Chapter 1: Introduction to PhysicsAnswers to Even-Numbered Conceptual Questions2. The quantity T + d does not make sense physically, because it adds together variables that have differentphysical dimensions. The quantity d/T does make sense, however; it could represent the distance d traveledby an object in the time T.4. (a) 107 s; (b) 10,000 s; (c) 1 s; (d) 1017 s; (e) 108 s to 109 s.Solutions to Problems and Conceptual Exercises1. Picture the Problem: This is simply a units conversion problem.Strategy: Multiply the given number by conversion factors to obtain the desired units.Solution: (a) Convert the units: $114,000,000 × 1 gigadollars1×109 dollars= 0.114 gigadollars(b) Convert the units again: $114,000,000 × 1 teradollars1×1012 dollars=1.14×10−4 teradollarsInsight: The inside back cover of the textbook has a helpful chart of the metric prefixes.2. Picture the Problem: This is simply a units conversion problem.Strategy: Multiply the given number by conversion factors to obtain the desired units.Solution: (a) Convert the units: 70 μm ×1.0×10−6 mμm= 7.0×10−5 m(b) Convert the units again: 70 μm ×1.0×10−6 mμm×1 km1000 m= 7.0×10−8 kmInsight: The inside back cover of the textbook has a helpful chart of the metric prefixes.3. Picture the Problem: This is simply a units conversion problem.Strategy: Multiply the given number by conversion factors to obtain the desired units.Solution: Convert the units: 0.3 Gms×1×109 mGm= 3×108 m/sInsight: The inside back cover of the textbook has a helpful chart of the metric prefixes.4. Picture the Problem: This is simply a units conversion problem.Strategy: Multiply the given number by conversion factors to obtain the desired units.Solution: Convert the units: 70.72 teracalculations×1×1012 calculationsteracalculation×1×10−6 sμs=7.072×107 calculations/μs© Copyright 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.1 – 1Chapter 1: Introduction to Physics James S. Walker, Physics, 4th EditionInsight: The inside back cover of the textbook has a helpful chart of the metric prefixes.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.1 – 2Chapter 1: Introduction to Physics James S. Walker, Physics, 4th Edition5. Picture the Problem: This is a dimensional analysis question.Strategy: Manipulate the dimensions in the same manner as algebraic expressions.Solution: 1. (a) Substitute dimensions for the variables: x =vtm=ms⎛⎝⎜⎞⎠⎟s( )=m ∴ The equation is dimensionally consistent.2. (b) Substitute dimensions for the variables: x =12at2m=12ms2⎛⎝⎜⎞⎠⎟s( )2=m ∴ dimensionally consistent3. (c) Substitute dimensions for the variables: t =2xa ⇒ s=mm s2= s2=s ∴ dimensionally consistentInsight: The number 2 does not contribute any dimensions to the problem.6. Picture the Problem: This is a dimensional analysis question.Strategy: Manipulate the dimensions in the same manner as algebraic expressions.Solution: 1. (a) Substitute dimensions for the variables: vt =ms⎛⎝⎜⎞⎠⎟s( )=m Yes2. (b) Substitute dimensions for the variables: 12at2=12ms2⎛⎝⎜⎞⎠⎟s( )2=m Yes3. (c) Substitute dimensions for the variables: 2at =2ms2⎛⎝⎜⎞⎠⎟s( )=ms No4. (d) Substitute dimensions for the variables: v2a=m s( )2m s2=m YesInsight: When squaring the velocity you must remember to square the dimensions of both the numerator (meters) and the denominator (seconds).7. Picture the Problem: This is a dimensional analysis question.Strategy: Manipulate the dimensions in the same manner as algebraic expressions.Solution: 1. (a) Substitute dimensions for the variables: 12at2=12ms2⎛⎝⎜⎞⎠⎟s( )2=m No2. (b) Substitute dimensions for the variables: at =ms2⎛⎝⎜⎞⎠⎟s( )=ms Yes3. (c) Substitute dimensions for the variables: 2xa=2mm s2=s No4. (d) Substitute dimensions for the variables: 2ax = 2ms2⎛⎝⎜⎞⎠⎟m( )=ms YesInsight: When taking the square root of dimensions you need not worry about the positive and negative roots; only the Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.1 – 3Chapter 1: Introduction to Physics James S. Walker, Physics, 4th Editionpositive root is physical. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.1 – 4Chapter 1: Introduction to Physics James S. Walker, Physics, 4th Edition8. Picture the Problem: This is a dimensional analysis question.Strategy: Manipulate the dimensions in the same manner as algebraic expressions.Solution: Substitute dimensions for the variables: v2=2axpms⎛⎝⎜⎞⎠⎟2=ms2⎛⎝⎜⎞⎠⎟m( )pm2=mp+1 therefore p=1Insight: The number 2 does not contribute any dimensions to the problem.9. Picture the Problem: This is a dimensional analysis question.Strategy: Manipulate the dimensions in the same manner as algebraic expressions.Solution: Substitute dimensions for the variables: a =2xtp[L][T]2=[L][T]p[T]−2=[T]p therefore p=−2Insight: The number 2 does not contribute any dimensions to the problem.10. Picture the Problem: This is a dimensional analysis question.Strategy: Manipulate the dimensions in the same manner as algebraic expressions.Solution: Substitute dimensions for the variables on both sides of the equation: v =v0+at[L][T]=[L][T]+[L][T]2[T][L][T]=[L][T] It is dimensionally consistent!Insight: Two numbers must have the same dimensions in order to be added or subtracted.11. Picture the Problem: This is a dimensional analysis question.Strategy: Manipulate the dimensions in the same manner as algebraic expressions.Solution: Substitute dimensions for the variables, where [M] represents the dimension of mass: F
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