Answer:
150.51 dB
Step-by-step explanation:
Data provided in the question:
decibel level of sound at 161 km distance = 180 dB
d₁ = 161 km
d₂ = 4800 km
I₁ = 180 db
The formula for intensity of sound is given as:
I = [tex]10\log(\frac{I_2}{I_1})[/tex]
and the relation between intensity and distance is given as:
I ∝ [tex]\frac{1}{d^2}[/tex]
or
Id² = constant
thus,
I₁d₁² = I₂d₂²
or
[tex]\frac{I_2}{I_1}=\frac{d_1}{d_2}[/tex]
therefore,
I = [tex]10\log(\frac{d_1}{d_2})^2[/tex]
or
I = [tex]10\times2\times\log(\frac{161}{4,800})[/tex]
or
I = 20 × (-1.474)
or
I = -29.49
Therefore,
the decibel level on Rodriguez Island, 4,800 km away
= 180 - 29.49
= 150.51 dB