11. ssm An airtight box has a removable lid of area 1.3 x 10-2 m² and negligible weight. The box is taken up a mountain where the air pressure outside the box is 0.85 x 105 Pa. The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box?
11. ssm An airtight box has a removable lid of area 1.3 x 10-2 m² and negligible weight. The box is taken up a mountain where the air pressure outside the box is 0.85 x 105 Pa. The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box?
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Physics Ch. 11 #11
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![**Problem Statement:**
An airtight box has a removable lid of area \(1.3 \times 10^{-2} \, \text{m}^2\) and negligible weight. The box is taken up a mountain where the air pressure outside the box is \(0.85 \times 10^5 \, \text{Pa}\). The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box?
**Explanation:**
For this problem, we'll use the concept of pressure and force to determine how much force is needed to remove the lid from the box when the inside is a vacuum and the outside air pressure is given.
**Key Information:**
- Area of the lid = \(1.3 \times 10^{-2} \, \text{m}^2\)
- External air pressure = \(0.85 \times 10^5 \, \text{Pa}\)
- Internal pressure = \(0 \, \text{Pa}\) (since the box is evacuated)
**Approach:**
- The force exerted by the air pressure on the lid can be calculated using the formula:
\[
F = P \times A
\]
where:
\(F\) is the force,
\(P\) is the pressure,
\(A\) is the area.
Plug in the given values to find the force required to pull the lid off:
\[
F = 0.85 \times 10^5 \, \text{Pa} \times 1.3 \times 10^{-2} \, \text{m}^2
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Transcribed Image Text:**Problem Statement:**
An airtight box has a removable lid of area \(1.3 \times 10^{-2} \, \text{m}^2\) and negligible weight. The box is taken up a mountain where the air pressure outside the box is \(0.85 \times 10^5 \, \text{Pa}\). The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box?
**Explanation:**
For this problem, we'll use the concept of pressure and force to determine how much force is needed to remove the lid from the box when the inside is a vacuum and the outside air pressure is given.
**Key Information:**
- Area of the lid = \(1.3 \times 10^{-2} \, \text{m}^2\)
- External air pressure = \(0.85 \times 10^5 \, \text{Pa}\)
- Internal pressure = \(0 \, \text{Pa}\) (since the box is evacuated)
**Approach:**
- The force exerted by the air pressure on the lid can be calculated using the formula:
\[
F = P \times A
\]
where:
\(F\) is the force,
\(P\) is the pressure,
\(A\) is the area.
Plug in the given values to find the force required to pull the lid off:
\[
F = 0.85 \times 10^5 \, \text{Pa} \times 1.3 \times 10^{-2} \, \text{m}^2
\]
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