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physics chapter 1

# physics chapter 1 - 1 Physics and Measurement CHAPTER...

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1 Physics and Measurement CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model-Building 1.3 Dimensional Analysis 1.4 Conversion of Units 1.5 Estimates and Order-of- Magnitude Calculations 1.6 Significant 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 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 Q1.4 No: A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. Yes: If an equation is not dimensionally correct, it cannot be correct. *Q1.5 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. *Q1.6 41 41 (1 L 1.3 )(1 qt 1 L)(1 gal 4 qt) (10 1.3) gal 8 gallons, answer (c) 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 significantly higher, so higher-density material must be down below the surface.

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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 . *P1.4 ρ = m V / and V r d d = = = ( / ) ( / ) ( / ) / 4 3 4 3 2 6 3 3 3 π π π where d is the diameter. Then ρ π π = = × × = × 6 6 1 67 10 2 4 10 2 3 3 27 15 3 m d / ( . ) ( . ) . kg m 10 17 3 kg/m 2.3 10 kg/m /(11.3 10 kg/m )= 17 3 3 3 × × it is 20 × 10 12 times the density of lead . P1.5 For either sphere the volume is V r = 4 3 3 π and the mass is m V r = = ρ ρ π 4 3 3 . We divide this equation for the larger sphere by the same equation for the smaller: m m r r r r s s s = = = ρ π ρ π 4 3 4 3 5 3 3 3 3 . Then r r s = = ( ) = 5 4 50 1 71 7 69 3 . . . cm cm . Section 1.2 Matter and Model-Building P1.6 From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a ﬂat plane. This diagonal distance may be obtained from the Pythagorean theorem, L L L diag = + 2 2 . Thus, since the atoms are separated by a distance L = 0 200 . nm, the diagonal planes are separated by 1 2 0 141 2 2 L L + = . nm .
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