Problem 5 A fan at a rock concert is 50.0{\ m m} from the stage, and at this poi

Question

Problem 5
A fan at a rock concert is 50.0{ m m} from the stage, and at this point the sound intensity level is 109{ m dB} . Sound is detected when a sound wave causes the tympanic membrane (the eardrum) to vibrate . Typically, the diameter of this membrane is about 8.40{ m mm} in humans

Part A
How much energy is transferred to her eardrums each second ?
answer in uj

Part B
How fast would a 2.00{ m mg} mosquito have to fly to have this much kinetic energy?
Answer in m/s

Can anyone help me understand this answer in detail i haven't recieved a correct answer yet.

Explanation / Answer

Sound intensity level L in dB and sound intensity are related as: L = 10·log10( I/I0 ) with I0 = 10?¹²W/m² threshold of hearing So the sound intensity at the edge of the stage is: I1 = I0 · 10^(L/10) = 10?¹²W/m² · 10^(124/10) = 2.51W/m² Sound waves intensity is inversely proportional to the squares distance to source: I ~ 1/r² Hence sound intensity experienced by the fan and sound intensity at edge of stage are related as I2/I1 = (r1/r2)² => I2 = I1·(r1/r2)² = 2.51W/m² · (1m / 33.4m)² = 2.25×10?³W/m² Multiply this by the area of the eardrum and you get the sound power transferred to it: P = I2 · A = I2 · p·(1/4)·d² = 2.25×10?³W/m² · p·(1/4) · (6.4×10?³m )² = 7.24×10?8 W That means there are 7.24×10?8 joules per second transferred to the eardrum

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