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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 measureboth mass and volume very accurately in order to use the density of water as a standard. (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms 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 No: A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. Yes: If anequation is not dimensionally correct, it cannot be correct.

Q1.2

*Q1.3

Q1.4

*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 41 € ≈ 41 € (1 L 1.3€)(1 qt 1 L)(1 gal 4 qt) ≈ (10 1.3) gal ≈ 8 gallons, answer (c) 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. 0.02(1.365) = 0.03. The result is (1.37 ± 0.03) × 107 kg. So (d) 3 digits are significant.

*Q1.6 *Q1.7

*Q1.8

SOLUTIONS TOPROBLEMS
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 kgm3. The average density of the Earth is significantly higher, so higher-density material must be down below the surface.

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ISMV1_5103_01.indd 1

10/27/06 4:33:21 PM

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Chapter 1

P1.2

With V = ( base area ) ( height ) V = (π r 2 ) h and ρ =

m , we have V

ρ=

⎛ 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 Let Vrepresent the volume of the model, the same in ρ = and ρgold = mgold V . Next, mgold ρgold and mgold = ρiron 9.35 kg 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 ⎟

*P1.4

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 −15m)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 For either sphere the volume is V = 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: m ρ 4π r 3 3 r 3 = = = 5. ms ρ 4π rs3 3 rs3 Then r = rs 3 5 = 4.50 cm (1.71) = 7.69 cm .

Section 1.2 P1.6

Matter andModel-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 P1.7 (a) (b) P1.8 (a)(b) (c)

Dimensional Analysis This is incorrect since the units of [ ax ] are m 2 s 2 , while the units of [ v ] are m s .
−1 This is correct since the units of [ y ] are m, and cos ( kx ) is dimensionless if [ k ] is in m .

Circumference has dimensions of L. Volume has dimensions of L3 . Area has dimensions of L2 . Expression (i) has dimension L ( L2 )
1/ 2

= L2 , so this must be...
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