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October 20, 2014

October 20, 2014

Posted by **Marie** on Monday, July 2, 2012 at 10:50pm.

The answer is 2.49e+05 Pa. I'm having trouble figuring out how to get this answer. Any help is greatly appreciated.

The

- Physics -
**ajayb**, Tuesday, July 3, 2012 at 2:56pmAssume:

P0-pressure at 4500m above sea level.

P1-pressure at sea-level

P2- pressure at depth of 20.6m

P2=P1+dgh

= P1 + 1024*9.8*20.6

= P1+2.06x10^5 Pa .......(1)

P0=P1-dgh

=P1 - 1.3*9.8*4500

=P1 - 0.57x10^5 Pa .......(2)

[It is assumed density of air near earth's surface is 1.3Kg/m^3 and is uniform in the atmosphere]

Change in pressure = P2 -P0

= (2.06+0.57)x10^5 Pa

= 2.63x10^5 Pa

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