if atmospheric pressure is 1.01×10^5 pascal then how much force does the air in a room exert on the inside of a window pane which is 50 cm × 100 cm ?
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Answered by
31
1 pascal = 1 n/m^2
so the force exerts on the window pane of area 5000*10^(-4) = 1.01*(10^5)*5000*(10^-4) =5.05*10^4N
so the force exerts on the window pane of area 5000*10^(-4) = 1.01*(10^5)*5000*(10^-4) =5.05*10^4N
Answered by
49
pressure = force/cross section area
Force = Pressure x Cross section area
= 1.01 x 10^5 N/m² x 50/100m x 100/100 m
= 1.01 x 10^5 x 1/2 N
= 0.505 x 10^5 N
= 5.05 x 10^4 N
Force = Pressure x Cross section area
= 1.01 x 10^5 N/m² x 50/100m x 100/100 m
= 1.01 x 10^5 x 1/2 N
= 0.505 x 10^5 N
= 5.05 x 10^4 N
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