 # Sound pressure level L in decibels is given by L=20log p/p_0 , where p is the sound pressure and p_0 is a reference pressure. The sound pressure level of a jet engine is 145 decibels, when p_0 is the sound pressure at the threshold of hearing, 2×10^−5 pascal. a. Write and solve an equation to find the sound pressure p of the jet engine to the nearest pascal. b. How many times the sound pressure at the threshold of hearing is the sound pressure of the jet engine? przesypkai4 2022-07-15 Answered
Sound pressure level L in decibels is given by $L=20log\frac{p}{{p}_{0}}$, where p is the sound pressure and ${p}_{0}$ is a reference pressure. The sound pressure level of a jet engine is 145 decibels, when ${p}_{0}$ is the sound pressure at the threshold of hearing, $2×{10}^{-5}$ pascal. a. Write and solve an equation to find the sound pressure p of the jet engine to the nearest pascal. b. How many times the sound pressure at the threshold of hearing is the sound pressure of the jet engine?
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a) We let $L=145,{p}_{0}=2×{10}^{-5}$ in the formula given:
$145=20\mathrm{log}\frac{p}{2×{10}^{-5}}$
Divide both sides by 20:
$7.25=\mathrm{log}\frac{p}{2×{10}^{-5}}$
Rewrite in exponential form:
${10}^{7.25}=\frac{p}{2×{10}^{-5}}$
Solve for p:
$p={10}^{7.25}\ast 2×{10}^{-5}$
=356Pa
b) This is given by the ratio $\frac{p}{{p}_{0}}:$
$\frac{p}{{p}_{0}}=\frac{356}{2×{10}^{-5}}=17,800,000$
Result:
a.356 Pascal
b. 17,8000,000 times greater

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