Recent content by kirbykirbykirby

Bump. I don't understand where this formula comes from. The answer in the textbook is 0.07 degrees celsius. Can I have some more hints please? Sorry, I don't understand. :rolleyes:

radou said something about not taking direction of momentum into account. I KNEW I LOOKED CRAZY :cry: Anyways, that's the right answer. Thanks so much. I am sastified customer.

Durr... I believe you two just gave me opposing information. and I calculate length to be l=0.459m but I lose marks for the wrong answer... durr... EDIT: Durr... the opposing info is gone and now I look like a crazy person. ;'( tear.

Durr... I am supposed to find how high the cart bounces back. Or maybe it is length. The 1m is apparently length. Edit: "After the cart bounces, how far does it roll back up the ramp?" Oh and I took the minus sign out because the final momentum is opposite the intial so the minus cancels?

Okay thanks, I did that \[mgh = 0.5mv^2\] \[g\sin30 = 0.5v^2\] v= 3.13 m/s So \[J= p_f - p_1\] \[2.67 = 0.5(v_f + 3.13)\] so \[v_f= 2.21 m/s\] then \[mgh = 0.5mv^2\] \[gh = 0.5(2.12)^2\] so h= 0.229m??! Looks like something went horribly wrong somewhere...

A 500 g cart is released from rest 1.0 m from the bottom of a frictionless, 30 degree ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. The figure shows the force during the collision. Here's what I did. Which is probably wrong for the very first part so...

Hi, I have the same question but where's the 3.13m/s coming from? Here I try to do it the fancy way. I did \[ v^2_f = v^2_i + 2ad \] \[ v^2_f = 0+ 2(mgsinx)(1)\] so /[v_f = 2.21 m/s \] I took area like the last guy expcept I got 2.67Ns Then I did whole bunch of other stuff to...

I found this on archive. I just post for 3 seconds. sorry... Hmm--I prefer to keep the static friction coefficients as \mu _{S,1} and \mu _{S,2} until the end, where \left\{ \begin{gathered} \mu _{S,1} = 0.60 \hfill \\ \mu _{S,2} = 0.20 \hfill \\ \end{gathered} \right\} and...

So.. why is arccotangent cot(1/x)... I don't understand.

What are the horizontal asymptotes of cot^(-1)x I don't understand because isn't that tangent and it has asymptotes?!

Thanks much! I also get what I was doing wrong. I was for some reason assuming the enterprise stopped so vf = 0. OH SNAP! I assumed vf for enterprise is equal to velocity of klingon at the end and it turned out right.

okay, I'll tell you how I got at it: d= d1 (which is 0) + vi(t) + .5at^2 so I did a= -vi/t in there and that's how I got it. my whole d(enterprise) = .5vi(t) http://dev.physicslab.org/Document.aspx?doctype=3&filename=Kinematics_DerivationKinematicsEquations.xml The end of equation 3...

The starship enterprise returns from warp drive to ordinary space with a forward speed of 50 km/s. To the crew’s shocking surprise, a Klingon ship is 100 km directly ahead, traveling in the same direction at a measly 20 km/s. Without evasive action, the Enterprise will overtake and collide with...

I am aware of that and I wasn't complaining. Maybe I should though because now I learned you're high. I'm only kidding, sorry you have a disease or mental problem or something. Thanks for the help, this is going to take me awhile. It took me 15 minutes to figure out what that last thing...

How do you solve: 600cosA = xcos35 600sinA - 100 = xsin35 The answer is supposed to be 42.9 or 90 - 42.9 for A in degrees. That equation might not be right though because I came up with it for Physics.