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(D) The energy incident on the surface per unit time is $U = I 	imes A$,where $I$ is the intensity (energy flux) and $A$ is the area.Given: $I = 75 \

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$10^{-5} \ kg m s^{-1}$

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(D) The energy incident on the surface per unit time is $U = I 	imes A$,where $I$ is the intensity (energy flux) and $A$ is the area.Given: $I = 75 	imes 10^4 \ W m^{-2}$,$A = 40 \ cm^2 = 40 	imes 10^{-4} \ m^2$,and $t = 1 \ s$.For a surface that completely absorbs the radiation,the momentum $p$ delivered is given by $p = \frac{U}{c} = \frac{I 	imes A 	imes t}{c}$,where $c = 3 	imes 10^8 \ m s^{-1}$ is the speed of light.Substituting the values:$p = \frac{75 	imes 10^4 	imes 40 	imes 10^{-4} 	imes 1}{3 	imes 10^8}$$p = \frac{75 	imes 40}{3 	imes 10^8} = \frac{3000}{3 	imes 10^8} = 1000 	imes 10^{-8} = 10^{-5} \ kg m s^{-1}$.

Electromagnetic waves of energy flux $75 \times 10^4 \ W m^{-2}$ are incident normally on a surface of area $40 \ cm^2$. If the surface absorbs the flux completely,the total momentum delivered to the surface in one second is:

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