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

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 14 - Fluids and Elasticity - Exercises and Problems: 8

Answer

(a) The volume of water is $6.32~m^3$ (b) The pressure at the bottom of the tank is $1.19\times 10^5~Pa$

Work Step by Step

(a) We can find the mass of the ethyl alcohol as: $M = \rho_a~V$ $M = (790~kg/m^3)(8.0~m^3)$ $M = 6320~kg$ We can find the volume of the water with this mass as: $V = \frac{M}{\rho_w}$ $V = \frac{6320~kg}{1000~kg/m^3}$ $V = 6.32~m^3$ The volume of water is $6.32~m^3$. (b) If the water tank is a cube, we can find the length $L$ of each side. $L = V^{1/3}$ $L = (6.32~m^3)^{1/3}$ $L = 1.85~m$ We can find the pressure at the bottom of the tank. $P = P_0 + \rho~g~h$ $P = 1.01\times 10^5~Pa + (1000~kg/m^3)(9.80~m/s^2)(1.85~m)$ $P = 1.19\times 10^5~Pa$ The pressure at the bottom of the tank is $1.19\times 10^5~Pa$.
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.