Answer:
Part a)
[tex]r = 9.4 \times 10^4 m[/tex]
Part b)
[tex]I = 0.25 \mu W/m^2[/tex]
Part c)
[tex]P = 112 kW[/tex]
Explanation:
Part a)
As we know that the intensity of sound for point source is given as
[tex]I = \frac{P}{4\pi r^2}[/tex]
so here we know that
[tex]\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}[/tex]
so we have
[tex]\frac{9.70}{1\times 10^{-3}} = \frac{r^2}{30.3^2}[/tex]
[tex]r = 9.4 \times 10^4 m[/tex]
Part b)
Now we know that intensity inversely varies with the distance
So if the friend house is at double distance then the intensity must be 1/4 times
so we have
[tex]I = \frac{1 mW/m^2}{4}[/tex]
[tex]I = 0.25 \mu W/m^2[/tex]
Part c)
Power received at the time of take off is given as
[tex]\frac{P}{4\pi r^2} = 9.70 W/m^2[/tex]
[tex]r = 30.3 m[/tex]
so we have
[tex]P = 112 kW[/tex]