Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 15 - Fluids - Problems and Conceptual Exercises - Page 531: 18

Answer

$1.2m$

Work Step by Step

We can find the required height difference as follows: $h=\frac{1}{\rho}(\frac{m_L}{\frac{\pi}{4}D_L^2}-\frac{m_R}{\frac{\pi}{4}D_R^2})$ This simplifies to: $\implies h=\frac{4}{\pi \rho}(\frac{m_L}{D_L^2}-\frac{m_R}{D_R^2})$ We plug in the known values to obtain: $h=\frac{4}{\pi(750)}(\frac{1.8}{(0.044)^2}-\frac{3.2}{(0.12)^2})$ $\implies h=1.2m$
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.