Física
1 . 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 x 103 m andl m -1 x 106 pm, 1 km = 103 m = 103 m 106 g mm = 109 m. The given measurement is 1. 0 km (two significant figures), which implies our result should be written as 1. 0 x 109 pm. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10—2 m, 1 cm = 10-2 m = 10-2 m 106 v m m = 104 V m. we conclude that the fraction of one centimeter equal to 1. 0 pm is 1. 0 x 10—4. c) Since I m/ft) = 0. 9144 m, 1. 0yd-(0. 91m) 106 p mm -9. 1 x 105 p m. rgaa to view nut*ge 2. (a) Using the conversion factors 1 inch – 2. 54 cm exactly and 6 picas – 1 inch, we obtain 1 inch 6 picas 1. 9 picas. 0. 80 cm = ( 0. 80 cm ) 2. 54 cm 1 inch (b) With 12 points 1 pica, we have in chains to be ( 4. 0 furlongs 201. 168 m furlong ) 20. 117 m chain = 40 chains. 4. The conversion factors 1 gry 1/10 line , 1 line=1/12 inch and 1 point — 1/72 inch imply that 1 gry (1 points) — 0. 60 point. Thus, 1 gry2 = (0. 60 point)2 = 0. 36 point2, which means that 0. 0 gry 2 = 0. 18 point 2 5. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earth as R – ( 6. 37 x 106 m )(10—3 km m) – 6. 37 x 103 km, its circumference is s = R = (6. 37 x 103 km) = 4. 00 X104 km. (b) The surface area of Earth is A 4TTR2 — 4TT( 6. 37 x 103 km ) = 5. 10 x 108km2. 2 (c) The volume of Earth is V – x 103 km 33 = 1. 08 x 1012 km? 6. From Figure 1. 6, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is eq – 156 Z. The of the semicircle is A — nr2/2, where r is the radius.
Therefore, the volume is ri V = r2z 2 where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have r – ( 2000 km ) 103 m Ikrn 102 cm = 2000 x 105 cm. Ir-n In these units, the thickness becomes z 3000 m 3000 m ) 102 cm 3000 x 102 cm 1m which yields V – n 2000 x 105 cm 2 10 cm = 1. 9 x 1022 cm3 8. We make use of Table 1-6. (a) We look at the first (“cahifl) column: 1 fanega is equivalent to What amount of cahiz? We note fram the already completed part of the table that 1 cahiz equals a dozen fanega. 1 Thus, 1 fanega = 12 cahiz, or 8. 3 x 10-2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartllla = 1 48 cahiZ, or 2. 08 x 10-2 cahiz. Continuing in this way, th P-AGF3 tries in the first column (“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. 00 almudes must be equal to 14. 0 medias. (f) Using the value (1 almude 6. 94 x 10-3 cahiz) found in part (a), we conclude that 7. 00 almudes is equivalent to 4. 86 x 10—2 cahiz. (g) Since each decimeter is 0. meter, then 55. 501 cubic decimeters is equal to 0. 055501 7. 00 7. 00 m3 or 55501 cm3. Thus, 7. 00 almudes 12 fanega = 12 (55501 cm3) = 3. 24 x 104 cm3. 9. We use the conversion factors found in Appendix D. 1 acre ft = (43,560 ft2 ) 43,560 ft3 Since 2 in. = (1/6) ft, the volume of ater that fell during the storm is V (26 km 2 ft) (26 km 2 2 (1/6 ft) – 4. 66 *107 ft3. Thus, 4. 66 x 107 11 x 103 acre • ft. 434. 3560x 10 ftacre •ft 10. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so 3. 7 m 106 p mm — 31 p m s.. 4 day 86400 s day gc gb is 1209600 s. By definition of the micro prefix, this is roughly 1. 21 x 1012 ps. 12. The metric prefixes (micro (p), pico, nano, are given for ready reference on the inside front cover of the textbook (also, Table 1-2). (a) 1 p century = (10—6 century) 1 century 365 day 1Y24h 1 day 60 min – 52. 6 min . h (b) The percent difference is therefore 52. 6 rnjn – 50 min = 4. 9%. 52. 6 mn 13. (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. 3. (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. 14. We denote the pulsar rotation rate f (for frequency). f – 1 rotation 1. 55780644887275 x 10-3 s (a) Multiplying f by the tme-interval t = 7. 00 days (which is equivalent to 604800 s, if we ignore significant figure onsiderations for a moment), we obtain the number of rotations: I rotation N = ( 604800 s ) = 388238218. 4 1. 55780644887275 x 10—3 s which should now be rounded to 3. 8 x 108 rotations since the time-interval was specified in the problem to three significant figures. (b) We note that the problem specifies the exact number of p problem to three significant figures. (b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or x 106 = 1 rotation t 1. 55780644887275 x 10-3 s hich Yields the result t = 1557. 80644887275 s (though students who do this calculation on their calculator might not obtain those last several digits). c) Careful reading of the problem shows that the time-uncertainty per revolution is ± 3 *10— 1 7s . We therefore expect that as a result of one million revolutions, the uncertainty should *10-17 106 XIO- 11 s. 15. The time on any of these clocks is a straight-line function of that on another, with slopes 1 and y-intercepts O. From the data in the figure we deduce tc = 2 594 , +77 = 33 662 . tA- 40 5 These are used in obtaining the following results. a) We find ‘ tB -ta = when t’A -tA= 600 s. (b) we obtain tC-tC=’22 tB – 4952 141 s. 7733 b (c) Ciock B reads te = – (662/5) 198 s when ciock A reads tA 400 s. (d) From tc = 15 + (594/7), we get te z -245 s. 16. Since a chan A reads tA — 400 s. (d) From tc 15 — (2/7)tg + (594/7), we get ta z 16. Since a change of longitude equal to 3600 corresponds to a 24 hour change, then one expects to change longitude by 3600 / 24 = 1 50 before resetting one’s watch by 1. 0 h 17. 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 uality 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 intervala 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. CLOCK A g C D E Sum -Mon. 16-3-58+67+70 Mon. -Tues. -16+5 -58 +67 +55 Tues. -wed. -15-10-58+67+2 Wed. -Thurs. -17+5-58+67+20 Thurs. -Fri. -15 +6-58+67+10 Fri. -Sat. —15 —7 —58 +67 +10 Clocks C and D are both good timekeepers in the sense that each is consis -Sat -15 -7-58+67+10 Clocks C and D are both good timekeepers in the sense that each IS consistent in its daily drlft (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. 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 first day.
Slnce the increase occurs unlformly, the umulative effect T is T = ( average increase in length of a day ) ( number of days ) = 0. 01 s day 365. 25 day y ( 2000 y) – 7305 s or roughly two hours. 19. When the Sun first disappears while lying down, your line of s ght to the top of the Sun is tangent to the Earth disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth’s surface at point A shown in the figure. As you stand, elevating your eyes by a height h, the line of s ght to the Sun is tangent to the Earth’s surface at point B. Let d be the distance from point B to your eyes.
From Pythagorean theorem, we have d 2 + r 2 = (r + h) 2 – r 2 2rh + h 2 or d 2 2rh + h 2 , where r is the radius of the Earth. Since rh the second term can be dropped, leading to d 2 2rh . Now the angle between the two radii to the two tangent points A and g is e, which is also the angle through which the Sun moves about Earth during the time interval t = 11. 1 s. The value of a can be obtained by using 9 ThiS Yields 360 0 t. 24h s) = 0,046250. (24 s/min) Using d – r tan , we have d 2 —r 2 tan 2rh , or 2h tan 2 e Using the above value for e and h = 1 . 7 m, we have r 20. The density of gold is P-aGFg = 5. nd mass m 27. 63 g, the volume of the leaf is found to be V = We convert the volume to SI units: m p = 1430 cm3 V – (1. 430 cm 1m 100 cm = 1. 430 x 10-6 rn3 . Since V = Az with z = 1 x 10-6 m (metric prefixes can be found in Table 1-2), we obtain A: 1430 x 10-6 m3 (b) The volume of a cylinder of length is V = A where the cross- section area is that of a circle: A = nr2. Therefore, with r = 2. 500 x 10—6 m and V — 1. 430 x 10—6 m3, we obtain V 7. 284 x 104 m 72. 84 km. n r2 21 . We introduce the notion of density: p= and convert to SI units: 1 g — x 10-3 kg. (a) For volume conversio PAGF -ir, = (1 X 10-2ffl)3