Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A

Answer 1

Answer:

$(a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A$

$(b)\ F_{No} = 771.125N$

Explanation:

Given

$d_D = 6000ft$ ---- Altitude of container in Denver

$A = 0.0155m^2$ -- Surface Area of the container lid

$P_D = 79000Pa$ --- Air pressure in Denver

$P_{No} = 100250Pa$ --- Air pressure in New Orleans

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Solving (a): The expression for $F_{No$

Force is calculated as:

$F = \triangle P * A$

The force in New Orleans is:

$F_{No} = \triangle P * A$

Since the inside pressure is half the pressure at sea level, then:

$\triangle P = P_{No} - \frac{P_{area}}{2}$

Where

$P_{area} = 101000Pa$ --- Standard Pressure

Recall that:

$F_{No} = \triangle P * A$

This gives:

$F_{No} = [P_{No} - \frac{P_{area}}{2}]* A$

Solving (b): The value of $F_{No$

In (a), we have:

$F_{No} = [P_{No} - \frac{P_{area}}{2}]* A$

Where

$A = 0.0155m^2$

$P_{No} = 100250Pa$

$P_{area} = 101000Pa$

So, we have:

$F_{No} = [100250 - \frac{101000}{2}] * 0.0155$

$F_{No} = [100250 - 50500] * 0.0155$

$F_{No} = 49750* 0.0155$

$F_{No} = 771.125N$