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1 Physics and Measurement CHAPTER OUTLINE 1.1 1.2 1.3 1.4 1.5 1.6 Standards of Length, Mass, and Time Matter and Model-Building Dimensional Analysis Conversion of Units Estimates and Order-ofMagnitude Calculations 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 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 dimensionally 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 kgⲐ2 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 LⲐ1.3 €)(1 qtⲐ1 L)(1 galⲐ4 qt) ≈ (10Ⲑ1.3) 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) × 107 kg. So (d) 3 digits are significant. SOLUTIONS TO PROBLEMS Section 1.1 P1.1 Standards of Length, Mass, and Time 3 4 4 Modeling the Earth as a sphere, we find its volume as π r 3 = π ( 6.37 × 10 6 m ) = 1.08 × 10 21 m 3. 3 3 m 5.98 × 10 24 kg Its density is then ρ = = = 5.52 × 10 3 kg m 3 . This value is intermediate 21 3 V 1.08 × 10 m between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kgⲐm3. The average density of the Earth is significantly higher, so higher-density material must be down below the surface. 1 ISMV1_5103_01.indd 1 10/27/06 4:33:21 PM 2 P1.2 Chapter 1 With V = ( base area ) ( height ) V = (π r 2 ) h and ρ = ρ= m , we have V ⎛ 10 9 mm 3 ⎞ m 1 kg = 2 2 π r h π (19.5 mm ) ( 39.0 mm ) ⎜⎝ 1 m 3 ⎟⎠ ρ = 2.15 × 10 4 kg m 3 . P1.3 and ρgold = *P1.4 m for both. Then ρiron = 9.35 kg V V ⎛ 19.3 × 10 3 kg/m 3 ⎞ = 23.0 kg . = 9.35 kg ⎜ ⎝ 7.86 × 10 3 kg/m 3 ⎟⎠ Let V represent the volume of the model, the same in ρ = mgold V . Next, mgold ρgold and mgold = ρiron 9.35 kg ρ = m / V and V = (4 / 3)π r 3 = (4 / 3)π (d / 2)3 = π d 3 / 6 where d is the diameter. Then ρ = 6 m / π d 3 = 6(1.67 × 10 −27 kg) = 2.3 × 1017 kg/m 3 π (2.4 × 10 −15 m)3 2.3 × 1017 kg/m 3 /(11.3 × 10 3 kg/m 3 ) = it is 20 × 1012 times the density of lead . P1.5 4 4 3 π r and the mass is m = ρV = ρ π r 3. We divide 3 3 this equation for the larger sphere by the same equation for the smaller: For either sphere the volume is V = mᐉ ρ 4π rᐉ3 3 rᐉ3 = = = 5. ms ρ 4π rs3 3 rs3 Then rᐉ = rs 3 5 = 4.50 cm (1.71) = 7.69 cm . Section 1.2 P1.6 Matter and Model-Building 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, Ldiag = L2 + L2 . Thus, since the atoms are separated by a distance 1 2 L = 0.200 nm, the diagonal planes are separated by L + L2 = 0.141 nm . 2 Section 1.3 Dimensional Analy