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agdgdgdgwngo Group Title

An ideal liquid with density ρ is poured into a cylindrical vessel with cross section A1 to a level of height h from the bottom, which has an opening of cross section A2. Find the time it takes for the liquid to flow out

  • 2 years ago
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  1. agdgdgdgwngo Group Title
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    quick analysis shows there's no need for the density

    • 2 years ago
  2. Ishaan94 Group Title
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    is it cylindrical or conical?

    • 2 years ago
  3. agdgdgdgwngo Group Title
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    says cylindrical right up there :-P

    • 2 years ago
  4. Ishaan94 Group Title
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    how could a cylinder have different cross sectional area? :-p

    • 2 years ago
  5. agdgdgdgwngo Group Title
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    I guess A1 = A2 then :-D

    • 2 years ago
  6. Ishaan94 Group Title
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    |dw:1324391145896:dw| This is the cylinder right?!

    • 2 years ago
  7. agdgdgdgwngo Group Title
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    it's not illustrated inn the problem, but that is a cylinder indeed!

    • 2 years ago
  8. agdgdgdgwngo Group Title
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    cmon!! IIT!!

    • 2 years ago
  9. Tomas.A Group Title
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    why u posted 2 same questions? and it's physics isn't it?

    • 2 years ago
  10. agdgdgdgwngo Group Title
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    it's actually physics.... so yeah :-D

    • 2 years ago
  11. agdgdgdgwngo Group Title
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    im stuck on the very first problem lol

    • 2 years ago
  12. shadowfiend Group Title
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    If you could, do try to keep physics questions in the physics group. You can use chat to get people to come to that group if you're not getting traction there :)

    • 2 years ago
  13. henkjan Group Title
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    You need to say that you use the 'dimensional analysis' .... It's just about getting the units of time (the unknown) equal to the units on the right side. So all the 'meters' on the right hand side should add up to 0. This is a tool to find a relation that COULD fullfill the requirements. The problem consists of density, h, A1, A2 and gravity g. Now you just write your solution in the form: t = rho^a*h^b*A1^c*A2^d*g^e You see that the only time dependent variable is g. So g should be (m/s^2) to the power of -1/2 to obtain seconds. The other variables like meter work in the same way, try to get rid of all of them by making a couple of easy assumptions

    • 2 years ago
  14. agdgdgdgwngo Group Title
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    I don't know how to make valid assumptions though :(

    • 2 years ago
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