Answer:
The distance of the second sound intensity level is 28.12 m.
Step-by-step explanation:
Given;
intensity of first sound level, I₁ = 110 dB
distance of first sound level, r₁ = 5 m
intensity of second sound level, I₂ = 95 dB
let the distance of second sound level = r₂
The intensity of the sound in W/m² is given as;
![dB = 10 Log[(I)/(I_o) ]](https://img.qammunity.org/2021/formulas/physics/college/zrv4kfgd990fy3dj3ug8tcytmmvcehtjh9.png)
where;
I₀ is threshold of hearing = 1 x 10⁻¹² W/m²
for 110 dB to W/m²
![110 = 10 Log[(I)/(1 \ * \ 10^(-12)) ]\\\\11 = Log[(I)/(1 \ * \ 10^(-12)) ]\\\\10^(11) = (I)/(1 \ * \ 10^(-12)) \\\\I = 1 \ * \ 10^(-12) \ * \ 10^(11)\\\\I = 10^(-1) \ W/m^2\\\\I = 0.1 \ W/m^2](https://img.qammunity.org/2021/formulas/physics/college/qblcuer8aep5hpmkv0l3muwbs81bkqcx32.png)
for 95 dB to W/m²
![95 = 10 Log[(I)/(1 \ * \ 10^(-12)) ]\\\\9.5 = Log[(I)/(1 \ * \ 10^(-12)) ]\\\\10^(9.5) = (I)/(1 \ * \ 10^(-12)) \\\\I = 1 \ * \ 10^(-12) \ * \ 10^(9.5)\\\\I = 10^(-2.5) \ W/m^2\\\\](https://img.qammunity.org/2021/formulas/physics/college/fpens7yaztpi5oa0vwit1tigxvmdehumul.png)
The intensity of sound is related to distance as follows;

Now, determine the distance of the second sound intensity level

Therefore, the distance of the second sound intensity level is 28.12 m.