physics
posted by Adam on .
A Ushaped tube is partly filled with water and partly filled with a liquid that does not mix with water. Both sides of the tube are open to the atmosphere. If h1 = 0.46 m and h2 = 0.25 m, what is the density of the liquid?
on the left side of the tube is the liquid h2= 0.25 m, and h1= 0.46, and on the right side is the water and its 0.45 m
How can you calculate the density of the liquid?
Thank you.

Supposing that ρ1 is the density of the unknown liquid, ρ is the density of the water, and g is the gravity acceleration; let's choose a random intersection face at the bottom of the Utube's bottom. Since the pressures from the both sides balance with each other, we can introduce the equation:
ρ1•h2•g+ ρ •(h1h2) •g= ρ•0.45•g.
Replace the symbols h1 and h2 with 0.46 m and 0.25 m, respectively,
we can conclude that
ρ1=0.96 ρ =0.96• 1000 = 960 kg/m^3. 
Thank you.I don't know why, but it says the answer is wrong even though i have to answer it as g/cm^3 which would be 960000 g/cm^3 i assume, but even that's wrong! :/ I am really confused I don't know why its wrong.

I'm sure that my solution is right.
Check your given data.
And more, 960 kg/m^3 = 960 000 g/m^3 =
=0.96 g/cm^3 (1 cm^3 =10^6 m^3) 
Thank you. It was because of the units my units should have been as g/cm^3, so I got it now. Thank you, I really appreciate it :)