(Parte 1 de 4)
1. Using the given conversion factors, we find (a) the distance d in rods to be
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 gry1/10 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 220.50 gry= 0.18 point.
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).
The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 109µm.
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−2 m,
(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain
()1inch6picas0.80cm = 0.80cm1.9picas. 2.54 cm 1 inch
(b) With 12 points = 1 pica, we have
5. Various geometric formulas are given in Appendix E. (a) Substituting intocircumference = 2πR, we obtain 4.0 × 104 km. (b) The surface area of Earth is
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 = 112cahiz, or 8.3 × 10−2cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = 148cahiz, or 2.08 × 10−2 cahiz. Continuing in this way, the remaining entries in the first column are 6.94 × 10−3and
(b) In the second (“fanega”) column, we similarly find 0.250, 8.3 × 10−2, and 4.17 × 10−2 for the last three entries.
(c) In the third (“cuartilla”) column, we obtain 0.3 and 0.167 for the last two entries.
(d) Finally, in the fourth (“almude”) column, we get 12 = 0.500 for the last entry.
(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.0 almudes must be equal to 14.0 medios.
(g) Since each decimeter is 0.1 meter, then 5.501 cubic decimeters is equal to 0.05501
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
Vr zπ =
wherez is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have
In these units, the thickness becomes
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,
(b) In units of Z,
9. We use the conversion factors found in Appendix D. 231 acreft = (43,560 ft)ft = 43,560 ft⋅⋅
ft 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).
(b) The percent difference is therefore
|seconds||Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix,|
1. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600 this is roughly 1.21 × 1012µs.
12. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so
day s day ms bg ch bg b gµ µ=
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. The clock reading can then easily be adjusted to give the correct interval. If the clock reading jumps around from one 24-h period to another, it cannot be corrected since it would impossible to tell what the correction should be. The following gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning.
Sun. Mon. Tues. Wed. Thurs. Fri. CLOCK -Mon. -Tues. -Wed. -Thurs. -Fri. -Sat.
Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best. The correction that must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to +10 s, for clock E it is in the range from -70 s to -2 s. After C and D, A has the smallest range of correction, B has the next smallest range, and E has the greatest range. From best to worst, the ranking of the clocks is C, D, A, B, E.
14. Since a change of longitude equal to 360°corresponds to a 24 hour change, then one expects to change longitude by360/2415°=° before resetting one's watch by 1.0 h.
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, but in the French system described in the problem, there would be 105 seconds. The ratio is therefore 0.864.
16. We denote the pulsar rotation rate f (for frequency).
3 1r otation
(a) Multiplying f by the time-interval t = 7.0 days (which is equivalent to 60480 s, if we ignore significant figure considerations for a moment), we obtain the number of rotations:
which should now be rounded to 3.8 × 108 rotations since the time-interval was specified in the problem to three significant figures.
(Parte 1 de 4)