Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - Problems: 10

Answer

$$1 \times 10^{5} \frac{km}{h}$$

Work Step by Step

Average speed is the distance divided by the time. $$1 \times 10^{9} \frac{km}{yr} \frac{1 year}{365.25 days} \frac{1 day}{24 h}= 1 \times 10^{5} \frac{km}{h}$$
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