Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems - Page 87: 56b

Answer

$8\ m/s^2$

Work Step by Step

We have: speed of earth $v=7.49\times 10^3 \ m/s$ mean radius $r =7010\times 10^3\ m$ Therefore: centripetal acceleration $a =\frac{v^2}{r}$ $a=\frac{(7.49\times 10^3 \ m/s)^2}{7010\times 10^3\ m}$ $a = 8\ m/s^2$
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