Respuesta :
Answer:
Energy transferred to eardrum while listening to this sound for 1.0 min is [tex]2.713 * 10^{-6}[/tex]
Explanation:
Given
Sound waves of intensity [tex]1.6 * 10^{-3} \frac{W}{m^2}[/tex]
Size of the ear drum i.e diameter [tex]6.0[/tex] millimeter
Radius of the ear drum
[tex]= \frac{D}{2} \\= \frac{6}{2} \\= 3[/tex]
Time for which the eardrum will listen to this sound [tex]= 1.0[/tex] minutes.
As we know that the energy transferred to eardrum can be calculated by using below given relation
[tex]E = I * A* t\\[/tex]
Where is the energy in Joules
I is the intensity in [tex]\frac{W}{m^2}[/tex]
A is the area in [tex]m^2[/tex]
and t is the time in seconds
Substituting the given values in above equation we get
[tex]E = 1.6 * 10^{-3} * (\pi r^2)* 60\\E = 1.6 * 10^{-3} * (3.14 * (3*10^{-3})^2)* 60\\E = 2.713 * 10^{-6}\\[/tex]
Energy transferred to eardrum while listening to this sound for 1.0 min is [tex]2.713 * 10^{-6}[/tex]
The energy transformed to eardrums in 1.0 min is [tex]2.71543 \times 10^{-6}[/tex]
Given to us
- The intensity of sound wave = [tex]1.6\times 10^{-3}\ \ W/m^2[/tex]
- Diameter of the eardrum = 6.0 mm = 0.006 m
- time = 1.0 min = 60 seconds
What is the radius of the eardrum?
[tex]Radius, r = \dfrac{Diameter}{2} = \dfrac{0.006}{2} = {0.003\ m}[/tex]
Therefore, the radius of the eardrum is 0.003 m.
Energy transformation
The energy transformed to the year drum can be written as,
[tex]E = I\times A\times t[/tex]
where
I is the intensity of the sound wave,
A is the area,
t is the time,
Substituting the values,
[tex]E = I\times A\times t\\E = 1.6 \times 10^{-3} \times (\pi r^2) \times 60\\E = 2.71543 \times 10^{-6}[/tex]
Hence, the energy transformed to eardrums in 1.0 min is [tex]2.71543 \times 10^{-6}[/tex].
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