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 280: 53



Work Step by Step

According to Bernoulli's principle $P_1+\frac{1}{2}\rho V_1^2=P_2+\frac{1}{2}\rho V_2^2$ This can be rearranged as: $P_2=P_1+\frac{1}{2}\rho(V_1^2-V_2^2)$ We plug in the known values to obtain: $P_2=16\times 10^3+\frac{1}{2}\times 1.06((35)^2-(175)^2)$ $P_2=16\times 10^3+(-15582\frac{g}{cm s^2})$ $P_2=16000-\frac{15582\times 10^{-3}Kg}{10^{-2}ms^2}$ This simplifies to: $P_2=14KPa$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.