Calculating Energy Absorption by an Eardrum Exposed to Sound Waves

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To calculate the energy absorbed by an eardrum exposed to a sound wave of 80.1 dB and an area of 0.7x10^-4 m² over 4 minutes, the correct approach involves converting the intensity in decibels to a linear scale. The formula for intensity in decibels is LI = 10 log(I / I0), where I0 is the reference intensity. After converting the intensity to watts per square meter, the power can be calculated using P = (Intensity)(Area). The energy absorbed is then found using E = (Power)(Time), leading to the correct answer of 1.72 Joules. Proper unit conversion is crucial for accurate results.
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A sound wave with intensity of 80.1 dB is incident on an eardrum of Area = 0.7x10^-4. How much energy is absorbed by the eardrum in 4 minutes?


I know this much... Power = (Intensity)(Area)..so...P=(80.1)(0.7x10^-4)

and Energy = (power)(time) so...E = (80.1)(0.7x10^-4)(240 sec) = 1.35


However, this is not the correct answer. Am i not doing the right conversions or what?? What looks wring here? (the correct answer is 1.72)
 
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E = (80.1) ( dB ) (0.7x10^-4)(240 sec)

I think that you must change unit db
 
Yea I thought about that but i would have no idea how to convert dB to Hz or whatever it needed to be.
Any thoughts??
 
I'm a little confused about how to use this equation. I looked at the example on yahoo over and over but can't seem to figure out where some of those numbers are coming from.
 
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