Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 15 - Exercises and Problems - Page 280: 41


a) 624 Pa b) 1,249 Pa

Work Step by Step

a) We use the equation for pressure to find: $P =\frac{F}{A}=\frac{mg}{4\pi r^2} = \frac{(.05)(9.81)}{4\pi (.005)^2} =624 \ Pa$ b) We use a similar process, now considering the mass of the water and the oil, to find: $P =\frac{F}{A}=\frac{mg}{4\pi r^2} = \frac{(.1)(9.81)}{4\pi (.005)^2} =1249 \ Pa$
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