Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 1 - Exercises - Page 41: 136


\[\underline{1\text{ cm}}\]

Work Step by Step

The relationship between the dyne and newton is as follows: \[1\text{ dyne}={{10}^{-5}}\text{ N}\] One newton is the force that produces a uniform acceleration of 1 meter per square second, when acting on a body of 1 kg mass. So, \[1\text{ dyne}=\left( 1\text{ kg} \right)\left( 1\text{ m/}{{\text{s}}^{2}} \right){{10}^{-5}}\] Now, \[1\text{ kg}=1000\text{ g}\] and \[1\text{ m}=100\text{ cm}\]. Thus, \[\begin{align} & 1\text{ dyne}=\left( 1000\text{ g} \right)\left( 100\text{ cm/}{{\text{s}}^{\text{2}}} \right){{10}^{-5}} \\ & =\left( 1\text{ g} \right)\left( 1\text{ cm/}{{\text{s}}^{\text{2}}} \right) \end{align}\] Thus, the unit of length used to define dyne is a centimeter. The unit of length in dyne is \[\underline{1\text{ cm}}\]. One dyne is the force that when applied on a mass of 1 g produces a uniform acceleration of 1 centimeter per square second.
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