Physics: Principles with Applications (7th Edition)

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

Chapter 22 - Electromagnetic Waves - Search and Learn - Page 643: 4

Answer

a) $4.244\times10^3\;\rm W/m^2$, $3.3\dot3\times10^{-9 }\;\rm N$ b) $8.488\times10^6\;\rm m/s^2$

Work Step by Step

a) We know that the pressure force is given by $$P=\dfrac{F}{A}$$ Thus, $$F=PA\tag 1$$ In our case, the pressure is given by $$P=\dfrac{I_{avg}}{c}$$ where $I_{avg}=\dfrac{P_{avg}}{A}$ Hence, $$P=\dfrac{P_{avg}}{Ac}\tag 2$$ Plugging into (1); $$F=\dfrac{P_{avg}}{Ac}A=\dfrac{P_{avg}}{c}$$ Plugging the known; $$F=\dfrac{1}{3\times10^8}=\color{red}{\bf 3.3\dot3\times10^{-9}}\;\rm N$$ The radiation pressure is given by equation (2). So we just need to plug the known into it. $$P=\dfrac{P_{avg}}{Ac}=\dfrac{P_{avg}}{\pi r^2c}$$ $$P=\dfrac{1}{3\times10^{8} \cdot \pi\cdot (5\times10^{-7})^2} =\color{red}{\bf4.244\times10^3}\;\rm W/m^2$$ Notice that $P$ is the pressure while $P_{avg}$ is the power, so do not be confused. --- b) According to Newton's second law, acceleration is given by $$a=\dfrac{F}{m}$$ and we know that the mass of this water cylinder is given by the density law; $m=\rho V=\rho Ah$. And in this case, the cross-sectional area is $A=\pi r^2$ and $h=r$. Therefore, $$a=\dfrac{F}{\rho \pi r^3}$$ Plugging the known; $$a=\dfrac{3.3\dot3\times10^{-9} }{1000\cdot \pi\cdot (5\times10^{-7})^3}$$ $$a=\color{red}{\bf8.488\times10^6}\;\rm m/s^2$$
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