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In a factory, three machines produce noise with intensity levels of 78 dB, 85 dB

ID: 2011487 • Letter: I

Question

In a factory, three machines produce noise with intensity levels of 78 dB, 85 dB, and 88 dB.
(a) When all three are running, what is the intensity level in dB?

b) How does this compare to running just the loudest machine?

Intensity level is less than 5% greater.
Intensity level is between 5% and 20% greater.
Intensity level is between 20% and 50% greater.
Intensity level is between 50% and 100% greater.
Intensity level is more than 100% greater.

could you please show all work/algebra to help me understand. thank you

Explanation / Answer

The loudness is given by = 10log(I/I0) the intensity I = I010/10 given loudness 1 = 78 dB I1 = I0101/10 I1 = I0107.8 2 = 85 dB I2 = I0102/10 I2 = I0108.5 3 = 88 dB I3 = I0103/10 I3 = I0108.8 The intensity when all machines are run together I = I1+I2 +I3 I = I0(107.8 + 108.5 + 108.8) The sound intensity level is = 10log(I/I0) = 10log(107.8 + 108.5 + 108.8) = 90.044 dB B) the fraction is given by /3 = 90.044/88 /3 = 1.023 /3 = 102.3 % Therefore the loudness is more than 100% 2 = 85 dB I2 = I0102/10 I2 = I0108.5 3 = 88 dB I3 = I0103/10 I3 = I0108.8 The intensity when all machines are run together I = I1+I2 +I3 I = I0(107.8 + 108.5 + 108.8) The sound intensity level is = 10log(I/I0) = 10log(107.8 + 108.5 + 108.8) = 90.044 dB B) the fraction is given by /3 = 90.044/88 /3 = 1.023 /3 = 102.3 % Therefore the loudness is more than 100% 3 = 88 dB I3 = I0103/10 I3 = I0108.8 The intensity when all machines are run together I = I1+I2 +I3 I = I0(107.8 + 108.5 + 108.8) The sound intensity level is = 10log(I/I0) = 10log(107.8 + 108.5 + 108.8) = 90.044 dB B) the fraction is given by /3 = 90.044/88 /3 = 1.023 /3 = 102.3 % Therefore the loudness is more than 100% B) the fraction is given by /3 = 90.044/88 /3 = 1.023 /3 = 102.3 % Therefore the loudness is more than 100%

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