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: 61


a) 98 percent less b) .17 meters

Work Step by Step

a) We know the following equation: $\Delta P = .5(v_1)^2(\rho)((\frac{A_1}{A_2})^2-1)$ Thus, we see that pressure varies by: $\Delta P = |((\frac{.04}{.0015})^2-1)|$ $ \Delta P = .98 = \fbox{98%}$ b) The lowest possible height would be as the flow speed approaches 0, so we plug in 0 for $v_1$ and solve: $\Delta h_{possible}=\frac{p_{boy}}{\rho g}$ Plugging in the known values, we find: $h_{possible}\approx .17 \ m$
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