Complete Question
When you ascend or descend a great deal when driving in a car yours ears "pop," which means that the pressure behind the eardrum is being equalized to that outside. If this did not happen, what would be the approximate force on an eardrum of area .50 cm2 if a change in altitude of 950 m takes place?
Answer:
The value is [tex]F = 0.60 \ N[/tex]
Explanation:
From the question we are told that
The area of the ear drum is [tex]A = 0.5 \ cm^2 = 0.50 *10^{-4} \ m^2[/tex]
The change in altitude is [tex]\Delta d = 950 \ m[/tex]
Generally the change in pressure is mathematically represented as
[tex]\Delta P = \frac{F}{A}[/tex]
This can also be mathematically represented as
[tex]\Delta P = \rho * g * \Delta d[/tex]
So
[tex]\frac{F}{A} = \rho * g * \Delta d[/tex]
=> [tex]F = \rho * g * \Delta d * A[/tex]
=> [tex]F = \rho * g * \Delta d * A[/tex]
Here [tex]\rho[/tex] is the density of dry air with value [tex]\rho = 1.29 \ kg /m^3[/tex]
So
[tex]F = 1.29 * 9.8 * 950 * 0.50 *10^{-4}[/tex]
=> [tex]F = 0.60 \ N[/tex]
