SOLUTION: The altitude of an aircraft,h, in miles, is given by h=-(100/9)log p/b where p= the outside pressure and B= the atmosperic pressure at sea level. Let B=31 inches of Mercury. What

Question 712195: The altitude of an aircraft,h, in miles, is given by
h=-(100/9)log p/b
where p= the outside pressure and B= the atmosperic pressure at sea level. Let B=31 inches of Mercury. What is the outside air pressure at a height of 2.7 miles?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The altitude of an aircraft,h, in miles, is given by
h=-(100/9)log p/b
where p= the outside pressure and B= the atmosperic pressure at sea level. Let B=31 inches of Mercury. What is the outside air pressure at a height of 2.7 miles?
--------
2.7 = -(100/9)log(p/31)
log(p/31) = -0.243
----
p/31 = 10^-0.243
p = 31*0.5715
----
p = 17.72 (outside air pressure)
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Cheers,
Stan H.
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