contestada

A partially evacuated airtight container has a tight-fitting lid of surface area 79.9 cm2 and negligible mass. If the force required to remove the lid is 388 N and the atmospheric pressure is 1.00 × 105 Pa, what is the air pressure in the container before it is opened?

Respuesta :

Answer:

[tex]P_1 = 5.14 \times 10^4 Pa[/tex]

Explanation:

As we know that the air container is fixed by the lid on it and it is removed by external force

F = 388 N

now this external force must be same as the force due to pressure difference of the object

so it is given as

[tex]F = P_2 A - P_1 A[/tex]

here we know that

[tex]A = 79.9 cm^2 = 79.9 \times 10^{-4} m^2[/tex]

here we have

[tex]P_2 = 1.00 \times 10^5 Pa[/tex]

now pressure is given as

[tex]P_1 = P_2 - \frac{F}{A}[/tex]

[tex]P_1 = 1.0 \times 10^5 - \frac{388}{79.9 \times 10^{-4}}[/tex]

[tex]P_1 = 5.14 \times 10^4 Pa[/tex]