College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 26 - Problems - Page 1013: 75


$E = 1326~MeV$

Work Step by Step

We can find the total energy $E$: $E^2 = E_0^2+(pc)^2$ $E^2 = (939.6~MeV)^2+[(935~MeV/c)(c)]^2$ $E^2 = (939.6~MeV)^2+(935~MeV)^2$ $E^2 = 1757073.16~(MeV)^2$ $E = \sqrt{1757073.16~(MeV)^2}$ $E = 1326~MeV$
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