How much force is exerted by the atmosphere on 10.8 km^2 of land at sea level?
The pressure acting on a surface is determined by the force acting on a unit area of the surface. At sea level the standard pressure exerted by the atmosphere or 1 atm = 101325 Pa.
Pascal is a unit of pressure in terms of the force in newton and the surface area in m^2.
The force exerted by the atmosphere on a surface area equal to 10.8 km^2 can be determined by first converting 10.8 km^2 to a value in m^2.
10.8 km^2 = 10.8/(1000)^2 m^2. As the pressure exerted is 101325 Pa, 101325 = F/10.8*10^-6
=> F = 101325*10.8*10^6 N
=> F = 1094310*10^6 N
The force exerted by the atmosphere on 10.8 km^2 of land at sea level is 1094310*10^6 N
At sea level the standard atmospheric pressure = 1 bar
1 bar = 100000 pa = 100000 N/m2 = 10^5 N/m2
Area = 10.8Km^2 = 10.8*10^6 m2
Focre = Pressure * Area = 10^5*10.8*10^6 N = 1.08*10^12 N
Force exerted by the atmosphere on 10.8 Km^2 area is equal to 1.08*10^12 Newton