Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 1 - Exercises and Problems - Page 12: 13

Answer

Please see the work below.

Work Step by Step

We can find the required distance as $d=vt$ We plug in the known values to obtain: $d=(3.00\times 10^8)(1\times 10^{-9})$ $d=0.300m$
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