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:
$d=(3.00\times 10^8)(1\times 10^{-9})$
$d=0.300m$
Essential University Physics: Volume 1 (4th Edition)
by Wolfson, Richard
Published by
Pearson
ISBN 10:
0-134-98855-8
ISBN 13:
978-0-13498-855-9
Chapter 1 - Exercises and Problems - Page 12: 13
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