Answer:
(a) [tex]I=8.84*10^{-9}W/m^{2}[/tex]
(b) [tex]\beta =39.46dB[/tex]
Explanation:
Given data
The source is 1.00 μW
The point is 3m away from the point source
For Part (a)
The intensity at a certain distance from a point source that emits sound wave is given as:
[tex]I=\frac{P_{s} }{4\pi r^{2} }\\I=\frac{1*10^{-6}W }{4\pi (3m)^{2} } \\I=8.84*10^{-9}W/m^{2}[/tex]
For Part (b)
The Sound level is given by
β=(10dB)×log(I/I₀)
Where
I₀=1×10⁻¹²W/m²
Substitute the given values to find Sound level
So
[tex]\beta =(10dB)log(\frac{8.84*10^{-9}W/m^{2} }{1*10^{-12}W/m^{2}} )\\\beta =39.46dB[/tex]
