Physics: Principles with Applications (7th Edition)

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

Chapter 10 - Fluids - Problems - Page 289: 70

Answer

$8.3\times 10^{-6}Kg$

Work Step by Step

We can find the required mass as follows: $m=\frac{12\pi r\gamma}{g}$ We plug in the known values to obtain: $m=\frac{12\pi(3\times 10^{-5})(0.72)}{9.81}$ $m=8.3\times 10^{-6}Kg$
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