What force must the seals of a passenger door resist at 10000m altitude?

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SUMMARY

At an altitude of 10,000 meters, the atmospheric pressure is approximately one-third of normal atmospheric pressure, equating to about 33,333.33 N/m². Given the dimensions of a passenger door at 1 meter by 2 meters, the area is 2 m². Therefore, the seals of the passenger door must resist a force of 66,666.66 N, calculated using the formula Force = Pressure x Area.

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a jet is kept at 70% of atmospheric pressure. at an altitude of 10000m pressure is 1/3 normal atmospheric pressure. what force must be resisted by seals of a passenger door dimensions 1m by 2m.
atmospheric pressure corresponds with 1.0 * 10^5Nm^-2
 
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Force = Pressure x Area

or conversely Pressure = Force / area
 

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