Respuesta :
Answer:
A) [tex]P=13.92\ J.s^{-1}[/tex]
B) [tex]v=3730.9912\ m.s^{-1}[/tex]
C) [tex]v=74.44\ mm.s^{-1}[/tex]
D) mosquitoes speed in part B is very much larger than that of part C.
Explanation:
Given:
- Distance form the sound source, [tex]s=50\ m[/tex]
- sound intensity level at the given location, [tex]\beta=114\ dB[/tex]
- diameter of the eardrum membrane in humans, [tex]d=8.4 \times 10^{-3}\ m[/tex]
- We have the minimum detectable intensity to the human ears, [tex]I_0=10^{-12}\ W.m^{-2}[/tex]
(A)
Now the intensity of the sound at the given location is related mathematically as:
[tex]\beta=10\ log(\frac{I}{I_0} )[/tex] ..........................................(1)
[tex]114=10\ log\ (\frac{I}{10^{-12}} )[/tex]
[tex]11.4=log\ I+12\ log\ 10[/tex]
[tex]I=0.2512\ W.m^{-2}[/tex]
As we know :
[tex]I=\frac{P}{A}[/tex]
[tex]0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }[/tex]
[tex]P=13.92\ J.s^{-1}[/tex] is the energy transferred to the eardrums per second.
(B)
mass of mosquito, [tex]m=2\times 10^{-6}\ kg[/tex]
Now the velocity of mosquito for the same kinetic energy:
[tex]KE=\frac{1}{2} m.v^2[/tex]
[tex]13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2[/tex]
[tex]v=3730.9912\ m.s^{-1}[/tex]
(C)
Given:
- Sound intensity, [tex]\beta = 20\ dB[/tex]
Using eq. (1)
[tex]20=10\ log\ (\frac{I}{10^{-12}} )[/tex]
[tex]2=log\ I+12\ log\ 10[/tex]
[tex]I=10^{-10}\ W.m^{-2}[/tex]
Now, power:
[tex]P=I.A[/tex]
[tex]P=10^{-10}\times \pi\times \frac{8.4^2}{4}[/tex]
[tex]P=5.54\times 10^{-9}\ J.s^{-1}[/tex]
Hence:
[tex]KE=\frac{1}{2} m.v^2[/tex]
[tex]5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2[/tex]
[tex]v=0.07444\ m.s^{-1}[/tex]
[tex]v=74.44\ mm.s^{-1}[/tex]
(D)
mosquitoes speed in part B is very much larger than that of part C.
