Answer:
the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB
Explanation:
Given the data in the question;
sound intensity reduced by the factor, m = 305
the sound intensity level experienced by the crew members wearing protective earplugs, L = 79 dB
Now, using the expression of sound intensity level;
L = 10log( [tex]I_0[/tex] )
where [tex]I_0[/tex] is the intensity at L level
so we substitute
79 = 10log( [tex]I_0[/tex] )
[tex]I_0[/tex] = [tex]10^{7.9[/tex]
Now, expression for actual intensity;
[tex]I[/tex] = m[tex]I_0[/tex]
where [tex]I[/tex] is the actual intensity
so we substitute
[tex]I[/tex] = 305 × [tex]10^{7.9[/tex]
Next, we write the expression of sound intensity level for reduced intensity;
L' = 10log( [tex]I[/tex] )
So we substitute
L' = 10log( 305 × [tex]10^{7.9[/tex] )
L' = 10log( 24227011159.09058 )
L' = 103.8 dB
Therefore, the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB
