Physics: Principles with Applications (7th Edition)

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

Chapter 17 - Electric Potential - Search and Learn - Page 500: 5

Answer

a) $100\;\rm hp$ b) $1.35\times10^9\;\rm J$ c) $3.74\times10^2\;\rm F$ d) $2700\;\rm kg$ e) No

Work Step by Step

a) We know that $\rm 1\;hp=746\;\rm W$, so the power of the car in hp is $$P=75\times 10^3\;\rm W\cdot \dfrac{1\;hp}{746\;W}=\color{red}{\bf1.0\times10^2}\;hp$$ b) We know that the power is given by $$P=\dfrac{E}{t}$$ Thus, the energy is given by $$E=Pt=75\times10^3\cdot 5\times 3600=\color{red}{\bf 1.35\times10^9}\;\rm J$$ c) We know that the stored energy in the capacitor is given by $$PE=\frac{1}{2}CV^2$$ So, the capacitance is given by $$C=\dfrac{2PE}{V^2} =\dfrac{2E}{V^2}=\dfrac{2\cdot 1.35\times10^8}{850^2}= \color{red}{\bf 3.74\times10^2}\;\rm F$$ e) The capacitance of 0.1 g of activated carbon is about 1 F (according to your textbook). And to find the amount of the activated carbon, we need first to find the capacitance in this case. Using the same formula we used in part (c); $$C =\dfrac{2E}{V^2}=\dfrac{2\cdot 1.35\times10^9}{10^2}= 2.7\times10^7 \;\rm F$$ Thus, $$m=C\cdot \rm \dfrac{0.1\;g}{1\;F}= 2.7\times10^7\;\rm F\cdot \dfrac{0.1\times10^{-3}\;g}{1\;F} = \color{red}{\bf2.7\times 10^3} \;\rm kg$$ e) This is not practical since the mass of the capacitor will be about 3 times of the car (an average car is about 1000 kg). So, the answer is: $\color{red}{\text{No, it is not}}.$
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