Respuesta :
Answer: The amount of energy transferred to each eardrum in one second is [tex]6.2 \times 10^{-9} J[/tex].
Explanation:
Given: Intensity = 0.10 [tex]W/m^{2}[/tex]
Now, at 30.0 m the intensity will be calculated as follows.
[tex]Intensity = \frac{(30.0 m)^{2}}{(1 m)^{2}}\\= 900[/tex]
This means that the intensity is 900 times less assuming that the wave is spherical.
Hence, the new intensity is calculated as follows.
[tex]\frac{0.10 W/m^{2}}{900}\\= 1.11 \times 10^{-4} W/m^{2}[/tex]
The area of ear drum is expressed as follows.
[tex]Area = \pi \times r^{2}\\= 3.14 \times (4.2 mm)^{2}\\= 5.54 \times 10^{-5} m^{2}\\[/tex]
Formula used to calculate energy is as follows.
[tex]P = I \times a[/tex]
where,
P = power or energy
I = intensity
a = area
Substitute the values into above formula as follows.
[tex]P = I \times a\\= 1.11 \times 10^{-4} W/m^{2} \times 5.54 \times 10^{-5} m^{2}\\= 6.2 \times 10^{-9} W[/tex]
Thus, we can conclude that amount of energy transferred to each eardrum in one second is [tex]6.2 \times 10^{-9} J[/tex].
