Essential University Physics: Volume 1 (4th Edition)

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

Chapter 1 - Exercises and Problems - Page 13: 43

Answer

$1.8\space ms$

Work Step by Step

We can write the increment of the length of the day $(\Delta T)$ as follows. $(\Delta T)= New\space length\space of\space the\space day\space-\space Previous\space length\space of\space the\space day $ $\space\space\space\space\space\space\space\space\space=86,400.0038\space s\space-\space 86,400.002\space s$ $\space\space\space\space\space\space\space\space\space =0.0018\space s$ $\space\space\space\space\space\space\space\space\space=0.0018\times\frac{10^{-3}\space s}{10^{-3}}$ $\space\space\space\space\space\space\space\space\space=\frac{0.0018\space ms}{10^{-3}}$ $\space\space\space\space\space\space\space\space\space=(\frac{0.0018\space}{10^{-3}})\times\frac{10^{3}}{10^{3}}\space ms$ $\space\space\space\space\space\space\space\space\space=1.8\space ms$ $$The\space increment\space of\space a\space day\space is\space 1.8\space ms$$
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