- #1
laminar
- 15
- 0
A building is 15.0m tall. By what percent is the air pressure at the floor greater than the air pressure at the ceiling? The floor is at sea level. -- I don't understand what this percentage relates to, and I got 0.17% greater.
A U-shaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is 12.9 cm deep. How far upward from its initial position does the mercury in the right arm rise? -- The pressure at one point is equal to the pressure at any other point if they are at the same depth.
So, P=Po+(density)(g)(h)
P=101300+(1000)(9.8)(0.129)
P=102564.2Pa
Then make this equal to the pressure in the arm with the mercury in it:
102564.2=101300+(13600)(9.8)(d)
I got d=0.95cm, and the computer says it isn't right. I hate this MasteringPhysics crap. Am I doing something wrong here?
A long cylindrical rod is partially submerged in water. 1.60m of the 6.80m long rod is out of the water. What is its density? -- Don't we have to know the radius, so we can get the area, so we can find the volume displaced by the submerged portion?
A U-shaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is 12.9 cm deep. How far upward from its initial position does the mercury in the right arm rise? -- The pressure at one point is equal to the pressure at any other point if they are at the same depth.
So, P=Po+(density)(g)(h)
P=101300+(1000)(9.8)(0.129)
P=102564.2Pa
Then make this equal to the pressure in the arm with the mercury in it:
102564.2=101300+(13600)(9.8)(d)
I got d=0.95cm, and the computer says it isn't right. I hate this MasteringPhysics crap. Am I doing something wrong here?
A long cylindrical rod is partially submerged in water. 1.60m of the 6.80m long rod is out of the water. What is its density? -- Don't we have to know the radius, so we can get the area, so we can find the volume displaced by the submerged portion?