Soluton_ManualFundamentals_of_Physics_Extended_10th_E dition

Soluton_ManualFundamentals_of_Physics_Extended_10th_E
dition
(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 = 12
1
cahiz, or 8.33  102 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already
completed part) implies that 1 cuartilla = 1
cahiz, or 2.08  102 cahiz. Continuing in this way,
48
the remaining entries in the first column are 6.94  103 and 3.47103 .



(b) In the second (“fanega”) column, we find 0.250, 8.33  102, and 4.17  102 for the last
three entries.



(c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries.



(d) Finally, in the fourth (“almude”) column, we get 1
2
= 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 medios.



(f) Using the value (1 almude = 6.94  103 cahiz) found in part (a), we conclude that 7.00
almudes is equivalent to 4.86  102 cahiz.



(g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 m3 or
55501 cm3. Thus, 7.00 almudes = 7.00 fanega = 7.00 (55501 cm3) = 3.24  104 cm3.
12 12

,2. 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 km 2 )(3281ft/km) 2 (1/6 ft)  4.66107 ft3.


Thus,
4.66  107 ft3
V   1.1  103 acre ft.
4.3560  10 ft acre ft
4 3




3. From Fig. 1-4, 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, we have
 258 W 
50.0 S  50.0 S  60.8 W
 
212 S
 


(b) In units of Z, we have
 156 Z 
50.0 S  50.0 S    43.3 Z
 180 S 


4. The volume of ice is given by the product of the semicircular surface area and the thickness.
The area of the semicircle is A = r2/2, where r is the radius. Therefore, the volume is

V  r2 z
2


where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have

,  103 m   102 cm  5
r  2000 km      2000  10 cm.
 1km   1m 

In these units, the thickness becomes
 102 cm  2

z  3000 m  3000 m    3000  10 cm
 1m 

2000  105 cm 3000  102 cm  1.9  1022 cm3.
2
which yields V 
2


5. 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.



6. (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.



7. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so
b3.7 mgc10 6
mm h
s dayg
 3.1  m s.



8. 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 33
t  t  594 , t  t  662 .
C
7 B 7
B
40 A 5


These are used in obtaining the following results.

, (a) We find
33
t  t 
B B
tA  t A   495 s
40
when t'A  tA = 600 s.



(b) We obtain t   t 
C
2
bt  t g 2 b495g141 s.
B B
C 7 7



(c) Clock B reads tB = (33/40)(400)  (662/5)  198 s when clock A reads tA = 400 s.



(d) From tC = 15 = (2/7)tB + (594/7), we get tB  245 s.



9. The metric prefixes (micro (), pico, nano, …) are given for ready reference on the inside
front cover of the textbook (also Table 1–2).


6
 100 y   365 day   24 h   60 min 
(a) 1 century  10 century       52.6 min .
 1 century  1y  1 day  1h 



(b) The percent difference is therefore



52.6 min  50 min  4.9%.
52.6 min


10. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600 seconds.
Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix, this is roughly 1.21
 1012 s.


11. THINK In this problem we’re given the radius of Earth, and asked to compute
its circumference, surface area and volume.


EXPRESS Assuming Earth to be a sphere of radius