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21. The force F acting on a body with mass m and velocity v is the rate of change of momentum: F = (d/dt)(mv). If m is constant, this becomes F = ma, where a = dv/dt is the acceleration. But in the theory of relativity the mass of a %3D particle varies with v as follows: m = mo//1 – v²/c²,where mo is the mass of the particle at rest and c is the speed of light. Show that moa F = (1 – v²/c?)/2 %3D

21. The force F acting on a body with mass m and velocity v is the rate of change of momentum: F = (d/dt)(mv). If m is constant, this becomes F = ma, where a = dv/dt is the acceleration. But in the theory of relativity the mass of a %3D particle varies with v as follows: m = mo//1 – v²/c²,where mo is the mass of the particle at rest and c is the speed of light. Show that moa F = (1 – v²/c?)/2 %3D

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Problem 21

 

21. The force F acting on a body with mass m and velocity v is the rate of change of momentum: F = (d/dt)(mv). If m is constant, this becomes F = ma, where a = dv/dt is the acceleration. But in the theory of relativity the mass of a %3D particle varies with v as follows: m = mo//1 – v²/c²,where mo is the mass of the particle at rest and c is the speed of light. Show that moa F = (1 – v²/c?)/2 %3D
Transcribed Image Text:21. The force F acting on a body with mass m and velocity v is the rate of change of momentum: F = (d/dt)(mv). If m is constant, this becomes F = ma, where a = dv/dt is the acceleration. But in the theory of relativity the mass of a %3D particle varies with v as follows: m = mo//1 – v²/c²,where mo is the mass of the particle at rest and c is the speed of light. Show that moa F = (1 – v²/c?)/2 %3D
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