Respuesta :
Answer:[tex]0.00130 W/m^2[/tex]
Explanation:
Given
[tex]\beta =107 db[/tex]
[tex]d_1=0.5 m[/tex]
[tex]d_2=3.1 m[/tex]
[tex]I_0=10^{-12}[/tex]
sound level
[tex]\beta =10log_{10}\frac{I_1}{I_0}[/tex]
[tex]107=10log_{10}\frac{I_1}{10^{-12}}[/tex]
[tex]I_1=0.05 w/m^2[/tex]
[tex]Intensity \propto \frac{1}{distance^2}[/tex]
[tex]\frac{I_1}{I_2}=\frac{d_2^2}{d_1^2}[/tex]
[tex]I_2=0.00130 W/m^2[/tex]
The intensity of the sound wave is inversely proportional to the distance of the object and source of sound wave.
The intensity of the sound heard by someone standing 3.1 m away is 0.00130 W/m squared.
What is the intensity of the sound?
Intensity of the sound is the intensity flowing per unit area which is perpendicular to the direction of sound wave travelling in.
The intensity of the sound wave is inversely proportional to the distance of the object and source of sound wave.
Given information-
The sound intensity of African cicada insect is 107 dB at a distance of 0.5 m.
The distance of object to hear the sound is 3.1 m.
The application of logarithm function can be used to find the sound level as,
[tex]D=10\log 10\dfrac{I}{I_n} \\[/tex]
Here, [tex]D[/tex] is the intensity of sound level in decibels.
[tex]107=10\log 10\dfrac{I}{10^{-12}} \\\\I=0.05 \rm \omega^2/m^2[/tex]
As, the intensity of the sound wave is inversely proportional to the distance of the object and source of sound wave. Thus,
[tex]\dfrac{I}{I_o} =\dfrac{3.1^2}{0.5^2} \\I_o=0.00130 \rm W/m^2[/tex]
Hence the intensity of the sound heard by someone standing 3.1 m away is 0.00130 W/m squared.
Learn more about the sound intensity here;
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