91Ó°ÊÓ

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.

The fan at a rock concert $30.0\text{″¾}$ above the stage is producing $110\text{ dµþ}$ sound at that point, the radius of the eardrum is$4.2×{10}^{{}^{−3}}\mathrm{m}$, the kinetic energy of the mosquito is$5.5\text{​â¶Ä‰W}$, the mass of the mosquito is$2\text{″¾g}$ , and the speed of the mosquito corresponding to whisper of $20\text{ dµþ}$ is $0.074\text{″¾}/\text{s}$.

Substitute$5.5\text{​â¶Ä‹â¶Ä‰}\mathrm{μ´³}$for E and $2\text{″¾g}$ for $\mathrm{m}$ in the equation$\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2$to find v.

$\begin{array}{c}(5.5\text{ }\mathrm{μ´³})\left(\frac{\mathrm{I}\text{ }\mathrm{J}}{{10}^6\text{ }\mathrm{μ´³}}\right)=\frac{1}{2}\left(2\text{ }\mathrm{mg}\right)\left(\frac{1\text{ }\mathrm{kg}}{{10}^6\text{ }\mathrm{mg}}\right)×{\mathrm{v}}^2\\ {\mathrm{v}}^2=5.5\text{ }{(\mathrm{m}/\mathrm{s})}^2\\ \mathrm{v}=2.3\text{ }\mathrm{m}/\mathrm{s}\end{array}$

Therefore, the speed of the mosquito in order to achieve kinetic energy of $5.5\text{ }\mathrm{μ´³}$ is $2.3\text{ }\mathrm{m}/\mathrm{s}$.

The relevant speed of the mosquito is $2.3\text{ }\mathrm{m}/\mathrm{s}$, according to the section 1 answer. It is assumed that the mosquito moves at a speed of $0.074\text{″¾}/\text{s}$. As a result, the mosquito's speed corresponds to a whisper of $110\text{ d±ð³¦¾±²ú±ð±ô²õ},$ which is far faster than the mosquito's speed corresponds to a whisper of $20\text{ d±ð³¦¾±²ú±ð±ô²õ}$.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$20\text{ d±ð³¦¾±²ú±ð±ô²õ}$ whisper.