College Physics (4th Edition)

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

Review & Synthesis: Chapters 9-12 - MCAT Review - Page 470: 5


The correct answer is: C. $6.0~m$

Work Step by Step

We can write two expressions for $L$: $L = \frac{n\lambda_n}{4} = \frac{(n)(8)}{4} = 2n$ $L = \frac{(n+2)~\lambda_{n+2}}{4} = \frac{(n+2)(4.8)}{4} = 1.2~n+2.4$ We can equate the two expressions to find $n$: $2n = 1.2n+2.4$ $0.8n = 2.4$ $n = 3$ Then: $~L = 2n = (2.0~m)(3) = 6.0~m$ The correct answer is: C. $6.0~m$
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