Q16E - University Physics with Modern Physics

Question

BIO Human Hearing. A fan at a rock concert is 30 m from the stage, and at this point the sound intensity level is 110 dB (a) How much energy is transferred to her eardrums each second? (b) How fast would a 2.0-mg mosquito have to fly (in mm/s) to have this much kinetic energy? Compare the mosquito’s speed with that found for the whisper in part (a) of Exercise 16.13.

Step-by-Step Solution

Answer

a) The speed of the mosquito in order to achieve kinetic energy of 5.5 PJ is 2.3 m/s.

b) The speed of the mosquito corresponding to a whisper of 110 dB is much greater than the speed of the mosquito corresponding to a whisper of 20 dB.

STEP 1 : Concept of the kinetic energy of the mosquito

The kinetic energy of the mosquito is, $\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$  were, E is the kinetic energy. m is the mass of the mosquito. v is the speed of the mosquito.

STEP 2 Calculate the kinetic energy of the mosquito

The fan at a rock concert  $30.0\text{\hspace{0.17em}m}$ above the stage is producing $110\text{\hspace{0.17em}dB}$  sound at that point, the radius of the eardrum is $4.2\times {10}^{{}^{-3}}\mathrm{m}$ , the kinetic energy of the mosquito is $5.5\text{\hspace{0.17em}W}$, the mass of the mosquito is $2\text{\hspace{0.17em}mg}$  , and the speed of the mosquito corresponding to whisper of  $20\text{\hspace{0.17em}dB}$  is $0.074\text{\hspace{0.17em}m}/\text{s}$ .

Substitute  $5.5\text{\hspace{0.17em}}\mathrm{\mu J}$ for E and  $2\text{\hspace{0.17em}mg}$ for $\mathrm{m}$  in the  equation $\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2$  to find v. 

 $\begin{array}{c}(5.5\mathrm{\mu J})\left(\frac{\mathrm{I}\mathrm{J}}{{10}^6\mathrm{\mu J}}\right)=\frac{1}{2}\left(2\mathrm{mg}\right)\left(\frac{1\mathrm{kg}}{{10}^6\mathrm{mg}}\right)\times {\mathrm{v}}^2\\ {\mathrm{v}}^2=5.5{(\mathrm{m}/\mathrm{s})}^2\\ \mathrm{v}=2.3\mathrm{m}/\mathrm{s}\end{array}$

Therefore, the speed of the mosquito in order to achieve kinetic energy of $5.5\mathrm{\mu J}$  is  $2.3\mathrm{m}/\mathrm{s}$.  

STEP 3 Speed of the mosquito

The relevant speed of the mosquito is $2.3\mathrm{m}/\mathrm{s}$ , according to the section 1 answer. It is assumed that the mosquito moves at a speed of $0.074\text{\hspace{0.17em}m}/\text{s}$ . As a result, the mosquito's speed corresponds to a whisper of $110\text{\hspace{0.17em}decibels},$ which is far faster than the mosquito's speed corresponds to a whisper of $20\text{\hspace{0.17em}decibels}$.As a result, the mosquito's speed equal to a whisper of 110 dB is substantially higher than the mosquito's speed, which corresponds to a $20\text{\hspace{0.17em}decibels}$   whisper.