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: 64

Answer

Please see the work below.

Work Step by Step

From equation of continuity $A.v=constant$ $\implies v_{\circ}\pi(\frac{d_{\circ}}{2})^2=v.\pi(\frac{d}{2})^2$ $d=d_{\circ}(\frac{v_{\circ}}{v})^\frac{1}{2}$ $\implies d=d_{\circ}(\frac{v_{\circ}}{(v_{\circ}^2+2gh)^{\frac{1}{2}}})^{\frac{1}{2}}$ $d=d_{\circ}(\frac{v_{\circ}^2}{v_{\circ}^2+2gh})^\frac{1}{4}$
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