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 287: 45

Answer

$4.12*10^{-3} m^3/s$

Work Step by Step

We find the velocity out of the pressure head as: $v^2= 2gh$ $v=\sqrt{2*9.8*12} = 15.96m/s$ To get $m^3$ we need to multiply by the area $v * A = 15.96 * (.0185/2)^2 *3.14) = 4.12*10^{-3} m^3/s$
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.