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: 4

Answer

(a) The average density of the mixture is $810~kg/m^3$ (b) The average density of the mixture is $840~kg/m^3$

Work Step by Step

(a) We can find the volume of the gasoline. $V_g = \frac{M_g}{\rho_g}$ $V_g = \frac{0.050~kg}{680~kg/m^3}$ $V_g = 7.35\times 10^{-5}~m^3$ We can find the volume of the water. $V_w = \frac{M_w}{\rho_w}$ $V_w = \frac{0.050~kg}{1.00\times 10^3~kg/m^3}$ $V_w = 5.0\times 10^{-5}~m^3$ We can find the average density of the mixture. $\rho = \frac{mass}{volume}$ $\rho = \frac{M_g+M_w}{V_g+V_w}$ $\rho = \frac{0.050~kg+0.050~kg}{(7.35\times 10^{-5}~m^3)+(5.0\times 10^{-5}~m^3)}$ $\rho = 810~kg/m^3$ The average density of the mixture is $810~kg/m^3$. (b) We can find the mass of the gasoline. $M_g = V_g~\rho_g$ $M_g = (50\times 10^{-6}~m^3)(680~kg/m^3)$ $M_g = 0.034~kg$ We can find the mass of the water. $M_w = V_w~\rho_w$ $M_w = (50\times 10^{-6}~m^3)(1.00\times 10^3~kg/m^3)$ $M_w = 0.050~kg$ We can find the average density of the mixture. $\rho = \frac{mass}{volume}$ $\rho = \frac{M_g+M_w}{V_g+V_w}$ $\rho = \frac{0.034~kg+0.050~kg}{(50\times 10^{-6}~m^3)+(50\times 10^{-6}~m^3)}$ $\rho = 840~kg/m^3$ The average density of the mixture is $840~kg/m^3$.
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