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: 36

Answer

7.24 m/s

Work Step by Step

Please see the attached image first. Here we use the Bernoulli's equation, $P+\frac{1}{2}\rho V^{2}+\rho gy = constant$ Where $P- Pressure$, $\frac{1}{2}\rho V^{2}-Kinetic\space energy\space per\space unit \space volume$, $\rho gy- Gravitational \space potential \space energy \space per\space unit \space volume.$ Let's plug known values into this equation. $P_{a}+0+\rho gh=P_{a}+\frac{1}{2}\rho V_{hole}^{2}+0$ $V_{hole}=\sqrt {2gh}=\sqrt {2\times9.8\space m/s^{2}\times2.68\space m}$ $V_{hole}=7.24\space m/s$
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