The sound intensity level of 100 physics professors talking simultaneously is equal to 72 decibel.
Given the following data:
Sound intensity can be defined as a measure of the power of a sound wave per unit area. Thus, sound intensity is a quantity that can be used to measure the energy of sound and its unit of measurement is Watts per square meter.
Mathematically, sound intensity level is given by this formula:
[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]
Note: The reference value of sound intensity is equal to [tex]1.0 \times 10^{-12}\;W/m^2[/tex]
Thus, the sound intensity for one (1) professor is given by:
[tex]I_1 = 1.0 \times 10^{-12} \times 10^{5.2}\\\\I_1=1.585 \times 10^{-7}\;W/m^2[/tex]
For 100 professors, the sound intensity is:
[tex]I_{100} = 1.585 \times 10^{-7} \times 100\\\\I_{100}=1.585 \times 10^{-5}\;W/m^2[/tex]
Substituting the parameters into the formula, we have;
[tex]\beta = 10 log(\frac{1.585 \times 10^{-5}}{1.0 \times 10^{-12}})\\\\\beta = 10 log(15.58 \times 10^{6})\\\\\beta = 10 \times 7.2\\\\\beta =72\;dB[/tex]
Read more on sound intensity here: https://brainly.com/question/17062836
Complete Question:
One physics professor talking produces a sound intensity level of 52 dB. It's a frightening idea, but what would be the sound intensity level of 100 physics professors talking simultaneously?
