Four tubes are set up in the lab at sea level as shown by the figure. The letter under each tube corresponds the question with the same letter.
1 atm = 101300 Pa = 101300 N/m2
ρwater = 1000 kg/m3
(a) What is the pressure pushing down on the surface of the water in tube (a) from the outside?
P = Pa *
101300 OK
(b) Tube (b) is also filled with water, A1 is 0.04 m2 and A2 is 0.22 m2 Two pistons are placed in both ends of the tube and forces, F1 and F2 are exerted on the pistons so that they remain at the same height. If F1 = 20 N what is F2?
F2 = N *
110 OK
(c) Tube (c) is again filled with water. A1 and A2 are the same as in part (b). Two pistons apply different forces to the water in the tube so that the water in the right side of the tube is a height h = 0.51 m above the height of the water in the left side of the tube. If F2 = 143 N what is F1?
F1 = N *
225.92 OK
HELP: What is the expression for the pressure difference?
(d) Now two fluids are placed in the same tube (d). Both sides are open to the atmosphere without pistons. One fluid is water and the other (on top of the water in the left branch of the tube) is an oil of unknown density. l = 111 mm and d = 5.7 mm. What is the density of the oil?
densityoil = kg/m3
HELP: Find two expressions for the pressure at the water/oil interface and solve for the density of the oil.
I got a, b, and c but I just need help with d
IF this is a U tube, the weights of the liquids has to be the same. Weight= density*height*area
so the height will be something like this (you workout the math)
height1/height2= density2/density1
Can you please explane C ???
Hey Sarah, can you explain how you got C??
Guys, here how she solved C:
first: Multiply the buoyancy, gravity, and height to get 4998.
2nd: take 143/.22= 650.
3rd: Take the sum= 5648.
4th: Multiply by .04 and BOOM!
FINAL ANSWER: 225.92
To solve part (d), we need to find the density of the oil in tube (d).
First, let's consider the pressures at the water/oil interface. The pressure at any point in a fluid depends on the depth of that point below the surface. Considering the water in the left branch of the tube, the pressure at the water/oil interface can be calculated using the equation:
Pwater = ρwater * g * h
Where:
Pwater = Pressure at the water/oil interface (Pa)
ρwater = Density of water (1000 kg/m^3)
g = Acceleration due to gravity (approximately 9.8 m/s^2)
h = Height difference between the water surface and the interface (in meters)
Next, let's consider the pressure in the oil. Since both sides of the tube are open to the atmosphere, the pressure in the oil must be equal to the atmospheric pressure. We can use the equation:
Patm = Pwater + Pdifference
Where:
Patm = Atmospheric pressure (101300 Pa)
Pdifference = Pressure difference between the water surface and the oil surface (Pa)
The pressure difference (Pdifference) can be calculated using:
Pdifference = ρoil * g * h'
Where:
ρoil = Density of the oil (unknown, to be determined) (kg/m^3)
h' = Height of the oil column above the water/oil interface (in meters)
Now, let's solve for the density of the oil:
Patm = Pwater + Pdifference
Substituting the equations for Pwater and Pdifference:
101300 = (ρwater * g * h) + (ρoil * g * h')
We also know the following from the given information:
l = 111 mm = 0.111 m (height of oil column)
d = 5.7 mm = 0.0057 m (diameter of oil column)
We can consider the oil column as a cylindrical column and use the formula for the volume of a cylinder to find the height of the oil column (h') in terms of l and d:
V = π * (d/2)^2 * h'
h' = V / (π * (d/2)^2)
h' = (π * (0.0057/2)^2 * 0.111) / (π * (0.0057/2)^2)
h' = 0.111
Substituting the values into the equation:
101300 = (1000 * 9.8 * h) + (ρoil * 9.8 * 0.111)
Simplifying the equation:
ρoil = (101300 - (1000 * 9.8 * h)) / (9.8 * 0.111)
Plug in the value of h from part (b) which is 0.51 m:
ρoil = (101300 - (1000 * 9.8 * 0.51)) / (9.8 * 0.111)
Calculating the expression:
ρoil ≈ 630 kg/m^3
Therefore, the density of the oil in tube (d) is approximately 630 kg/m^3.