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 3 years ago
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
 3 years ago
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

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agdgdgdgwngo
 3 years ago
Best ResponseYou've already chosen the best response.1quick analysis shows there's no need for the density

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0is it cylindrical or conical?

agdgdgdgwngo
 3 years ago
Best ResponseYou've already chosen the best response.1says cylindrical right up there :P

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0how could a cylinder have different cross sectional area? :p

agdgdgdgwngo
 3 years ago
Best ResponseYou've already chosen the best response.1I guess A1 = A2 then :D

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1324391145896:dw This is the cylinder right?!

agdgdgdgwngo
 3 years ago
Best ResponseYou've already chosen the best response.1it's not illustrated inn the problem, but that is a cylinder indeed!

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.0why u posted 2 same questions? and it's physics isn't it?

agdgdgdgwngo
 3 years ago
Best ResponseYou've already chosen the best response.1it's actually physics.... so yeah :D

agdgdgdgwngo
 3 years ago
Best ResponseYou've already chosen the best response.1im stuck on the very first problem lol

shadowfiend
 3 years ago
Best ResponseYou've already chosen the best response.0If 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 :)

henkjan
 3 years ago
Best ResponseYou've already chosen the best response.1You 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

agdgdgdgwngo
 3 years ago
Best ResponseYou've already chosen the best response.1I don't know how to make valid assumptions though :(
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