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If the atmosphere is assumed to be composed of a layer of air of uniform density, 1.23...
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We know that we can express the pressure by the following equation.
`P = hrhog` where;
P = pressure
h = height
`rho` = density of air
g = gravitational acceleration
For our question;
`P = 1.01xx10^5pa`
`h = ?`
`rho = 1.23(kg)/(m^3)`
`g = 9.81m/(s^2)`
`h = P/(rhog)`
`h = (1.01xx10^5)/(1.23xx9.81)`
`h = 8.37xx10^3`
So the height of the atmosphere that produces `1.01xx10^5pa` pressure at earth surface is 8370m.
Posted by jeew-m on June 16, 2013 at 3:23 AM (Answer #1)
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