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 281: 69


$ t = \frac{A_0}{A_1}\sqrt{\frac{2h}{g}}$

Work Step by Step

We know that the volume is changing at a rate of $A\frac{dh}{dt}$, and that the volume through the orifice is changing at a rate of $A_1\sqrt{2gh}$. Thus, setting these equal, we find: $A\frac{dh}{dt}=A_1\sqrt{2gh} \\ \frac{Adh}{\sqrt{2gh}}=A_1dt$ Taking the integral of both sides and simplifying, we find: $ t = \frac{A_0}{A_1}\sqrt{\frac{2h}{g}}$
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