Answer:
the required distance is 89.125 m
Explanation:
Given the data in the question;
we know that, sound intensity B in decibels of sound is;
β(dB) = 10log₁₀( [tex]I[/tex] / [tex]I_0[/tex] )
where intensity [tex]I[/tex] = power / area carried by wave
[tex]I_0[/tex] = 10⁻¹² W/m² { minimum threshold intensity }
Now,
intensity [tex]I[/tex] = power / area carried by wave = P/A = P/4πr² { spherical }
given that; β = 80.0 dB and P = 10 W
so
β(dB) = 10log₁₀( [tex]I[/tex] / [tex]I_0[/tex] )
we substitute
80 = 10log₁₀( P / 4πr²× [tex]I_0[/tex])
80 = 10log₁₀( 10 / 4πr²× 10⁻¹² )
8 = log₁₀(10) - log₁₀( 4πr²× 10⁻¹² )
8 = 1 - log₁₀( 4πr²× 10⁻¹² )
8 - 1 = -log₁₀( 4πr²× 10⁻¹² )
7 = -log₁₀( 1.2566 × 10⁻¹¹ × r² )
7 = -[ log₁₀( 1.25 × 10⁻¹¹) + log₁₀( r² ) ]
7 = -[ -10.9 + log₁₀( r² ) ]
7 = 10.9 - log₁₀( r² )
-log₁₀( r² ) = 7 - 10.9
-log₁₀( r² ) = - 3.9
log₁₀( r² ) = 3.9
2log₁₀r = 3.9
log₁₀r = 3.9 /2
log₁₀r = 1.95
r = 89.125 m
Therefore, the required distance is 89.125 m
