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: 36


a) $1.91\ m/s$ b) $30.6\ m/s$

Work Step by Step

We know that the velocity is equal to the flow rate divided by the cross sectional area times the density of water. Thus, we find: a) $ v = \frac{15}{\pi(.05)^2(1000)} = 1.91\ m/s$ b) We do the same process to find: $ v = \frac{15}{\pi(.0125)^2(1000)} = 30.6\ m/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.