Here's the question you clicked on:
rosho
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 one-third of the sea level pressure. At what elevation is the pressure half of the sea-level pressure? At what elevation is it 1% of the sea-level pressure?
that's already done for us.
okay 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 sea-level 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?
this is a calc 2 question, but it's more physics related so i thought i tried.
you don't need calculus for this
sure, just go ahead and do it your way. i'll compare it to the answer i have in the book.
because i don't understand this problem at all.
don't bother i'm just going into my tutoring session and asking the tutor. because i just don't understand it. thanks for trying.
The 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?