SOLUTIONS TO THINK-PAIR-SHARE AND ACTIVITIES

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.1 Physics and Measurement

CHAPTER OUTLINE

1.1 Standards of Length, Mass, and Time 1.2 Modeling and Alternative Representations 1.3 Dimensional Analysis 1.4 Conversion of Units 1.5 Estimates and Order-of-Magnitude Calculations 1.6 Significant Figures

• An asterisk indicates a question or problem new to this edition.

SOLUTIONS TO THINK-PAIR-SHARE AND ACTIVITIES

TP1.1 (a) The fourth experimental point from the top is a circle: this point lies just above the best-fit curve that passes through the point (400 cm 2 , 0.20 g). The interval between horizontal grid lines is 1 space = 0.05 g.An estimate from the graph shows that the circle has a vertical separation of 0.3 spaces = 0.015 g above the best-fit curve.Physics for Scientists and Engineers with Modern Physics, 10e Raymond Serway, John Jewett (Solutions Manual All Chapters, 100% original verified, A+ Grade) All Chapters Download Link At The End Of This File 1 / 4

• Physics and Measurement

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.(b) The best-fit curve passes through 0.20 g, so the percentage difference is 0.015 g 0.20 g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

×100=7.5%

(c) The best-fit curve passes through the origin and the point (600 cm 2 , 0.32 g). Therefore, the slope of the best-fit curve is slope= 0.32 g−0 600 cm 2 −0

=5.3×10

−4 g/ cm 2 =5.3 g/ m 2

(d) For shapes cut from this copy paper, the mass of the cutout is

proportional to its area: m = aA. The proportionality constant a is

5.3 g/m 2 .(e) This result is to be expected if the paper has thickness and density that are uniform within the experimental uncertainty.(f) The slope is the areal density of the paper, its mass per unit area.]

Answers: Answers will vary, depending on how the graph is read

*TP1.2 Answer: All results should be close to 2.54, representing the

conversion factor 2.54 cm/in.

*TP1.3 Solution: The difference is due to the average density, which is related

to the composition of the penny. Before 1982, U.S. pennies were 95% copper and 5% zinc. After that date, they are 97.5% zinc, with a coating of 2.5% copper. Both copper and zinc pennies were produced in 1982.Perhaps a measurement of the mass of a sample of 1982 pennies would be interesting.

Answer: Pre-1982 pennies have an average mass of 3.1 g, while post-

1982 pennies have a lower average mass of 2.5 g

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Chapter 1 3 © 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SOLUTIONS TO END-OF-CHAPTER PROBLEMS

Section 1.1 Standards of Length, Mass, and Time

P1.1 (a) Modeling the Earth as a sphere, we find its volume as Its density is then (b)This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2000 to 3000 kg/m 3 . The average density of the Earth is significantly higher, so higher-density material must be down below the surface.P1.2 (a) where d is the diameter.Then (b) P1.3 For sphere the volume is and the mass is We divide this equation for the larger sphere by the same equation for

the smaller:

Then 3 / 4

• Physics and Measurement

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.P1.4 The volume of a spherical shell can be calculated from

From the definition of density, , so

*P1.5 Let us find the angle subtended by the width of the Great Wall at the height of the spacecraft orbit. From the description of a subtended angle in the problem statement, we obtain θ= w idthofGreatWall distance to Great Wall = 7m 200 000 m

=3.5×10

−5 rad The angle subtended by the width of the Great Wall at a height of 200 km is 3.5 × 10 –5 rad, which is smaller than the normal visual acuity of the eye by about a factor of ten. Therefore, despite its great length, its width cannot be seen. In the same way, a single human hair cannot be seen from several meters away, despite its length. Your argument should be based on this calculation.]

Answer: The angle subtended by the Great Wall is less than the visual

acuity of the eye.

Section 1.2 Matter and Model Building P 1.6 Figure P1.6 suggests a right triangle where, relative to angle θ, its adjacent side has length d and its opposite side is equal to width of the river, y; thus,

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