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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
 2 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
 2 years ago
 2 years ago

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

Ishaan94Best ResponseYou've already chosen the best response.0
is it cylindrical or conical?
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.1
says cylindrical right up there :P
 2 years ago

Ishaan94Best ResponseYou've already chosen the best response.0
how could a cylinder have different cross sectional area? :p
 2 years ago

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

Ishaan94Best ResponseYou've already chosen the best response.0
dw:1324391145896:dw This is the cylinder right?!
 2 years ago

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

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

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

agdgdgdgwngoBest ResponseYou've already chosen the best response.1
im stuck on the very first problem lol
 2 years ago

shadowfiendBest ResponseYou've already chosen the best response.0
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

henkjanBest ResponseYou've already chosen the best response.1
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

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