Question

It's a frightening idea, but what would be the sound intensity level of 100 physics professors talking simultaneously

Answer 1

The sound intensity level of 100 physics professors talking simultaneously is equal to 72 decibel.

Given the following data:

• Sound intensity = 52 dB.

What is sound intensity?

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.

How to calculate the sound intensity level.

Mathematically, sound intensity level is given by this formula:

$\beta = 10 log(\frac{I}{I_{ref}} )$

Note: The reference value of sound intensity is equal to $1.0 \times 10^{-12}\;W/m^2$

Thus, the sound intensity for one (1) professor is given by:

$I_1 = 1.0 \times 10^{-12} \times 10^{5.2}\\\\I_1=1.585 \times 10^{-7}\;W/m^2$

For 100 professors, the sound intensity is:

$I_{100} = 1.585 \times 10^{-7} \times 100\\\\I_{100}=1.585 \times 10^{-5}\;W/m^2$

Substituting the parameters into the formula, we have;

$\beta = 10 log(\frac{1.585 \times 10^{-5}}{1.0 \times 10^{-12}})\\\\\beta = 10 log(15.58 \times 10^{6})\\\\\beta = 10 \times 7.2\\\\\beta =72\;dB$

Read more on sound intensity here: 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?