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: 1.13

Answer

$V = 4.2 \times 10^{-12}~cm^3$ $A = 1.3 \times 10^{-5}~mm^2$

Work Step by Step

The radius is half the diameter, so the radius $r$ is $1.0 ~\mu m$ We can convert the radius to units of cm. $r = (1.0 ~\mu m)(\frac{10^{-4} ~cm}{1 ~\mu m}) = 1.0 \times 10^{-4} ~cm$ We can use the radius to find the volume. $V = \frac{4}{3}\pi ~r^3 = \frac{4}{3}\pi ~(1.0 \times 10^{-4} ~cm)^3$ $V = 4.2 \times 10^{-12}~cm^3$ We can convert the radius to units of mm. $r = (1.0 ~\mu m)(\frac{10^{-3} ~mm}{1 ~\mu m}) = 1.0 \times 10^{-3} ~mm$ We can use the radius to find the surface area. $A = 4\pi ~r^2 = 4 \pi ~(1.0 \times 10^{-3} ~mm)^2$ $A = 1.3 \times 10^{-5}~mm^2$
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