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rosho
 2 years ago
how do you solve this
The pressure of earth's atmosphere at sea level is approximately 1000 millibars and decreases exponentially with elevation. At an elevation of 30,000ft, the pressure is onethird of the sea level pressure. At what elevation is the pressure half of the sealevel pressure? At what elevation is it 1% of the sealevel pressure?
rosho
 2 years ago
how do you solve this The pressure of earth's atmosphere at sea level is approximately 1000 millibars and decreases exponentially with elevation. At an elevation of 30,000ft, the pressure is onethird of the sea level pressure. At what elevation is the pressure half of the sealevel pressure? At what elevation is it 1% of the sealevel pressure?

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rosho
 2 years ago
Best ResponseYou've already chosen the best response.0that's already done for us.

nincompoop
 2 years ago
Best ResponseYou've already chosen the best response.1okay I am used to the atm or pascal. 1 atm is 760 torr and 1 bar is 750 torr if we have 1 bar at sealevel and 1/3 at 30,000 ft, then look for the elevation where it is 1/2 or 50% do you want to set up a dimensional analysis?

rosho
 2 years ago
Best ResponseYou've already chosen the best response.0this is a calc 2 question, but it's more physics related so i thought i tried.

nincompoop
 2 years ago
Best ResponseYou've already chosen the best response.1you don't need calculus for this

rosho
 2 years ago
Best ResponseYou've already chosen the best response.0sure, just go ahead and do it your way. i'll compare it to the answer i have in the book.

rosho
 2 years ago
Best ResponseYou've already chosen the best response.0because i don't understand this problem at all.

rosho
 2 years ago
Best ResponseYou've already chosen the best response.0don't bother i'm just going into my tutoring session and asking the tutor. because i just don't understand it. thanks for trying.

gleem
 2 years ago
Best ResponseYou've already chosen the best response.2The above answer is not correct. The problem states that the pressure decreases exponentially with height. that means tha\[p=p _{0}e ^{\alpha h}\] where po is the pressure at sea level and h is the height aboce sea level and alpha is a constant which we determine from the info provided above. If we take the logrithm of the equation above we get \[\log p=\log p _{0}  \alpha h \] or \[\log(p/p _{0} ) = \alpha h \] Use the info that at 30000 ft the pressure is 1/3 of 1000mb to find alpha. Then for p/p0 = 0.50 solve for h with the alpha you determined and the same for p/po = 0.01. OK?
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