- #1
warrior2014
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Pressure of a lid!
A lid is put on a box that is 10cm long, 20cm wide and 8cm tall and the box is then evacuated until its inner pressure is 72000 Pa.
1.How much force is required to lift the lid at sea level?
2. How much force is required to lift the lid in Denver on a day when the atmospheric pressure is 67.5 kPa (2/3 the value at sea level)?
P=F/A
A=L x W
I got #1 where I did the following:
A= L x W= 0.1 x 0.2= 0.02m^2
P=F/A
(101325 - 72000)= F/ 0.02
586.5 N= F
I attempted to do #2 as shown below and got both methods wrong:
so first I did: P=F/A
(67500- 72000)= F/0.02
-90N= F (wrong answer)
so then I did: P=F/A
67500= F/0.02
1350N= F (wrong answer again)
I am unsure how to answer this question, so any help is appreciated! It's online homework so there's no given answer either. thanks!
Homework Statement
A lid is put on a box that is 10cm long, 20cm wide and 8cm tall and the box is then evacuated until its inner pressure is 72000 Pa.
1.How much force is required to lift the lid at sea level?
2. How much force is required to lift the lid in Denver on a day when the atmospheric pressure is 67.5 kPa (2/3 the value at sea level)?
Homework Equations
P=F/A
A=L x W
The Attempt at a Solution
I got #1 where I did the following:
A= L x W= 0.1 x 0.2= 0.02m^2
P=F/A
(101325 - 72000)= F/ 0.02
586.5 N= F
I attempted to do #2 as shown below and got both methods wrong:
so first I did: P=F/A
(67500- 72000)= F/0.02
-90N= F (wrong answer)
so then I did: P=F/A
67500= F/0.02
1350N= F (wrong answer again)
I am unsure how to answer this question, so any help is appreciated! It's online homework so there's no given answer either. thanks!