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Use the following information for determining sound intensity. The number of decibels β of a sound with an intensity of I watts per square meter is given by β = 10 log(I/I0), where I0 is an intensity of 10−12 watt per square meter, corresponding roughly to the faintest sound that can be heard by the human ear. Find the number of decibels β of the sound.

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amnamurad28748

Answer:

120 decibels

Explanation:

we know sound in decibels is given by.

[tex]\beta =10log(\frac{I_{} }{I_{0} } )[/tex]

Where I is intensity of the sound and [tex]I_{0}=10^-12W/m^{2 }[/tex] is reference intensity.

substituting All of this in our decibel formula with I =1W/[tex]m^2[/tex]

gives us 120 decibels.

abidemiokin

The number of decibels β of the sound is 120 decibels

Given the number of decibels β of a sound with an intensity of I watts per square meter is given by;

[tex]\beta=10log(\frac{I}{I_0} )[/tex]

Given the following parameters

  • [tex]I_0[/tex] =[tex]10^{-12}watt/m^2[/tex]
  • [tex]I = 1watt/m^2[/tex]

Substitute the given parameters into the formula will give;

[tex]\beta=10log(\frac{1}{10^{-12}} )\\\beta =10log10^{12}\\\beta = 120log10\\\beta=120decibel[/tex]

Hence the number of decibels β of the sound is 120 decibels

Learn more about sound intensity here: https://brainly.com/question/15922588

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