An airtight box has a removable lid of area 1.19E-2 m2 and negligible weight. The box is taken up a mountain where the air pressure outside the box is 8.67E+4 Pa. The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box?

Well F = P/A

I guess that's what they're looking for.

To find the magnitude of the force required to pull the lid off the box, we need to consider the pressure difference between the inside and outside of the box. The force exerted on the lid is equal to the pressure difference multiplied by the area of the lid.

First, we need to calculate the pressure inside the box. Since the inside of the box is completely evacuated, the pressure inside is essentially zero.

The force required to pull the lid off the box is then given by:

Force = Pressure difference * Area of the lid

To calculate the pressure difference, we subtract the pressure inside the box from the pressure outside the box:

Pressure difference = Pressure outside - Pressure inside

Given:
Area of the lid (A) = 1.19E-2 m^2
Pressure outside (P) = 8.67E+4 Pa
Pressure inside (P) = 0

Substituting the values into the formula, we have:

Pressure difference = (8.67E+4 Pa) - (0 Pa) = 8.67E+4 Pa

Finally, we can calculate the force:

Force = (8.67E+4 Pa) * (1.19E-2 m^2)

Now we can calculate the force required to pull the lid off the box using these values:

Force = 8.67E+4 Pa * 1.19E-2 m^2 = 1034.13 N

Therefore, the magnitude of the force required to pull the lid off the box is approximately 1034.13 Newtons.