Answer:
0.0032 W/m²
95.0514997832 dB
Explanation:
r = Distance
I = Intensity = 2 W/m²
[tex]I_0[/tex] = Threshold intensity = [tex]10^{-12}\ W/m^2[/tex]
Intensity is given by
[tex]I=\dfrac{P}{4\pi r^2}\\\Rightarrow P=I4\pi r^2\\\Rightarrow P=2\times 4\pi\times 2^2\\\Rightarrow P=32\pi[/tex]
at r = 50
[tex]I=\dfrac{P}{4\pi r^2}\\\Rightarrow I=\dfrac{32\pi}{4\pi 50^2}\\\Rightarrow I=0.0032\ W/m^2[/tex]
The sound intensity is 0.0032 W/m²
Sound intensity levele is given by
[tex]\beta=10log\dfrac{I}{I_0}\\\Rightarrow \beta=10log\dfrac{0.0032}{10^{-12}}\\\Rightarrow \beta=95.0514997832\ dB[/tex]
The sound intensity level is 95.0514997832 dB
(a) The sound intensity for the person watching from 50 m away is 0.0032 W/m².
(b) The sound intensity level is 95.1 dB.
The given parameters;
The power of the sound is calculated as follows;
[tex]I= \frac{P}{A} \\\\P = IA\\\\P = I(4\pi r^2)\\\\P = 2 \times 4 \times \pi \times 2^2\\\\P = 32 \pi \ W[/tex]
The sound intensity for the person watching from 50 m away is calculated as follows;
[tex]I = \frac{P}{4\pi r^2} \\\\I = \frac{32\pi }{4\pi \times 50^2} \\\\I = 0.0032 \ W/m^2[/tex]
The sound intensity level is calculated as follows;
[tex]\beta = 10\ log\frac{I}{I_o} \\\\\beta = 10 \times log (\frac{0.0032}{1\times 10^{-12}} )\\\\\beta = 95.1 \ dB[/tex]
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