Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 14 - Fluids - Problems - Page 409: 38b

Answer

$V = 2720~cm^3$

Work Step by Step

In part (a), we found that the density of the ball is $~~1.5~g/cm^3$ When the density of the liquid is zero, the kinetic energy after falling $4.0~cm$ is $1.6~J$ We can find the speed after falling $4.0~cm$: $v_f^2 = v_0^2+2ad$ $v_f^2 = 0+2ad$ $v_f = \sqrt{2ad}$ $v_f = \sqrt{(2)(9.8~m/s^2)(0.040~m)}$ $v_f = 0.8854~m/s$ We can find the mass of the ball: $K = \frac{1}{2}mv^2 = 1.60~J$ $m = \frac{(2)(1.60~J)}{v^2}$ $m = \frac{(2)(1.60~J)}{(0.8854~m/s)^2}$ $m = 4.082~kg$ We can find the volume of the ball: $V = \frac{m}{\rho}$ $V = \frac{4.082~kg}{1500~kg/m^3}$ $V = 0.00272~m^3$ $V = 2720~cm^3$
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