## College Physics (4th Edition)

The star radiates EM energy at a rate of $8.82\times 10^{26}~W$
We can convert the distance $r$ to units of meters: $r = (14\times 10^6~ly) \times \frac{9.46\times 10^{15}~m}{1~ly}$ $r = 1.3244\times 10^{23}~m$ We can find the power of the source: $I = \frac{P}{A}$ $P = I~A$ $P = I~(4\pi~r^2)$ $P = (4\times 10^{-21}~W/m^2)(4\pi)~(1.3244\times 10^{23}~m)^2$ $P = 8.82\times 10^{26}~W$ The star radiates EM energy at a rate of $8.82\times 10^{26}~W$