College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 1 - Problems - Page 22: 72


a) $3.3\times 10^{-8}m$ b) $3.3\times 10^{-2}micron$ c) $1.3\times 10^{-6}in$

Work Step by Step

Circumference of a viroid is approximately 300 times 0.35nm. The diameter is given by $C=pi\times d$ so, $d=C/pi$ a) d=$\frac{300\times 0.35nm\times10^{-9}m}{pi\times 1nm}=3.3\times 10^{-8}m$ b) d=$\frac{300\times 0.35nm\times10^{-3}m}{pi\times 1nm}=3.3\times 10^{-2}micron$ c) d=$\frac{300\times 0.35nm\times10^{-7}m}{pi\times 1nm}\times\frac{1 in}{2.54cm}=1.3\times 10^{-6}in$
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