Essential University Physics: Volume 1 (4th Edition) Clone

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

Chapter 15 - Exercises and Problems - Page 291: 67

Answer

$14.6 \ kg$

Work Step by Step

We find the following equation for the mass it can support: $ m = \frac{\mu P\pi d^2}{4g}$ Plugging in the known values, we obtain: $ m = \frac{(.72)(1.01\times10^5)(\pi)(.05)^2}{4(9.81)}=14.6 \ kg$
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