Respuesta :
Answer: 2.113785 * 10^(-7) watts.
Explanation:
To solve this problem, we need to use the formulas for sound intensity level and power.
The formula for sound intensity level (L in dB) is:
L = 10 * log10(I/I₀)
Where I₀ is the reference intensity which is 1 * 10^(-12) W/m^2.
Now we can calculate the initial power received by one eardrum:
Given:
Noise level (L₁) = 90 dB
Eardrum Diameter (d) = 8.4 mm = 0.0084 m
First, let's convert the eardrum diameter to radius (r):
r = d/2 = 0.0084/2 = 0.0042 m
Next, we need to calculate the surface area of the eardrum (A):
A = π * r^2 = 3.14159 * (0.0042)^2 = 0.00005538038 m^2
Now, we can calculate the initial sound intensity (I₁) received by one eardrum:
L₁ = 10 * log10(I₁/I₀)
90 = 10 * log10(I₁/(1 * 10^(-12)))
9 = log10(I₁/(1 * 10^(-12)))
Using logarithmic properties, we can rewrite the equation as:
I₁/(1 * 10^(-12)) = 10^9
I₁ = (1 * 10^(-12)) * 10^9
I₁ = 10^(-3) W/m^2
Now, let's calculate the initial power (P₁) received by one eardrum:
P₁ = I₁ * A
P₁ = 10^(-3) * 0.00005538038
P₁ = 5.538038 * 10^(-8) W
Now, we need to calculate the power received by one eardrum after using the earplugs:
Given:
Noise reduction (ΔL) = 26 dB
To calculate the new power (P₂) received by one eardrum, we need to adjust the sound intensity (I) using the noise reduction:
ΔL = 10 * log10(I₂/I₁)
26 = 10 * log10(I₂/(10^(-3)))
Dividing by 10 and isolating I₂, we have:
I₂/(10^(-3)) = 10^(26/10)
I₂/(10^(-3)) = 10^(2.6)
I₂ = 10^(2.6) * (10^(-3))
Now, let's calculate the new power (P₂) received by one eardrum:
P₂ = I₂ * A
P₂ = 10^(2.6) * (10^(-3)) * 0.00005538038
P₂ = 2.113785 * 10^(-7) W
Therefore, the power received by one eardrum after using the earplugs is approximately 2.113785 * 10^(-7) watts.
