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At a classic auto show, a 840-kg 1955 Nash Metropolitan

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 76P Chapter 8

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 76P

At a classic auto show, a 840-kg 1955 Nash Metropolitan motors by at 9.0 m/s, followed by a 1620-kg 1957 Packard Clipper purring past at 5.0 m/s. (a) Which car has the greater kinetic energy? What is the ratio of the kinetic energy of the Nash to that of the Packard? (b) Which car has the greater magnitude of momentum? What is the ratio of the magnitude of momentum of the Nash to that of the Packard? (c) Let ?F?N be the net force required to stop the Nash in time ?t?, and let ?F?p be the net force required to stop the Packard in the same time. Which is larger: ?F?N or ?F?P? What is the ratio ?F?N/?F?P of these two forces? (d) Now let ?F?N be the net force required to stop the Nash in a distance d, and let ?F?P be the net force required to stop the Packard in the same distance. Which is larger: ?F?N or ?F?P? What is the ratio ?F?N/?? ?

Step-by-Step Solution:

Solution 76P Problem (a) Step 1: Mass of Nash Metropolitan m = 840.0 Kg 1 Mass of Packard Clipper m = 1620.2Kg Speed of Nash Metropolitan v = 9.0 m1 Speed of Packard Clipper v = 5.0 2 Step 2: 1 2 Kinetic energy of Nash Metropolitan K = 1 2 m1 1 1 2 K1 2 *840*9 K = 34.02 KJ 1 Step 3: 1 2 Kinetic energy of Packard Clipper K = 2 2 m2 2 K2 1 *1620*5 2 2 K = 20.20 KJ 2 Kinetic energy of Nash Metropolitan is greater. Step 4: Ratio of the kinetic energies K 1 = K 2 = 34.02 K 20.20 K = 1.68 Ratio of the kinetic energy of the Nash to that of the Packard is 1.68 Problem (b) Step 1: Momentum of Nash Metropolitan P1 m 1 1 P = 840*9 1 P1 7560 Kg.m/s Step 2: Momentum of Packard Clipper P2= m 2 2 P2= 1620*5 P2= 8100 Kg.m/s Momentum of Packard Clipper is greater Step 3: Ratio of Momentum P1 = P 2 7560 = 8100 = 0.93 Ratio of the Momentum of the Nash to that of the Packard is 0.93 Problem (c) Step 1: Force required to stop the Nash in time t = F N Force required to stop the Packard in time t = F P P 1 FN t P 2 FP t Since force is directly proportional to rate of change in momentum F is greatP than F N F > F P N Step 2: Ratio of forces F N P 1 F P = P 2 or F N P 1 F P = P 2 = 0.93 or F = 0.93F N P Problem (d) Step 1: Force is directly proportional to work done to move the car to the distance d. Kinetic energy does a work to move the car to distance d. And the force required to stop the car is also directly proportional to kinetic energy Hence Nash Metropolitan car requires greater force to stop it. F > F N P

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Chapter 8, Problem 76P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

University Physics was written by and is associated to the ISBN: 9780321675460. This full solution covers the following key subjects: Nash, packard, Net, ratio, required. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The answer to “At a classic auto show, a 840-kg 1955 Nash Metropolitan motors by at 9.0 m/s, followed by a 1620-kg 1957 Packard Clipper purring past at 5.0 m/s. (a) Which car has the greater kinetic energy? What is the ratio of the kinetic energy of the Nash to that of the Packard? (b) Which car has the greater magnitude of momentum? What is the ratio of the magnitude of momentum of the Nash to that of the Packard? (c) Let ?F?N be the net force required to stop the Nash in time ?t?, and let ?F?p be the net force required to stop the Packard in the same time. Which is larger: ?F?N or ?F?P? What is the ratio ?F?N/?F?P of these two forces? (d) Now let ?F?N be the net force required to stop the Nash in a distance d, and let ?F?P be the net force required to stop the Packard in the same distance. Which is larger: ?F?N or ?F?P? What is the ratio ?F?N/?? ?” is broken down into a number of easy to follow steps, and 168 words. The full step-by-step solution to problem: 76P from chapter: 8 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. Since the solution to 76P from 8 chapter was answered, more than 255 students have viewed the full step-by-step answer.

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At a classic auto show, a 840-kg 1955 Nash Metropolitan

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