Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 15 - Exercises and Problems - Page 291: 68

Answer

The proof is below.

Work Step by Step

From the Bernoulli equation, we know: $ P_B + 0 $ and $P_A+\frac{1}{2}\rho v^2$ are both equal. Thus, we find: $P_B =P_A+\frac{1}{2}\rho v^2 \\ P_B -P_A=\frac{1}{2}\rho v^2 \\\Delta P =\frac{1}{2}\rho v^2 \\ v = \sqrt{\frac{2\Delta P}{\rho}}$
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