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: 60

Answer

$ \rho = 6 \times 10^2 \ kg/m^2$

Work Step by Step

We simplify the equation for flow speed in order to isolate density. Doing this, we obtain: $ \rho = \frac{2\Delta P}{((\frac{A_1}{A_2})^2 - 1)v^2}$ We know that $\frac{A_1}{A_2}=4$. Plugging this in, the pressure as $16\times 10^3$, and the velocity as 1.9 meters per second, we obtain: $ \rho = 6 \times 10^2 \ kg/m^2$
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