Answer
(a) The intensity is $0.050~W/m^2$.
(b) The energy emitted in one hour is $2.2\times 10^4~J$.
Work Step by Step
(a) We can find the intensity $I_2$ at a distance of 3.1 meters from the source.
$\frac{r_2^2}{r_1^2} = \frac{I_1}{I_2}$
$I_2 = \frac{I_1~r_1^2}{r_2^2}$
$I_2 = \frac{(0.026~W/m^2)(4.3~m)^2}{(3.1~m)^2}$
$I_2 = 0.050~W/m^2$
The intensity is $0.050~W/m^2$.
(b) We can find the power of the source.
$P = I_1~A_1$
$P = I~(4\pi~r_1^2)$
$P = (0.026~W/m^2)~(4\pi)(4.3~m)^2$
$P = 6.04~W$
We can find the energy emitted in one hour.
$E = P~t$
$E = (6.04~W)(3600~s)$
$E = 2.2\times 10^4~J$
The energy emitted in one hour is $2.2\times 10^4~J$.
