University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 1 - Units, Physical Quantities, and Vectors - Problems - Exercises - Page 28: 1.11

Answer

r=9.0cm

Work Step by Step

density=$\frac{mass}{volume}=\frac{m}{v}=19.5g/cm^{3}$ $m_{critical}$=60.0kg Volume of sphere=v=$\frac{4}{3}πr^{3}$ v=$\frac{m_{critical}}{density}=(\frac{60.0kg}{19.5g/cm^{3}})(\frac{1000g}{1.0kg})=3080cm^{3}$ r=$\sqrt[3] \frac{3v}{4π}$=$\sqrt[3]\frac{3(3080cm^{3})}{4π}$=9.0 cm
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