Answer:
The power is
![P = 2.88*10^(-4 ) \ W](https://img.qammunity.org/2021/formulas/physics/college/uyljgtf5l7e36b9udne9ydnjevn1ehp9ey.png)
Step-by-step explanation:
From the question we are told that
The distance from the siren is
![d = 17 \ m](https://img.qammunity.org/2021/formulas/physics/college/e7m92w145n7s4hqrfod79uyjo5q4dzhbhu.png)
The intensity level is
![\beta = 49\ dB](https://img.qammunity.org/2021/formulas/physics/college/y5q3b7181pcrosyhnl3dhgogaryc1mp2y6.png)
The threshold of hearing is
![I_0 = 1.0 *10^(-12) \ W/m^2](https://img.qammunity.org/2021/formulas/physics/college/viwyezwngqj1bk8ub0nl9gxekeccakdoy7.png)
Generally the intensity level is mathematically represented as
![\beta = 10dB * log [(I)/(I_o) ]](https://img.qammunity.org/2021/formulas/physics/college/9ao1v0p30m93pwbpa0x0ez4tjv1gdnp11h.png)
Where I is the intensity at which the siren radiates the sound
substituting values
![49 = 10 * log [(I)/(1.0 *10^(-12)) ]](https://img.qammunity.org/2021/formulas/physics/college/24vd3vfnxf89nx980s9n4p5g21cozyt9vt.png)
=>
![I = 7.94*10^(-8) W/m^2](https://img.qammunity.org/2021/formulas/physics/college/kx6t644c2335gipstosq091t51yzdj9hi3.png)
Now the amount of power the siren put out is mathematically evaluated as
![P= IA](https://img.qammunity.org/2021/formulas/physics/college/d14sttw7kizqooeh73qa2qrkx3iwa84yq7.png)
Where A is the area of the siren which is taken as a sphere and it is mathematically evaluated as
So
![P = I * 4 \pi d^2](https://img.qammunity.org/2021/formulas/physics/college/pg6tovetmxdsaydrulq18i0jy3sspqy6dw.png)
substituting values
![P = 7.94 *10^(-8) * 4 * 3.142 * (17)^2](https://img.qammunity.org/2021/formulas/physics/college/j69fqo92ctgewr5kww7wuol9rwz1c5md1h.png)
![P = 2.88*10^(-4 ) \ W](https://img.qammunity.org/2021/formulas/physics/college/uyljgtf5l7e36b9udne9ydnjevn1ehp9ey.png)