I need help please!!!

Question

i have never seen this befoer in my textbook can someone explain this to me???

During normal conversation, the amplitude of a pressure wave is 0.022 Pa.

(a) If the area of the eardrum is 0.53 cm2, what is the force on the eardrum?

__ N

(b) The mechanical advantage of the bones in the inner ear is 1.5. What force is exerted on the oval window?

__ N

(c) The area of the oval window is 0.026 cm2. What is the pressure increase transmitted to the liquid in the cochlea?

__ Pa

Answer 1

During normal conversation, the amplitude of a pressure wave is 0.022 Pa.

(a) If the area of the eardrum is 0.53 cm2, what is the force on the eardrum?

1 Pa is 1N/m²

so 0.022 Pa is 0.022 N/m²

0.022 N/m² x 0.53 cm² x 1m²/100²cm² = 1.166e-6 N

(b) The mechanical advantage of the bones in the inner ear is 1.5. What force is exerted on the oval window?

I guess 1.5 times 1.166e-6 N = 1.74e-6  N

(c) The area of the oval window is 0.026 cm2. What is the pressure increase transmitted to the liquid in the cochlea?

I'm lost in the anatomy part, but sounds like:

1.74e-6  N / 0.026 cm² x 100²cm²/1m² = 0.669 Pa

.

Answer 2

Force  = Pressure * Area

1 Pa = 1 N/m2 = 0.0001 N/cm2

A = 0.53 cm2

P = 0.022 Pa

F = P*A = 0.53 * 0.022 * 0.0001 = 1.166 N * 10^(-6)

Mechanical advantage is the the lever arm multiplier of the force; so the force on the oval window would be 1.5 * F = 1.749 * 10^(-6)

P = F / A; you do the math for the last one.