WEBVTT

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This is Giancoli Answers
with Mr. Dychko.

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Atmospheric pressure of
this column of air

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is gonna equal the air's density
times <i>g</i> times <i>Δh</i>.

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We can use this formula only when
you have a fluid of uniform density

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and we are told to let's assume that air

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fits that description which really isn't true
because air is highly compressible

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so its density is changing as you go
lower and lower

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but we are told to approximate
that it's true by saying

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uniform density is equal to half the density
at the sea level down here.

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So how high would this column of air be?

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Now there's no atmospheric pressure
applied to the top of it,

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I mean this is the atmosphere so
out here there's just outer space.

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So the total pressure at the bottom
then is just this

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and we can solve this for <i>Δh</i> by
multiplying both sides by 2

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and dividing by <i>g</i> times density of
air at sea level

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and we have <i>Δh</i> is 2 times

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1.103 times 10 to the 5 newtons
per square meter

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divided by 1.29 kilograms per cubic meter
times 9.8 newtons per kilogram

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which gives about 16 kilometers or
1.60 times to the 4 meters.