MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK NEXT Chapter 28, Problem

Question

MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK NEXT Chapter 28, Problem 22 Go Three particies are listed in the table. The mass and speed of each partice are given as multiples of the variables m and v, which have the values m 2.29x10 5 kg and v 0.200c. The speed of light in a vacuum is c 3.00x108 m/s. Determine the momentum for each particle according to special relativity. Particle Mass Speed Momentum kg m/s 1.40 b m/2 2v kg m/s a kg m/s B m/4 Question Attempts: Unlimited SAVE FOR LATER SUBHzrANSMER copyright o 2000-2017 by John wiey & Sons, Inc. or related companies All rights reserved. prvacy Policy I 20oo-A12 xobn wley s sons Inc. All Rights Reserved. A Division of zehn wary Asons, Inc.

Explanation / Answer

Momentum P = MV

For a. P1 = MV

P1 = [2.29 * 10^(-8)] * [ 0.200 * 3*10^(8)]

P1 =1.374 kg*m/s

For b. P2 =(m/2) * 2V

P2 = MV = 1.374 kg*m/s

For c. P3 = (m/4) * 4V

P3 = MV = 1.374 kg*m/s

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