Answer:
The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰
Explanation:
Given;
rock concert sound intensity level, β₁ = 120 dB
whisper sound intensity level, β₂ = 20 dB
The sound intensity level is given as;
[tex]\beta = 10Log(\frac{I}{I_o} )\\\\[/tex]
where;
I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²
I is the sound intensity
Intensity of sound at rock concert ;
[tex]120 = 10Log(\frac{I}{10^{-12}} )\\\\12 = Log(\frac{I}{10^{-12}} )\\\\10^{12} = \frac{I}{10^{-12}}\\\\I = 10^{12} * 10^{-12}\\\\I = 10^0\\\\I = 1 \ W/m^2[/tex]
The intensity of sound of a whisper;
[tex]20 = 10Log(\frac{I}{10^{-12}} )\\\\2 = Log(\frac{I}{10^{-12}} )\\\\10^{2} = \frac{I}{10^{-12}}\\\\I = 10^{2} * 10^{-12}\\\\I = 10^{-10} \ W/m^2\\\\[/tex]
Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper
[tex]\frac{I_{Concert}}{I_{whisper}} = \frac{1}{10^{-10}} \\\\\frac{I_{Concert}}{I_{whisper}} = 1 * 10^{10}\\\\I_{Concert} = 1 * 10^{10}*I_{whisper}[/tex]
Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰
