Recent content by Albeaver89

Well I don't know how to or what an array is, but I guess I can look it up. Thanks! :) I'll fix the wrong value thing to loop until the correct value is given. Thank you again :)

Homework Statement So, we have a project to make a tic-tac-toe game and for bonus points we can make sure the player(s) can not enter the same position more than once. I can't think of a way to do this. I'm sure I need to use a 'for loop' or a 'while loop'. I am using eclipse to write the...

Well what I mean is, we have all kinds of formulas for rigid bodies, can we just apply them to a non-rigid or fluid body, or are there different mechanics for non-rigid/fluid bodies?

Is there a way to predict the motion of a non-rigid body?

If you had melted copper into the stainless steel can you then remove the copper?

I heard a rumor from some people at a scrap yard that it's impossible to sepperate copper from stainless steel, and that the copper ruins the stainless steels structual integrety. Is there any truth to this?

Sorry that's what I thought i had...my bad :redface:

This isn't a homework question, but I felt it was appropriate... Proof that √(x^2)/x=sin(x)

So I got V=D/sqrt(2*h/g) Then plug that into: m*g*H=1/2*m*(D/sqrt(2*h/g))^2 Since H=x+h we get m*g*(x+h)=1/2*m*(D/sqrt(2*h/g))^2 then solve for x giving us: x=((.25*d^2)/h)-h Does that seem correct? I got -2.975? Never mind I forgot to add in the mgh to the end of the equation silly me thanks...

Wait...Don't you need to know how fast it's going off the ramp?

omg...I don't see how I didn't see it before...Thank you very much!

Homework Statement A skier (m=55.00 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. (A)If h = 6.80 m and D = 10.20 m, find H.(B)Find her total kinetic energy as she reaches the ground. Homework Equations KEi+PEi=KEf+PEf (law of...

Homework Statement A 3.5 kg block starts at rest and slides a distance d down a frictionless 28.0° incline, where it runs into a spring. The block slides an additional 21.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant is 435 N/m. A)What is the...

Ok. I figured it out!you add the energy from the spring and get this equation: 1/2kx^2=1/2mv^2+1/2ky^2 Where y= the distance the spring travels with the cork i.e. after the equilibrium Then solve for v and get: v=(√(20)*√(.082^2-.04^2))/√(.00376) and you get v=5.22 m/s.

Sorry I forgot to post the picture...