A person drives to the top of a mountain. On the way up, the person's ears fail to pop, or equalize the pressure of the inner ear with the outside atmosphere. The pressure of the atmosphere drops from 1.010 x 105 Pa at the bottom of the mountain to 0.998 x 105 Pa at the top. Each eardrum has a radius of 0.40 cm. What is the pressure difference between the inner and outer ear at the top of the mountain?

Question

27 March, 09:58

+4

Answers (1)

Know the Answer?

Margaret Gentry · Answer 1

The pressure on the inner ear is 1.010x10^5 -.998x10^5 = 1200Pa outward

The net force = P*A = 1200Pa*π * (0.40x10^-2m) ^2 = 0.0603N