A person rides up a lift to a mountaintop, but the persons ears fail to pop –that is, the pressure of the inner ear does not equalize with the outside atmosphere. The radius of each ear drum is 0.40×10ˆ-2m. The pressure of the atmosphere drops from 1.010×10ˆ5 Pa at the bottom of the lift to 0.998×10ˆ5Pa at the top. What is the pressure on the inner ear at the top of the mountain?
You want the difference in pressure, subtract.
To find the pressure on the inner ear at the top of the mountain, we need to calculate the difference in pressure between the inner ear and the atmosphere at the bottom of the lift.
Given:
Radius of each ear drum (r) = 0.40 × 10^(-2) m
Pressure at the bottom of the lift (P_bottom) = 1.010 × 10^5 Pa
Pressure at the top of the mountain (P_top) = 0.998 × 10^5 Pa
We can use the equation for pressure difference in a liquid or gas:
ΔP = ρgh
Where:
ΔP = pressure difference
ρ = density of the liquid or gas
g = acceleration due to gravity
h = height or depth
In this case, we can assume that the density of the air (ρ) remains constant. Therefore, we can solve for ΔP by finding the difference in height between the bottom and top of the lift.
Let's solve for the height difference:
Δh = (P_bottom - P_top) / (ρg)
Assuming ρ and g remain constant, we can substitute the given values:
ρ ≈ 1.2 kg/m³ (density of air at sea-level)
g = 9.8 m/s² (acceleration due to gravity)
Δh = (1.010 × 10^5 - 0.998 × 10^5) / (1.2 × 9.8)
Δh = 1.2 × 10^3 / 11.76
Δh ≈ 102.04 meters
Now, we need to calculate the pressure on the inner ear at the top of the mountain. Since we assume the ear drum to be a sphere, we can use the formula for pressure on a surface of a sphere due to a fluid:
P_inner_ear = P_top + (2/3) * ΔP
P_inner_ear = 0.998 × 10^5 + (2/3) * (ΔP)
Let's substitute the values:
P_inner_ear ≈ 0.998 × 10^5 + (2/3) * (1.2 × 10^3)
P_inner_ear ≈ 0.998 × 10^5 + 800
P_inner_ear ≈ 1.006 × 10^5 Pa
Therefore, the pressure on the inner ear at the top of the mountain is approximately 1.006 × 10^5 Pa.
To calculate the pressure on the inner ear, we need to use the concept of Pascal's law, which states that the pressure is the same in all directions in a fluid. In this case, the fluid is the air.
First, let's calculate the change in pressure between the bottom and top of the lift:
ΔP = P(top) - P(bottom)
= 0.998×10^5 Pa - 1.010×10^5 Pa
= -1.2×10^3 Pa
We have a negative value because the pressure decreases as the person rides up the lift.
Next, we need to calculate the area of each ear drum. The formula for the area of a circle is A = πr^2, where r is the radius.
For each ear drum:
A = π * (0.40×10^-2 m)^2
= π * (0.16×10^-4 m^2)
= 0.503×10^-4 m^2
Now, we can use the formula for pressure force:
Force = Pressure * Area
Since the pressure force is equal to the force exerted by the air on the inner ear, we can set up the equation:
Force(inner ear) = Force(atmosphere)
P(inner ear) * A(ear drum) = P(atmosphere) * A(ear drum)
Now, we can substitute the known values:
P(inner ear) * 0.503×10^-4 m^2 = -1.2×10^3 Pa * 0.503×10^-4 m^2
Simplifying the equation:
P(inner ear) = -1.2×10^3 Pa
Therefore, the pressure on the inner ear at the top of the mountain is -1.2×10^3 Pa.
Please put the subject of your question in the subject line. We have no idea what the subject "garyson" is about.
The radius of the ear drum is not needed for the answer.
Your question seems to already contain the answer. They tell you the atmospheric pressure on the ear at the top, and then they ask you what it is.