Q: Aperson rides up a lift to a mountain top, but the person\'s earsfail to \"po

Question

Q: Aperson rides up a lift to a mountain top, but the person's earsfail to "pop" - that is, the pressure of the inner ear does NOTequalize with the outside atmosphere. The radius of each eardrum is0.4 cm. The pressure of the atmosphere drops from 1.01 x 10^5 Pa atthe bottom of the lift to 0.988 x 10^5 Pa at the top. What is thepressure on the inner ear at the top of the mountain?

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Hi, I amconfused as to what formula (or formulas…) to use for thisproblem. I know Pressure is equal to Force over Area, but since Ididn't know for either side I don't know if I can use that. I alsoknow that F1 / A1 = F2 / A2, but again I'm not sure how tocalculate force or if I even need to calculate force.

Cansomeone show a step by step solution to solve this?

I knowthe answer is 1.2 x 10^3 Pa…but I can't arrive at this.Thanks for reading.

Explanation / Answer

at normal height, the atmosphere inside ear and outside equal. so atmosphere inside ear is 1.001*105Pa. net force F=P*S where S=r^2=*0.4*10-2 m2 P=1,001*105-0,989*105 Pa solve for f

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