###### tags: `physics solution`
# measurement
## Problem 1-6
1. Earth is approximately a sphere of radius $6.37\times 10^6m$. what are (a) its circumference in kilometers, (b) its surface area in square kilometers,and (ｃ) its volume in cubic kilometers?
> formula : $circumference=2r\pi、sureface\ area=4r^2\pi、volume=\frac{4}{3}r^3\pi$
> Ans:(a)$4.002\times 10^4km$、(b)$5.099\times 10^8km^2$、(ｃ)$1.083\times 10^{12}$
2. A $gry$ is an old English measure for length, defined as 1/10 of a $line$, where $line$ is another English measure for length, defined as 1/12 inch. A common measure for length in the publishing business is a $point$, defined as 1/72 inch. What is an area of 0.75 $gry^2$ in $points\ squared$ ($points^2$)
> $1\ gry=1/10\ line=1/120\ inch=3/5\ point\rightarrow 0.75\ gry^2=0.75*(3/5)^2points^2=0.27points^2$
> Ans:$0.27points^2$
3. The micrometer($1\mu m$) is often called the micron. (a)How many mictons make up $1.0km$? (b)What fraction of a centimeter equals $1.0\mu m$? (ｃ)How many microns are in 1.0 yd ($0.9144m$)?
> Ans:(a)$1.0\times 10^9$、(b)$1.0\times 10^{-4}$、(c)$9.144\times 10^5$
4. Spacing in this book was generally done in units of points and picas: 12 points = 1 pica, and 6 picas = 1 inch. If a figure was misplaced in the page proofs by $0.70cm$, what was the misplacement in (a)picas and (b)points?
> $1\ inch=2.54\ cm\rightarrow1cm=1/2.54\ inch\rightarrow 0.7cm =0.7/2.54\times 6picas\simeq1.653picas$ then $0.7cm=0.7/2.54\times 72points\simeq19.842points$
> Ans:(a)$1.653picas$、(b)$19.842points$
5. Horses are to race over a certain English meadow for a distance of $4.0$ furlong. What is the race distance in (a)rods and (b)chains?($1\ furlong=201.168m;1\ rod =5.0292m$, and $1\ chain = 20.117m$)
> $1 furlong : 1 rod : 1 chain=201.168:5.0292:20.117\rightarrow 4.0furlong=160rod=40chain$
> Ans:(a)$160rod$、(b)$40chain$
6. You can easily convert common unit and measures electronically. but you still should be able to use a conversion table, such as those in Appendix D. Table 1-6 is part of a conversion table for a system of volume measure once common in Spain; a volume of $1fanega$ is equivalent to $55.501dm^3$(cubic decimeters). To complete the table, what numbers(to three significant figures) should be enterd in (a) the cahiz column, (b)the fanega column, ( c )the cuartilla column, and (d)the almude column, starting with the top blank? Express $7.0$ almudes in (e)medios, (f)cahizes, and (g)cubic centimeters($cm^3$) 
> Ans:

| table 1-6 | cahiz | fenega | cuartilla | almude | medio |
| ------------ | ----- | ------ | --------- | ------ | ----- |
| 1 cahiz= | 1 | 12 | 48 | 144 | 288 |
| 1 fanega= | 1/12 | 1 | 4 | 12 | 24 |
| 1 cuartilla= | 1/48 | 1/4 | 1 | 3 | 6 |
| 1 almude= | 1/144 | 1/12 | 1/3 | 1 | 2 |
| 1 medio= | 1/288 | 1/24 | 1/6 | 1/2 | 1 |
> (e)$14.0medio$、(f)$7/144cahizes$、(g)$32.376\times 10^3 cm^3$
## Problem 7-17

7. Ans:
> $1 acre-foot=1acre \times 1foot \\ \rightarrow 26km^2 \times 2.0in=26\times 247.105acre\times 2.0 \times 1/12ft=1070.788acre-foot$
8. Ans:
> $212s=258w$、$180s=156z$,$\rightarrow 50.0 smoots=60.849w=43.333z$
9. Ans:
> $V=area\times Length=\frac{1}{2}(2\times 10^6m)^2\pi\times 3\times 10^3m=6\times 10^{21}\pi\ cm^2\simeq1.9\times 10^{22}$
10. Ans:
> $360^\circ/24=15^\circ$
11. Ans:
> (a)$1day=1day\rightarrow 1French\ decimal\ week(10fdw)=10day\rightarrow 1fdw=\frac{10}{7}sw$
(b)$86400s=100000fds\rightarrow 1fds=0.864s$
Thus, (a)$\frac{10}{7}$(b)$0.864$
12. Ans:
>　$3.7m/14day=3700ms/14\times 86400\times 1000=3.059\times 10^{-6}$
13. Ans:
> $200A(s)=165B(s)、175B(s)=50C(s)\rightarrow 600A(s)=495B(s)-(a)=141.429C(s)-(b)$
$312+88A(s)=125+x,x=88\times 165/200=72.6\rightarrow 125+72.6=197.6B(s)-(c)$
$15C(s)=92-77C(s)=25-269.5B(s)=-244.5B(s)-(d)$
14. Ans:
> $10^{-6} century=10^{-6}\times (100\times 365+24)24\times 60=52.595min-(a)$
$\frac{52.595-50}{52.595}\times 100\%=4.934\%-(b)$
15. Ans:
> $2\times 7\times 86400\times 10^3s=1.2096\times 10^{12}\mu s$
16. Ans:
> $(a)8\times 86400\times 10^3/1.57780644887275=438,076,546.394$
$(b)1.57780644887275\times 10^6=1577806.44887275ms$
$(c)\pm 3\times 10^{-17}s$
17. Ans:
> CDABE
important criterion is the consistency of the daily variation, not its magnitude
## Problem 18-31

18. Ans:
> $1.0ms\times 30 =30ms$
19. Ans:
> 
> $\frac{r}{r+h}=cos(\theta), \theta=360^\circ\times\frac{11.1}{86400}\rightarrow \frac{r}{r+h}=0.999999674\rightarrow r=0.999999674h/0.000000326=5.215\times 10^6 m$
20. Ans:
> (a)$193\times 3.785=730.505L=730.505\times 1000cm^3=0.7305 million\ cm^3$
> (b)$0.7305kL\times 1000kg/m^3=730.505kg=730505g\rightarrow 730505/1.5=487,003.333min$
21. ans:
> $5.98\times 10^{24}/40\times 1.66\times 10^{-27}=9.006\times 10^{49}$
22. ans:
> (a)$V=area\times l\rightarrow \frac{29.34}{19.32\times 1.0\times 10^{-4}}\simeq 15,186.335cm^2$
> (b)$V=area\times l\rightarrow\frac{29.34}{19.32\times (2.5\times 10^{-4})^2\pi}\simeq 7,734,343.643cm$
23. ans:
> (a)$1g/cm^3=0.001kg/(0.01)^3m=1000kg/m^3$
> (b)$5700\times 1000/10\times 60\times 60\simeq158.333kg/s$
24. ans:
> $6/(1.2\times 10^{-4})^2\pi\times\frac{4}{3}(6\times 10^{-5})^3\pi\times 2600=0.312kg$
25. ans:
> $V=area\times l\rightarrow 2500\times 800\times 2 /400\times 400=25\rightarrow 25\times 4.0\times 1900=1.9\times 10^5kg$
26. ans:
> $50\thicksim 500/cm^3\rightarrow$
> (a)$10^3\times 10^3\times\pi\times 3\times 10^3=3\pi\times 10^9m^3=3\pi\times 10^{15}cm^3 \\ \rightarrow(lower):4.712\times 10^{17} water\ drops;(hight):4.712\times 10^{18} water\ drops$
> (b)、$(c)$ $4.712\times 10^{17}\times\frac{4}{3}(10^{-5})^3\pi=1.974\times 10^3m^3=1.974\times 10^6kg=1.974\times 10^6 L$
27. ans:
> (a)$9.27\times 10^{-26}\times 1000/7.87=1.178\times 10^{-23}cm^3$
> (b)$V=\frac{4}{3}r^3\pi=1.178\times 10^{-23}\rightarrow r=1.411\times 10^{-8}\rightarrow d=2r=2.822\times 10^{-8}cm$
28. ans:
> ???
29. ans:
> $0.3799\times 10\times 10\times 16\times 100\times 28.9=1756.657kg$
30. ans:
> (a)$\frac{d}{dt} 5.0t^{0.8}-3.00t+20.00=4t^{-0.2}-3=0\rightarrow t=(\frac{4}{3})^5=4.213$
> (b)$t=(\frac{4}{3})^5\rightarrow 5.0t^{0.8}-3.00t+20.00=5\times(\frac{4}{3})^4-3\times 4.213+20.00=23.163g$
> $(c)(d)$ $4t^{\frac{-1}{5}}-3.00\rightarrow \\ t=3,v=0.211g/s=1.266\times 10^{-3}kg/min ; t=5,v=-6.05\times 10^{-3}kg/min$
31. ans:
> $14\times 17\times0.25\times 0.02/(5\times 10^{-2})=23.8g/s=1.428kg/min$