physics

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

  1. 6
asked by Sarah
  1. 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

    posted by bobpursley
  2. Can you please explane C ???

    posted by Taras
  3. Hey Sarah, can you explain how you got C??

    posted by John
  4. 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

    posted by Chukwuka

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