Exam (elaborations) TEST BANK FOR Serway & Jewett’s Physics for Scientists and Engineers 9th Edition VOLUME 1 By Raymond A. Serway and John W. Jewett (Study Guide with Student Solutions Manual)

Exam (elaborations) TEST BANK FOR Serway & Jewett’s Physics for Scientists and Engineers 9th Edition VOLUME 1 By Raymond A. Serway and John W. Jewett (Study Guide with Student Solutions Manual) Student Solutions Manual and Study Guide Prepared by Vahe Peroomian University of California at Los Angeles John R Gordon Emeritus, James Madison University Raymond A. Serway Emeritus, James Madison University John W. Jewett Emeritus, California State Polytechnic University, Pomona Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States Physics for Scientists and Engineers Volume 1 NINTH EDITION Raymond A. Serway Emeritus, James Madison University John W. Jewett Emeritus, California State Polytechnic University, Pomona Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN iii © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Table of Contents Preface v Acknowledgments vi Suggestions for Study vii Chapter Page 1. Physics and Measurement 1 2. Motion in One Dimension 14 3. Vectors 40 4. Motion in Two Dimensions 59 5. The Laws of Motion 89 6. Circular Motion and Other Applications of Newton’s Laws 120 7. Energy of a System 143 8. Conservation of Energy 167 9. 195 10. Rotation of a Rigid Object About a Fixed Axis 225 11. Angular Momentum 259 12. Static Equilibrium and Elasticity 282 13. Universal Gravitation 303 14. Fluid Mechanics 326 15. Oscillations and Mechanical Waves 348 16. Wave Motion 377 17. Sound Waves 395 18. Superposition and Standing Waves 416 Linear Momentum and Collisions Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN iv Table of Contents © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 19. Temperature 441 20. The First Law of Thermodynamics 462 21. The Kinetic Theory of Gases 486 22. Heat Engines, Entropy, and the Second Law of Thermodynamics 508 Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 1 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1 Physics and Measurement EQUATIONS AND CONCEPTS The density of any substance is defined as the ratio of mass to volume. Density is an example of a derived quantity. ρ ≡ m V (1.1) The SI units of density are kg/m3. SUGGESTIONS, SKILLS, AND STRATEGIES A general strategy for problem solving will be described in Chapter 2. Appendix B of your textbook includes a review of mathematical techniques including: • Scientific notation: using powers of ten to express large and small values. • Basic algebraic operations: factoring, handling fractions, and solving quadratic equations. • Fundamentals of plane and solid geometry: graphing functions, calculating areas and volumes, and recognizing equations and graphs of standard figures (e.g., straight line, circle, ellipse, parabola, and hyperbola). • Basic trigonometry: definition and properties of functions (e.g., sine, cosine, and tangent), the Pythagorean Theorem, and basic trigonometry identities. REVIEW CHECKLIST You should be able to: • Describe the standards which define the SI units for the fundamental quantities length (meter, m), mass (kilogram, kg), and time (second, s). Identify and properly use prefixes and mathematical notations such as the following: ∝ (is proportional to), < (is less than), ≈ (is approximately equal to), Δ (change in value), etc. (Section 1.1) Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 2 Physics and Measurement © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. • Convert units from one measurement system to another (or convert units within a system). Perform a dimensional analysis of an equation containing physical quantities whose individual units are known. (Sections 1.3 and 1.4) • Carry out order-of-magnitude calculations or estimates. (Section 1.5) • Express calculated values with the correct number of significant figures. (Section 1.6) ANSWERS TO SELECTED OBJECTIVE QUESTIONS 3. Answer each question yes or no. Must two quantities have the same dimensions (a) if you are adding them? (b) If you are multiplying them? (c) If you are subtracting them? (d) If you are dividing them? (e) If you are equating them? Answer. (a) Yes. Three apples plus two jokes has no definable answer. (b) No. One acre times one foot is one acre-foot, a quantity of floodwater. (c) Yes. Three dollars minus six seconds has no definable answer. (d) No. The gauge of a rich sausage can be 12 kg divided by 4 m, giving 3 kg/m. (e) Yes, as in the examples given for parts (b) and (d). Thus we have (a) yes, (b) no, (c) yes, (d) no, and (e) yes. ☐ ☐ ☐ ANSWERS TO SELECTED CONCEPTUAL QUESTIONS 3. Suppose the three fundamental standards of the metric system were length, density, and time rather than length, mass, and time. The standard of density in this system is to be defined as that of water. What considerations about water would you need to address to make sure the standard of density is as accurate as possible? Answer. There are the environmental details related to the water: a standard temperature would have to be defined, as well as a standard pressure. Another consideration is the quality of the water, in terms of defining an upper limit of impurities. A difficulty with this scheme is that density cannot be measured directly with a single measurement, as can length, mass, and time. As a combination of two measurements (mass and volume, which itself involves three measurements!), a density value has higher uncertainty than a single measurement. ☐ ☐ ☐ Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN Chapter 1 3 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. SOLUTIONS TO SELECTED END-OF-CHAPTER PROBLEMS 9. Which of the following equations are dimensionally correct? (a) vf = vi + ax (b) y = (2 m)cos(kx), where k = 2 m–1 Solution Conceptualize: It is good to check an unfamiliar equation for dimensional correctness to see whether it can possibly be true. Categorize: We evaluate the dimensions as a combination of length, time, and mass for each term in each equation. Analyze: (a) Write out dimensions for each quantity in the equation vf = vi + ax. The variables vf and vi are expressed in units of m/s, so [vf] = [vi] = LT –1 The variable a is expressed in units of m/s2: [a] = LT –2 The variable x is expressed in meters. Therefore [ax] = L2 T –2 Consider the right-hand member (RHM) of equation (a): [RHM]=LT –1+L2 T –2 Quantities to be added must have the same dimensions. Therefore, equation (a) is not dimensionally correct.  (b) Write out dimensions for each quantity in the equation y = (2 m) cos (kx) For y, [y] = L for 2 m, [2 m] = L and for (kx), [kx] = 2 m–1 ( )x ⎡⎣ ⎤⎦ = L–1L Therefore we can think of the quantity kx as an angle in radians, and we can take its cosine. The cosine itself will be a pure number with no dimensions. For the left-hand member (LHM) and the right-hand member (RHM) of the equation we have [LHM] = [y] = L [RHM] = [2m][cos (kx)] = L These are the same, so equation (b) is dimensionally correct.  Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 4 Physics and Measurement © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Finalize: We will meet an expression like y = (2 m)cos(kx), where k = 2 m–1, as the wave function of a wave. 15. A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kilograms per cubic meter). Solution Conceptualize: From Table 14.1, the density of lead is 1.13 × 104 kg/m3, so we should expect our calculated value to be close to this value. The density of water is 1.00 × 10 3 kg/m3, so we see that lead is about 11 times denser than water, which agrees with our experience that lead sinks. Categorize: Density is defined as ρ = m/V. We must convert to SI units in the calculation. Analyze: ρ = 23.94 g 2.10 cm3 ⎛⎝ ⎜ ⎞⎠ ⎟ 1 kg 1 000 g ⎛⎝ ⎜ ⎞⎠ ⎟ 100 cm 1 m ( )3 = 23.94 g 2.10 cm3 ⎛⎝ ⎜ ⎞⎠ ⎟ 1 kg 1 000 g ⎛⎝ ⎜ ⎞⎠ ⎟ 1 000 000 cm3 1 m3 ( ) = 1.14 × 104 kg/m3  Finalize: Observe how we set up the unit conversion fractions to divide out the units of grams and cubic centimeters, and to make the answer come out in kilograms per cubic meter. At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference from the tabulated values is possibly due to measurement uncertainty and does not indicate a discrepancy. 17. A rectangular building lot has a width of 75.0 ft and a length of 125 ft. Determine the area of this lot in square meters. Solution Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A ≈ (25 m)(40 m) = 1 000 m2. Categorize: We will use the geometrical fact that for a rectangle Area = Length × Width; and the conversion 1 m = 3.281 ft. Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN Chapter 1 5 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Analyze: A =  × w = (75.0 ft) 1 m 3.281 ft ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (125 ft) 1 m 3.281 ft ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 871 m2  Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m2. Unit conversion is a common technique that is applied to many problems. Note that one square meter is more than ten square feet. 25. One cubic meter (1.00 m3) of aluminum has a mass of 2.70 × 103 kg, and the same volume of iron has a mass of 7.86 × 10 3 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 2.00 cm on an equal-arm balance. Solution Conceptualize: The aluminum sphere must be larger in volume to compensate for its lower density. Its density is roughly one-third as large, so we might guess that the radius is three times larger, or 6 cm. Categorize: We require equal masses: mA1 = mFe or ρA1VA1 = ρFeVVe Analyze: We use the volume of a sphere. By substitution, ρA1 4 3 π rA1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ρFe 4 3 π (2.00 cm)3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Now solving for the unknown, rA1 3 = ρFe ρA1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (2.00 cm)3 = 7.86 × 103 kg/m3 2.70 × 103 kg/m3 ⎛ ⎝ ⎜⎞ ⎠ ⎟ (2.00 cm)3 = 23.3 cm3 Taking the cube root, rA1= 2.86 cm.  Finalize: The aluminum sphere is 43% larger than the iron one in radius, diameter, and circumference. Volume is proportional to the cube of the linear dimension, so this excess in linear size gives it the (1.43)(1.43)(1.43) = 2.92 times larger volume it needs for equal mass. Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 6 Physics and Measurement © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 27. One gallon of paint (volume = 3.78 × 10–3 m3) covers an area of 25.0 m2. What is the thickness of the fresh paint on the wall? Solution Conceptualize: We assume the paint keeps the same volume in the can and on the wall. Categorize: We model the film on the wall as a rectangular solid, with its volume given by its “footprint” area, which is the area of the wall, multiplied by its thickness t perpendicular to this area and assumed to be uniform. Analyze: V = At gives t = V A = 3.78 × 10–3 m3 25.0 m2 = 1.51 × 10–4 m  Finalize: The thickness of 1.5 tenths of a millimeter is comparable to the thickness of a sheet of paper, so this answer is reasonable. The film is many molecules thick. 29. (a) At the time of this book’s printing, the U.S. national debt is about $16 trillion. If payments were made at the rate of $1 000 per second, how many years would it take to pay off the debt, assuming no interest were charged? (b) A dollar bill is about 15.5 cm long. How many dollar bills attached end to end would it take to reach the Moon? The front endpapers give the Earth-Moon distance. Note: Before doing these calculations, try to guess at the answers. You may be very surprised. Solution (a) Conceptualize: $16 trillion is certainly a large amount of money, so even at a rate of $1 000/second, we might guess that it will take a lifetime (~100 years) to pay off the debt. Categorize: The time interval required to repay the debt will be calculated by dividing the total debt by the rate at which it is repaid. Analyze: T = $16 trillion $1000 / s = $16 × 1012 ($1000 / s)(3.156 × 107 s/yr) = 507 yr  Finalize: Our guess was a bit low. $16 trillion really is a lot of money! (b) Conceptualize: We might guess that 16 trillion bills would reach from the Earth to the Moon, and perhaps back again, since our first estimate was low. Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN Chapter 1 7 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Categorize: The number of bills is the distance to the Moon divided by the length of a dollar. Analyze: N = D  = 3.84 × 108 m 0.155 m = 2.48 × 109 bills  Finalize: Sixteen trillion dollars is larger than this two-and-a-half billion dollars by more than six thousand times. The ribbon of bills comprising the debt reaches across the cosmic gulf thousands of times, so again our guess was low. Similar calculations show that the bills could span the distance between the Earth and the Sun sixteen times. The strip could encircle the Earth’s equator nearly 62 000 times. With successive turns wound edge to edge without overlapping, the dollars would cover a zone centered on the equator and about 4.2 km wide. 31. Find the order of magnitude of the number of table-tennis balls that would fit into a typical-size room (without being crushed). [In your solution, state the quantities you measure or estimate and the values you take for them.] Solution Conceptualize: Since the volume of a typical room is much larger than a Ping- Pong ball, we should expect that a very large number of balls (maybe a million) could fit in a room. Categorize: Since we are only asked to find an estimate, we do not need to be too concerned about how the balls are arranged. Therefore, to find the number of balls we can simply divide the volume of an average-size living room (perhaps 15 ft × 20 ft × 8 ft) by the volume of an individual Ping-Pong ball. Analyze: Using the approximate conversion 1 ft = 30 cm, we find VRoom = (15 ft)(20 ft)(8 ft)(30 cm/ft)3 ≈ 6 × 107 cm3 A Ping-Pong ball has a diameter of about 3 cm, so we can estimate its volume as a cube: Vball = (3 cm)(3 cm)(3 cm) ≈ 30 cm3 Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 8 Physics and Measurement