Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems - Page 239: 47


(a)$K=6.7\times 10^{-5}\frac{N}{m}$ (b)$m=3.4\times 10^{-10}Kg$

Work Step by Step

(a)We know that $K=\frac{F}{x}$ We plug in the known values to obtain: $K=\frac{1\times 10^{-12}}{15\times 10^{-9}}=6.7\times 10^{-5}\frac{N}{m}$ (b) We can find the mass as follows: $m=\frac{K}{\omega^2}$ $\implies m=\frac{6.667\times 10^{-5}}{2\pi(70)^2}$ $m=3.4\times 10^{-10}Kg$
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