An airtight box has a removable lid of area 2.87 × 10-2 m2 and negligible weight. The box is taken up a mountain where the air pressure outside the box is 7.20 × 104 Pa. The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box?
To find the magnitude of the force required to pull the lid off the box, we need to consider the difference in pressure between the inside and outside of the box.
The force required to pull the lid off the box is equal to the pressure difference multiplied by the area of the lid.
First, let's calculate the pressure difference. The pressure inside the box is zero, as it is completely evacuated. The pressure outside the box is given as 7.20 × 10^4 Pa.
Now, let's calculate the force:
Force = Pressure difference x Area
Given:
Pressure difference = 7.20 × 10^4 Pa
Area of the lid = 2.87 × 10^-2 m^2
Plugging in the values, we can calculate the force:
Force = (7.20 × 10^4 Pa) x (2.87 × 10^-2 m^2)
= 2.0684 N
Therefore, the magnitude of the force required to pull the lid off the box is 2.0684 N.