Question

Air pressure, P , decreases exponentially with the height, h , in meters above sea level: where Po is the air pressure at sea level. P=Poe^-.00012h

(a) At the top of Mount McKinley, height 6194 meters (about 20320 feet), what is the air pressure, as a percent of the pressure at sea level?
Round your answer to one decimal place.The air pressure is about of sea level pressure.
(b) The maximum cruising altitude of an ordinary commercial jet is around 12000 meters (39000 feet). At that height, what is the air pressure, as a percent of the sea level value?

Answer 1

Explanation:

GIven that P at any time t is exponential as

`P=P_o e^(-.00012h)` where P0 is pressure when t=0

a) Height of Mt. Mckinley = 6194 m
`P=P_o e^(-.00012*6194)\\=P_0 (0.4756)\\`

P/P0 in percent = 47.56%

b) When h = 12000 m

`P=P_o e^(-.00012*12000)\\(P)/(P_0) =0.2370\\=23.70%`

Answer 2

(A) the air pressure as a percent at sea level is P1 = Po e^(-0.00012*6194) then P1 is 0.4755Po the percent of sea level's pressure is equals to 47.55%

(B)the maximum cruising altitude of an ordinary commercial jet is 12000 meters let the equation be P2 = P0 e^(-0.00012*12000) = 0.236 *100% the answer is 23.6%