Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 25 - The Electric Potential - Exercises and Problems: 11

Answer

$4.37 \times 10 ^{4} m/s$

Work Step by Step

By the conservation of energy; Loss in energy = Gain in energy $q \Delta V=\frac{1}{2}m_{p}v_{p}^{2}$ $v_{p}=[\frac{2q\Delta V}{m}]^{1/2}$ $v_{p}=[\frac{2\times 1.6 \times 10^{-19} \times 10^{3}}{1.67\times 10^{-27}}]^{1/2}= 4.37 \times 10 ^{4} m/s$
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