Essential University Physics: Volume 1 (4th Edition)

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

Chapter 15 - Exercises and Problems - Page 290: 43

Answer

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: $P =\frac{F}{A}=\frac{mg}{4\pi r^2} = \frac{(.1)(9.81)}{4\pi (.005)^2} =1249 \ Pa$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.