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Unformatted text preview: 1 Physics and Measurement CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time 1.2 Matter and ModelBuilding 1.3 Dimensional Analysis 1.4 Conversion of Units 1.5 Estimates and Orderof Magnitude Calculations 1.6 Signifi cant Figures ANSWERS TO QUESTIONS * An asterisk indicates an item new to this edition. Q1.1 Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard. Q1.2 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms *Q1.3 In the base unit we have (a) 0.032 kg (b) 0.015 kg (c) 0.270 kg (d) 0.041 kg (e) 0.27 kg. Then the ranking is c = e > d > a > b Q1.4 No: A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimension ally correct. Yes: If an equation is not dimensionally correct, it cannot be correct. *Q1.5 The answer is yes for (a), (c), and (f ). You cannot add or subtract a number of apples and a number of jokes. The answer is no for (b), (d), and (e). Consider the gauge of a sausage, 4 kg / 2 m, or the volume of a cube, (2 m) 3 . Thus we have (a) yes (b) no (c) yes (d) no (e) no (f ) yes *Q1.6 41 € ≈ 41 € (1 L / 1.3 € )(1 qt / 1 L)(1 gal / 4 qt) ≈ (10 / 1.3) gal ≈ 8 gallons, answer (c) *Q1.7 The meterstick measurement, (a), and (b) can all be 4.31 cm. The meterstick measurement and (c) can both be 4.24 cm. Only (d) does not overlap. Thus (a) (b) and (c) all agree with the meterstick measurement. *Q1.8 0.02(1.365) = 0.03. The result is (1.37 ± 0.03) × 10 7 kg. So (d) 3 digits are signifi cant. 1 SOLUTIONS TO PROBLEMS Section 1.1 Standards of Length, Mass, and Time P1.1 Modeling the Earth as a sphere, we fi nd its volume as 4 3 4 3 6 37 10 1 08 10 3 6 3 21 3 π π r = × ( ) = × . . m m . Its density is then ρ = = × × = × m V 5 98 10 1 08 10 5 52 10 24 21 3 3 3 . . . kg m kg m . This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kg / m 3 . The average density of the Earth is signifi cantly higher, so higherdensity material must be down below the surface. ISMV1_5103_01.indd 1 ISMV1_5103_01.indd 1 10/27/06 4:33:21 PM 10/27/06 4:33:21 PM P1.2 With V = ( )( ) base area height V r h = ( ) π 2 and ρ = m V , we have ρ π π = = ( ) ( ) m r h 2 2 9 1 19 5 39 0 10 1 kg mm mm mm m 3 3 . . ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × ρ 2 15 10 4 3 . . kg m P1.3 Let V represent the volume of the model, the same in ρ = m V for both. Then ρ iron kg = 9 35 . V and ρ gold gold = m V . Next, ρ ρ gold iron gold kg = m 9 35 . and m gold 3 3 3 kg 19.3 10 kg/m kg/m = × × ⎛ 9 35 7 86 10 3 . . ⎝ ⎜ ⎞ ⎠ ⎟ = 23 0 . kg ....
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This note was uploaded on 11/09/2010 for the course PHYS 208 taught by Professor Smith during the Spring '10 term at CUNY City Tech.
 Spring '10
 Smith
 Physics, Mass

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