Respuesta :
Answer
Given,
Sound Power, P = 600 W
a) Intensity of sound,
r = 5 m
we know,
[tex]I = \dfrac{P}{A}[/tex]
[tex]I = \dfrac{P}{4\pi r^2}[/tex]
[tex]I = \dfrac{600}{4\pi\times 5^2}[/tex]
I = 1.91 W/m²
b) Sound intensity
[tex]\beta = 10 log(\dfrac{I}{I_0})[/tex]
I₀ = 10⁻¹²
[tex]\beta = 10 log(\dfrac{1.91}{10^{-12}})[/tex]
[tex]\beta = 123\ dB[/tex]
c) Sound experienced by the Phil
β₁ = 123 - 23 = 100 \ dB
distance from the sound
again using sound intensity formula
[tex]\beta = 10 log(\dfrac{I}{I_0})[/tex]
[tex]100= 10 log(\dfrac{I}{10^{-12}})[/tex]
[tex]I = 10^{-12}\times 10^{10}[/tex]
where, I = P/A
[tex]\dfrac{P}{4\pi r^2} = 10^{-2}[/tex]
[tex] r^2 = \dfrac{600}{0.04\times \pi}[/tex]
r = 69.09 m
Distance where Phil will experience same sound intensity without earplug is equal to r = 69.09 m
