Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 21 - Electric Current and Direct-Current Circuits - Problems and Conceptual Exercises - Page 754: 7

Answer

(a) $1.5KC$ (b) $8.5 y$

Work Step by Step

(a) We know that $I=\frac{\Delta Q}{\Delta t}$ This can be rearranged as: $\Delta Q=I\Delta t$ We plug in the known values to obtain: $\Delta Q=(0.42A.h)(\frac{3600A.s}{A.h})$ $\Delta Q=1512As$ $\implies \Delta Q=1512C$ $\Delta Q=1.5\times 10^3C=1.5KC$ (b) As $\Delta t=\frac{\Delta Q}{I}$ We plug in the known values to obtain: $\Delta t=\frac{1512}{5.6\times 10^{-6}}$ $\Delta t=2.7\times 10^8$ $\Delta t=(2.7\times 10^8)(\frac{min}{60})(\frac{h}{60min})(\frac{day}{24h})(\frac{year}{365 days})$ $\Delta t=8.5 y$
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