8m. 1cubit 0.43m = 3.9m. 1cubit
(b) Sim
38. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain
1 ken 2 1.97 2 m 2 = = 3.88. 1 m2 1 m2
(b) Similarly, we find 1 ken 3 197 3 m3 . = = 7.65. 3 1m 1 m3 (c) The volume o
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
28. Table 7 can be completed as follows: (a) It should be clear that the first column (under wey) is the reciprocal of the first 9 3 row so that 10 = 0.900, 40 = 7.50 102, and so forth. Thus, 1 pott
27. Abbreviating wapentake as wp and assuming a hide to be 110 acres, we set up the ratio 25 wp/11 barn along with appropriate conversion factors:
hide acre ( 25 wp ) ( 100wp ) ( 110hide ) ( 4047 m 1
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
15. (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio of weeks is simply 10/7 or (to 3 significant figures) 1.43. (b) In a regular day, there are 86400 seconds, bu
16. 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
22. (a) We find the volume in cubic centimeters
193 gal = (193 gal ) 231 in 3 1 gal 2.54 cm 1in
3
= 7.31 105 cm3
and subtract this from 1 106 cm3 to obtain 2.69 105 cm3. The conversion gal in3 i