Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 15 - Fluids and Elasticity - Exercises and Problems - Page 436: 22


The diameter of the hose is 2.0 cm

Work Step by Step

We find the volume flow rate of the water; $flow~rate = \frac{V}{t}$ $flow~rate = \frac{600~L}{(8.0~min)(60~s/min)}$ $flow~rate = 1.25~L/s$ We convert the flow rate to units of $m^3/s$; $flow~rate = (1.25~L/s)(\frac{1~m^3}{1000~L})$ $flow~rate = 1.25\times 10^{-3}~m^3/s$ The flow rate is equal to the cross-sectional area of the hose times the speed of the water. We can find the radius of the hose. $flow~rate = v~A = 1.25\times 10^{-3}~m^3/s$ $v~\pi~r^2 = 1.25\times 10^{-3}~m^3/s$ $r^2 = \frac{1.25\times 10^{-3}~m^3/s}{v~\pi}$ $r = \sqrt{\frac{1.25\times 10^{-3}~m^3/s}{v~\pi}}$ $r = \sqrt{\frac{1.25\times 10^{-3}~m^3/s}{(4.0~m/s)~(\pi)}}$ $r = 0.009974~m = 0.9974~cm$ We then find the diameter of the hose: $d = 2r = (2)(0.9974~cm) = 2.0~cm$ The diameter of the hose is 2.0 cm.
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