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Complete Solution Manual Fundamentals of Physics Extended 10th Edition Halliday Questions & Answers with rationales

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Fundamentals of Physics Extended 10th Edition Halliday Solutions Manual Complete Solution Manual Fundamentals of Physics Extended 10th Edition Halliday Questions & Answers with rationales PDF File All Pages All Chapters Grade A+

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  • June 19, 2023
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  • 2022/2023
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  • fundamentals of physics extended 10th edition
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  • Complete Solution Manual Fundamentals of Physics Extended 10th Edition Halliday Questions & Answers with rationales
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1 Chapter 1 1. 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   6 3 36.37 10 m 10 km m 6.37 10 km,ER    the corresponding circumference, surface area and volume are: 23 42 , 4 ,3E E E C R A R V R   . The geometric formulas are given in Appendix E. ANALYZE (a) Using the formulas given above, we find the circumference to be 342 2 (6.37 10 km) 4.00 10 km.ECR     (b) Similarly, the surface area of Earth is  22 3 8 24 4 6.37 10 km 5.10 10 kmEAR    
, (c) and its volume is  33 3 12 3 446.37 10 km 1.08 10 km .33EVR     LEARN From the formulas given, we see that ECR , 2
EAR
, and 3
EVR . The ratios of volume to surface area, and surface area to circumference are / /3EV A R and /2E A C R
. 2. The conversion factors are: 1 gry 1/10 line , 1 line 1/12 inch and 1 point = 1/72 inch. The factors 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). CHAPTER 1 2 (a) Since 1 km = 1  103 m and 1 m = 1  106 m,   3 3 6 91km 10 m 10 m 10 m m 10 m.   The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0  109 m. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 102 m,   2 2 6 41cm =10 m = 10 m 10 m m 10 m. We conclude that the fraction of one centimeter equal to 1.0 m is 1.0  104. (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,   651.0yd = 0.91m 10 m m 9.1 10 m.  4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain  1inch 6 picas0.80 cm = 0.80 cm 1.9 picas.2.54 cm 1inch        (b) With 12 points = 1 pica, we have  1inch 6 picas 12 points0.80 cm = 0.80 cm 23 points.2.54 cm 1inch 1pica           5. THINK This problem deals with conversion of furlongs to rods and chains, all of which are units for distance. EXPRESS Given that 1 furlong 201.168 m,
1rod 5.0292 m and 1chain 20.117 m , the relevant conversion factors are 1 rod1.0 furlong 201.168 m (201.168 m) 40 rods,5.0292 m   and 1 chain1.0 furlong 201.168 m (201.168 m) 10 chains20.117 m  
. Note the cancellation of m (meters), the unwanted unit. ANALYZE Using the above conversion factors, we find (a) the distance d in rods to be  40 rods4.0 furlongs 4.0 furlongs 160 rods,1 furlongd   3 (b) and in chains to be  10 chains4.0 furlongs 4.0 furlongs 40 chains.1 furlongd   LEARN Since 4 furlongs is about 800 m, this distance is approximately equal to 160 rods (
1 rod 5 m ) and 40 chains (
1 chain 20 m ). So our results make sense. 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 = 1
12 cahiz, or 8.33  102 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = 1
48 cahiz, or 2.08  102 cahiz. Continuing in this way, the remaining entries in the first column are 6.94  103 and 33.47 10
. (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) Usin
g 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
12 fanega = 7.00
12 (55501 cm3) = 3.24  104 cm3. 7. We use the conversion factors found in Appendix D. 231 acre ft = (43,560 ft ) ft = 43,560 ft Since 2 in. = (1/6) ft, the volume of water that fell during the storm is 2 2 2 7 3(26 km )(1/6 ft) (26 km )(3281ft/km) (1/6 ft) 4.66 10 ft .V    Thus, V
   466 10
43560 1011107
43 .
..ft
ftacreftacre ft.3
3 CHAPTER 1 4 8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is e
quivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z. (a
) In units of W, we have  258 W50.0S 50.0S 60.8 W212S  (b) In units of Z, we have  156 Z50.0S 50.0S 43.3 Z180S  9. 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 2
2V r z where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have  32
5 10 m 10 cm2000km 2000 10 cm.1km 1mr             In these units, the thickness becomes  2
2 10 cm3000m 3000m 3000 10 cm1mz    which yields    25 2 22 32000 10 cm 3000 10 cm 1.9 10 cm .2V     10. 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. 11. (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. 12. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so

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