Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 10 - Fluids - Problems: 22

Answer

a. $1.29 \times 10^5 N/m^2$ b. $9.7 \times 10^4 N/m^2$

Work Step by Step

The pressure in the tank is atmospheric pressure, plus the pressure difference indicated by the column of mercury. Use equation 10–3c for the open-tube manometer. Note that 1040 mbar = $1.04 \times 10^5 N/m^2$. a. $P_t=1.04 \times 10^5 N/m^2+(13.6 \times 10^3 kg/m^3)(9.80 m/s^2)(0.185m)=1.29 \times 10^5 N/m^2$ b. $P_t=1.04 \times 10^5 N/m^2+(13.6 \times 10^3 kg/m^3)(9.80 m/s^2)(-0.056m)=9.7 \times 10^4 N/m^2$
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