Calculation of atmospheric presure

AI Thread Summary
To estimate atmospheric pressure in London at 550m above sea level, the formula patm = rho g h was used, with air density assumed constant at 1.2 kg/m^3. The calculation yielded a pressure of 66,648 Pa, which was then subtracted from the sea level pressure of 100,000 Pa, resulting in an estimated atmospheric pressure of 93,352 Pa or 92.5 kPa. The discussion highlighted the importance of significant figures in reporting results. The assumption of constant air density with height was briefly acknowledged as a simplification. Overall, this method provides a reasonable estimate for atmospheric pressure at that elevation.
Derek1997
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Homework Statement


Taking atmospheric pressure at sea level to be 1x10^5 pa CPa and the density of air at sea level to be
about 1.2kg/m^3,estimate the atmospheric pressure in London 550m above sea level. For this calculation, it is easiest to assume that air density doesn’t change with height – briefly discuss
the validity of this assumption

Homework Equations


patm= rho g h

The Attempt at a Solution


Well, I did p=pgh- 1.2x9.8x550 and substracted the answer by the atmosperhic pressure of the sea to get the answer? Am I right?
 
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Derek1997 said:
Am I right?
Don't know, you haven't stated your result. It's certainly one way to "estimate" atmospheric pressure in London.
 
Bystander said:
Don't know, you haven't stated your result. It's certainly one way to "estimate" atmospheric pressure in London.
Well alright thanks for reply.. here the results i got above
so for p atmospheric i got- 1.2kg/m^3x9.8m/s^2x 550m= 6648pa then i substractfrom the very given pressure that is 1x10^5pa-6648pa= 93352pa pressure of london?
 
Derek1997 said:
1x10^5pa-6648pa= 93352pa pressure of london?
You might want, in addition, to pay attention to significant figures.
 
Bystander said:
You might want, in addition, to pay attention to significant figures.
you mean i convert to kpa?
 
Bystander said:
to significant figures.
Not units --- significant figures.
 
I don't understand. 93352pa= 92.5kpa?
 
Yes. I'll re-emphasize the "might," I've no idea how picky your instructor "might" be.
 
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Oh ok, thanks i gave u a like.
 
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:wink:
 
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