Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 18 - Electric Currents - General Problems - Page 524: 78

Answer

$A=4.39\times10^{-8}m^2$ $l=39.7m$

Work Step by Step

$15.2\Omega=(1.68\times10^{-8}\Omega \cdot m)\frac{l}{A}$ $lA=\frac{0.0155kg}{8900kg/m^3}$ $l=\frac{15.2\Omega A}{1.68\times10^{-8}\Omega \cdot m}$ $\frac{15.2\Omega A^2}{1.68\times10^{-8}\Omega \cdot m}=\frac{0.0155kg}{8900kg/m^3}$ $A=4.39\times10^{-8}m^2$ $l=39.7m$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.