Answer:
The sound level of the 26 geese is [tex]Z_{26}= 96.15 dB[/tex]
Explanation:
From the question we are told that
The sound level is [tex]Z_1 = 81.0 \ dB[/tex]
The number of geese is [tex]N = 26[/tex]
Generally the intensity level of sound is mathematically represented as
The intensity of sound level in dB for one goose is mathematically represented as
[tex]Z_1 = 10 log [\frac{I}{I_O} ][/tex]
Where I_o is the threshold level of intensity with value [tex]I_o = 1*10^{-12} \ W/m^2[/tex]
[tex]I[/tex] is the intensity for one goose in [tex]W/m^2[/tex]
For 26 geese the intensity would be
[tex]I_{26} = 26 * I[/tex]
Then the intensity of 26 geese in dB is
[tex]Z_{26} = 10 log[\frac{26 I }{I_o} ][/tex]
[tex]Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )[/tex]
[tex]Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )[/tex]
From the law of logarithm we have that
[tex]Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ][/tex]
[tex]= 14.15 + 82[/tex]
[tex]Z_{26}= 96.15 dB[/tex]