Recent content by xannaxiero

Do you think it could be because the tube is open to the atmosphere? Would that impact my formula at all?

The answer ended up being .563 g/mL. Any idea why this is? There's no explanation with the homework :(

Online homework...they're being graded as incorrect :'(

Oops you're right, I'm sleepy >.< Its g/mL. My bad. That doesn't make any difference in the equation though does it? And yes, I'm assuming separation is in the bottom...this is supposed to be a pretty simple, standard u-tube problem, no tricks...

Homework Statement A U-shaped 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 = 0.52 m and h2 = 0.16 m, what is the density of the liquid? Homework Equations ρ_{u}=h_{1}/h_{2} *...

velocity increases due to acceleration... a = change in velocity/change in time so we have a change from zero to 66.4 over the course of 9 seconds meaning acceleration is 7.4 m/s ? and i know v = vi + at so i feel like i could do v= 0 + 7.4*9 giving me 66.4, which is wrong.

ah yes. okay so that means 400 = uy*9 + 1/2 *9.8*9^2 so uy = .34 and i guess i did ux wrong, so ux is actually 600 = ux*9, giving ux = 67 but from here i still don't know how to find the final velocity

t = 9 sec so i did the following: 600 = uy + 1/2*9.8*9^2 solving for uy I got 22.6

u_{y} = 22.6 so if i do square root of 66.4^2 + 22.6^2 I should get my answer but i get 70.1 instead. how do you find the final based on the x and y components?

Homework Statement A cannon ball is fired horizontally off the top of a cliff that is 400 m high above the ocean. The ball hits the water 600 m away. With what speed does the cannon ball enter the water? Homework Equations I've tried using d_{x} = u_{x}*t d_{y} = u_{y}*t +...