Recent content by notagenius08

Ok, so I have messed around a bunch and I think the answer for part a) is: v=sqrt of 2*g(h-y) Can anyone tell me if I am on the right track?

Homework Statement A water tank of height h has a small hole at height y. The water is replenished to keep h from changing. The water squirting from the hole has range x. The range approaches zero as y goes to 0 because the water squirts right onto the table. The range also approaches zero as...

Ok, nevermind, it is because I had Vi and Vf on the wrong sides of the equation, it does work out to be the second equation! Thanks again.

I always have trouble setting up the before and after, with conservation... OK, PE+KE=KE so solving for Vi gives Vi=sqrt of Vf^2 -2gh but it should be Vi=sqrt of Vf^2+2gh But yes the second equation gives the right answer. Thank you very much.

No angle given. In the drawring it is not linear, but curved like a slightly flattened 's', or a local min concave up, inflection point, then concave down local max, if that makes sense. Yes 2.0m is the top of the ramp.

Homework Statement A water skier is going to glide up a 2.0m frictionless ramp then sail over a 5.0m wide tank filled with sharks, she'll drop the tow rope at the base of the ramp. What minimum speed must she have as she reaches the ramp in order for her to clear the sharks.Homework Equations...

Thanks for your help. It actually does simplify down, and so you don't have to use the quadratic equations. Thanks again for your help!

Homework Statement A proton is traveling to the right at 2.0x10^7 m/s. It has a head on perfectly elastic collision with a carbon atom. The Mass of the Carbon atom is 12 times the mass of the proton. What are the speed and direction of each after collision? Homework Equations...

Finally solved it! Thanks for your help. Still don't understand why the book listed it under nonuniform circular motion... Thanks again.

Homework Statement This homework problem is under the Nonuniform circular motion problem, but seems more like a elementary kinematics. You push the 100 g object and release it 1.60 m from the table edge. Unfortunately, you push a little too hard. The object slides across, sails off the edge...

First solve for time: t=sqrt of 2*height/g t=sqrt of 2(.823)/9.8 = .43 secs. This is how long it takes to fall vertically, the horizontal and vertical are tied through time. Given that the object travels 1.44m horizontally in .43 sec we can divide 1.44m by .43 secs to get 3.348 m/s or 3.5...