Answer:
[tex](a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]
[tex](b)\ F_{No} = 771.125N[/tex]
Explanation:
Given
[tex]d_D = 6000ft[/tex] ---- Altitude of container in Denver
[tex]A = 0.0155m^2[/tex] -- Surface Area of the container lid
[tex]P_D = 79000Pa[/tex] --- Air pressure in Denver
[tex]P_{No} = 100250Pa[/tex] --- Air pressure in New Orleans
See comment for complete question
Solving (a): The expression for [tex]F_{No[/tex]
Force is calculated as:
[tex]F = \triangle P * A[/tex]
The force in New Orleans is:
[tex]F_{No} = \triangle P * A[/tex]
Since the inside pressure is half the pressure at sea level, then:
[tex]\triangle P = P_{No} - \frac{P_{area}}{2}[/tex]
Where
[tex]P_{area} = 101000Pa[/tex] --- Standard Pressure
Recall that:
[tex]F_{No} = \triangle P * A[/tex]
This gives:
[tex]F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]
Solving (b): The value of [tex]F_{No[/tex]
In (a), we have:
[tex]F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]
Where
[tex]A = 0.0155m^2[/tex]
[tex]P_{No} = 100250Pa[/tex]
[tex]P_{area} = 101000Pa[/tex]
So, we have:
[tex]F_{No} = [100250 - \frac{101000}{2}] * 0.0155[/tex]
[tex]F_{No} = [100250 - 50500] * 0.0155[/tex]
[tex]F_{No} = 49750* 0.0155[/tex]
[tex]F_{No} = 771.125N[/tex]
