Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 4 - Exercises and Problems: 36

Answer

Please see the work below.

Work Step by Step

We know that $t=\sqrt{\frac{2h}{g}}$ We plug in the known values to obtain: $t=\sqrt{\frac{2(1.20)}{9.8}}=0.4949s$ The acceleration of the earth is given as $a=\frac{F}{M_e}$ $a=\frac{mg}{M_e}$ $a=\frac{65(9.8)}{5.97\times 10^{24}}=1.067\times 10^{-22}\frac{m}{s^2}$ The displacement of the earth is given as $d=\frac{1}{2}at^2=\frac{1}{2}(1.067\times 10^{-22})(0.4949)^2=1.3\times 10^{-23}m$
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