Answers to Even Numbered Conceptual QuestionsAnswers to Even Numbered ProblemsProblem SolutionsChapter 1 Introduction Answers to Even Numbered Conceptual Questions 2. Atomic clocks are based on the electromagnetic waves that atoms emit. Also, pulsars are highly regular astronomical clocks. 4. (a) ~ ≈-10.5 lb 0.25 kg or ~10 kg (b) ≈0~4 lb 2 kg or ~10 kg (c) ~ ≈34000 lb 2000 kg or ~10 kg 6. Let us assume the atoms are solid spheres of diameter 10 . Then, the volume of each atom is of the order of 10 . (More precisely, volume = −10 m-30 3 mππ341 = 36r3d.) Therefore, since 1 , the number of atoms in the solid is on the order of =3-6 cm 10 m3−−6243010= 1010 atoms. A more precise calculation would require knowledge of the density of the solid and the mass of each atom. However, our estimate agrees with the more precise calculation to within a factor of 10. 8. Realistically, the only lengths you might be able to verify are the length of a football field and the length of a housefly. The only time intervals subject to verification would be the length of a day and the time between normal heartbeats. 10. On the average, a typical person drives about 10 000 miles per year or about 30 miles per day. (A reasonable estimate would be in the range of 5 to 50 miles per day.) 12. In general, the manufacturers are using significant figures correctly. The length and width of the aluminum foil both include three significant figures and the cited value of their product (the area) correctly includes three significant figures in both the metric and English versions. The manufacturer of the tape also gives three significant figures in the dimensions of the tape (metric version) and retains three significant figures in their product (area). However, the width of the tape is given in English units as ″12 , which gives no indication of the accuracy of the measurement. To be consistent with the three significant figure accuracy cited in the metric width dimension, the width should be given as . ″0.50012 CHAPTER 1 Answers to Even Numbered Problems 2. (a) 2LT (b) L 4. All three equations are dimensionally incorrect. 6. (a) (b) -2MLT ⋅2kg m s 8. ±22209 cm 4 cm10. (a) ×8 3.00 10 m s (b) ×8 2.997 9 10 m s (c) 8 2.997 924 10 m s× 12. (a) (b) ±2346 m 13 m2±2266.0 m 1.3 m14. (a) (b) ×92.96 10−×26.876 10 16. (a) , , ×25.60 10 km ×55.60 10 m ×75.60 10 cm (b) , , 0.4912 km 491.2 m ×44.912 10 cm (c) , 6.1 , 6.192 km ×392 10 m ×56.192 10 cm (d) , , 2.499 km ×32.499 10 m ×52.499 10 cm18. 9.2 nm s 20. ×9 310yr22. , ×23 2.9 10 m ×83 2.9 10 cm24. ×632.57 10 m26. (a) =1 mi h 1.609 km h (b) 88 km h (c) 16 km h 28. It would require about 47.6 yr to count the money. We advise against it. 30. of beef, ~ head of cattle (assumes 0.25 lb per burger and a net of 300 lb of meat per head of cattle) 10~10 lb71032. (assumes 7~10 blades21 16 in per blade) 34. 10~10 cans yr , 5~10 ton yr (Assumes an average of 1 can per person each week, a population of 250 million, and 0.5 oz of aluminum per can) 36. 2.2 mIntroduction 3 38. 8.1 cm 40. 2.33 m42. (a) 1.50 m (b) 2.60 m 44. 8.60 m 46. 70.0 m 48. (a) 30.677 g cm (b) ×17 24.63 10 ft50. (a) (b) 2~10 kg3~10 kg52. (a) (b) 2~10 yr4~10 times54. (a) 4 (b) 84 CHAPTER 1 Problem Solutions 1.1 (a) The units of volume, area and height are: , [] , and =3[] LV =2LA=[] We then observe that Lh=3LL2L or =[] Thus, the equation [][]VAh=V is dimensionally correct. Ah (b) ()ππ== =22cylinderhRhVR , where Ahπ=2AR ()==AA rectangular boxVwhw=hAh, where == ×AAwlengthwidth 1.2 (a) From xB, we find that =2t=2xBt. Thus, B has units of ==2[][]xt[]B2LT (b) If ()π= sin 2 ftxA , then ()π= [][sin2 ][]Axft But the sine of an angle is a dimensionless ratio. Therefore, ==][] [Ax L 1.3 Substituting in dimensions, we have ==22LT TLT=T Thus, the dimensions are consistent 1.4 In the equation =+2211022mv mv mgh , == =222202LM[][]MTTmv mvL while ==122LMMLTTmghL. Thus, the equation is dimensionally incorrect In , =+20vv at ==0L[ ]Tvv[] but ()==22 22L[][] T LTat a t=[] . Hence, this equation is d y incorrectimensionallIntroduction 5 In the equation , we see that =2ma v== =2LM[][]MTTma m a2L[] while ==2222LL[]TTv Therefore, this equation is also dimensionally incorrect )22∝aF1.5 (()⋅2kgkg m=smG, so cross-multiplying gives the units of G as ⋅32mkg s 1.6 (a) Given that m, we have ∝Fma . Therefore, the units of force are those of ma, ()== =2[] [ ] MLTFma =[ ][]ma-2MLT (b) newton = ⋅2kg ms 1.7 (a) has ±78.9 0.2 3 significant figures (b) 3.7 has ×988 10 4 significant figures (c) 2.4 has −×66 10 3 significant figures (d) 0 has −=×3.0032 3.2 10 2 significant figures 1.8 . Multiplying out this product of binominals gives ()()== ± ±A 21.3 0.2 cm 9.8 0.1 cmAw() ()()()+20.2 0.1 cm=±±21.3 9.8 21.3 0.1 0.2 9.8A= The first term gives the best value of the area. The second and third terms add together to give the uncertainty and the fourth term is negligible in comparison to the other terms. The area and its uncertainty are found to be A±22 209 cm 4 cm 1.9 (a) The sum is rounded to 797 because 756 in the terms to be added has no positions beyond the decimal. (b) ()×=××-3.0032 356.3 3.2 10 356.3=1.140160 must be rounded to 1.1 because has only two significant figures. ×-33.2 106 CHAPTER 1 (c) π×.620 5 must be rounded to 17.66 because 5.620 has only four significant figures. 1.10 =×82.997 924 574 10 m sc (a) Rounded to 3 significant figures: c = ×8 3.00 10 m s (b) Rounded to 5 significant figures: c = ×8 2.997 9 10 m s (c) Rounded to 7 significant figures: c = 8 2.997 924 10 m s× 1.11 The distance around is +++=38.44 m 19.5 m 38.44 m 19.5 m 115.88 m , but this answer must be rounded to 115.9 m .5 m because the distance 19 carries information to only one place past the decimal. 1.12 (a) ( )()()()()ππ π== ± = ± +22210.5 m 0.2 m 10.5 m 2 10.5 m 0.2 m 0.2 mAr=2 giving A±22346 m 13 m (b) ()ππ== ±2 2 10.5 m 0.2 m =Cr ±66.0 m 1.3 m 1.13 Adding the two lengths together, we get 228.76 cm. However, 135.3 cm has only one decimal place.