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phy 202
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** Question Form_labelMessages **
test 2 ?
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Problem Number 2
One sound is 7500 times louder than a sound which measures 45 decibels. What is the decibel level of this sound?
I am not sure how to attempt this? I slightly remember something like D=10 log (I/10^_12) ?? Could you help me?
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You can use the equation to find the intensity I of the 45 decibel sound.
You can then multiply this intensity by 7500 to get the intensity of the louder sound.
The equation can then be used to find the decibel level of the louder sound.
To solve the equation for I you will divide both sides by 10 to get
D / 10 = log( I / (10^-12 watts / meter^2) ).
Since 10^x and log(x) are inverse functions, your next step is
10^(D / 10) = 10^(log(I / (10^-12 watts / meter^2))
Applying the above-mentioned inverse property you get
I / (10^-12 watts / meter^2) = 10^(D / 10)
so that
I = 10^(D/10) * 10^-12 watts / meter^2.
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