Answer:
64.57
Explanation:
We are given that
For decreasing measured sound intensity level=36.2 dB
We have to find the factor by which the distance increase.
Let initial distance=x
Final distance=x'
According to question
[tex]36.2=10log(\frac{x'^2}{x^2})[/tex]
[tex]36.2=10log(\frac{x'}{x})^2[/tex]
[tex]36.2=10\times 2log\frac{x'}{x}[/tex]
[tex]log\frac{x'}{x}=\frac{36.2}{20}=1.81[/tex]
[tex]\frac{x'}{x}=10^{1.81}[/tex]
[tex]x'=10^{1.81}x=64.57x[/tex]
Hence, the distance is increases by factor of 64.57
