## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 14 - Fluids and Elasticity - Exercises and Problems - Page 384: 27

#### Answer

The fluid's speed in the circular part of the tube is $1.27~v_0$.

#### Work Step by Step

The volume flow rate is constant throughout the tube. The volume flow rate is the speed of the fluid times the cross-sectional area of the tube. We can find the fluid's speed $v_2$ in the circular part of the tube as: $v_2~A_2 = v_0~A_0$ $v_2~\pi~(\frac{L}{2})^2 = v_0~L^2$ $v_2 = \frac{4~v_0~L^2}{\pi~L^2}$ $v_2 = \frac{4~v_0}{\pi}$ $v_2 = 1.27~v_0$ The fluid's speed in the circular part of the tube is $1.27~v_0$.

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