Respuesta :
I found this question online and we should calculate the efficiency of this device.
To do this we need to calculate the power of the sound wave this device outputs. We can do this by using acoustic power level definition.
[tex]L_w=10log_{10} \frac{I}{I_0} [/tex]
I is sound intensity and [tex]I_0[/tex] is reference sound intensity.
[tex]I_0=1 \frac{pW}{m^2} =10^{-12} \frac{W}{m^2} [/tex]
We are given power level of the sound wave and we can use that information to find sound intensity.
[tex]94=10log_{10} \frac{I}{I_0} [/tex]
[tex]10^{9.4}= \frac{I}{10^{-12}} [/tex]
[tex]I=10^{9.4}\cdot10^{-12}[/tex]
[tex]I=2.5\cdot10^{-3} \frac{W}{m^2} [/tex]
To get the power we need to multiply intentisy with the surface through witch the wave is passing through. This would be the surface of the sphere.
[tex]P=IA[/tex]
[tex]A=4 \pi r^2[/tex]
[tex]A=4\pi(250)^2[/tex]
[tex]A=785398.2m^2[/tex]
[tex]P=785398.2\cdot2.5\cdot10^{-3} =1963.5 W[/tex]
Now we just divide this with input power to find effiency.
[tex]e= \frac{P}{P_{in}} = \frac{1963.5 }{40000}= 0.049=4.9\% [/tex]
To do this we need to calculate the power of the sound wave this device outputs. We can do this by using acoustic power level definition.
[tex]L_w=10log_{10} \frac{I}{I_0} [/tex]
I is sound intensity and [tex]I_0[/tex] is reference sound intensity.
[tex]I_0=1 \frac{pW}{m^2} =10^{-12} \frac{W}{m^2} [/tex]
We are given power level of the sound wave and we can use that information to find sound intensity.
[tex]94=10log_{10} \frac{I}{I_0} [/tex]
[tex]10^{9.4}= \frac{I}{10^{-12}} [/tex]
[tex]I=10^{9.4}\cdot10^{-12}[/tex]
[tex]I=2.5\cdot10^{-3} \frac{W}{m^2} [/tex]
To get the power we need to multiply intentisy with the surface through witch the wave is passing through. This would be the surface of the sphere.
[tex]P=IA[/tex]
[tex]A=4 \pi r^2[/tex]
[tex]A=4\pi(250)^2[/tex]
[tex]A=785398.2m^2[/tex]
[tex]P=785398.2\cdot2.5\cdot10^{-3} =1963.5 W[/tex]
Now we just divide this with input power to find effiency.
[tex]e= \frac{P}{P_{in}} = \frac{1963.5 }{40000}= 0.049=4.9\% [/tex]