Recent content by Lopina

Well, I was not sure of this because the solution looks very awful, and I had no one to verify this. But if it seems OK to you, I guess it's correct then. Thanks

Homework Statement We are given a bar, with length of d, and it's densitiy is given by this formula: \lambda=\lambda_{0}+2ax, where x is the distance from one side of the bar and a is a constantHomework Equations \lambda=\lambda_{0}+2ax \vec{r}_{CM}=\frac{\sum\vec{r}_{i}\Delta m_{i}}{m} The...

Ok, here's the picture before collision. You should be able to visualize what I was trying to say to you. You need to look at this matter in 2D, because the cars are not on the same track. http://img8.imageshack.us/img8/2686/beforeead.jpg

Ok. Let's imagine you're in this rotating thing. The force which is pressing you to the wall is centrifugal. So the Fn has the same amount but opposite direction of centrifugal force. Now you have realized which force is Fn, you can determine the Fk. If you want to be motionless in this rotating...

Hint: Both velocity and momentum are vectors and should be dealt with accordingly. You have one car moving north with speed 14m/s, so his speed in 2D is 14j m/s The other car is moving west, so his speed is -14i m/s.

If you could provide us a picture, I'm sure we could work some things out. :D

By definition, if there is an elastic collision, both objects will bounce one of each other and each will go in their own direction with their own speed. You are given a inelastic collision, so there will be no bouncing and the 2 cars could be observers as a single object which goes in a...

This isn't a 1D problem. If you take a better look at it you will see that it's in 2D, so try to draw yourself a picture. And one more thing. If it's an inelastic collision then the cars will continue to move as one with one speed which will be common for both of them.

Ty for your help, David. It sure brightens things up a bit for me

Here's the new picture I hope it's better than ASCII one http://img15.imageshack.us/img15/6522/pictureaor.jpg The reason why I think I'm wrong is the following. If I rotate the square on the upper picture 45° in any direction around z-axis then the x-axis no longer lies on a diagonal of...

You're welcome :D

Well, I'd solve it this way \frac{mv_{1}^{2}}{2}+mgh=\frac{mv_{2}^{2}}{2} \frac{v_{1}^{2}}{2}+gh=\frac{v_{2}^{2}}{2} h=\frac{v_{2}^{2}-v_{1}^{2}}{2g} d=\frac{h}{sin(\alpha)}

Dunno if I'm right, but this is how I see this problem. First, I calculate the distance car has traveled in that 0.25 seconds d=v*t=\frac{100m}{9s}*\frac{1s}{4}=\frac{25m}{9} After that, the car slows down v^{2}=v_{0}^{2}+2ad 0=\frac{100}{9}^{2}-2*8d d=\frac{100*100}{81*2*8} Now...

Homework Statement Calculate the moment of inertia of a straight homogenous plate with mass m shaped like a square where the axis of rotation goes through the diagonal of the plate. ^ |y | /|\ / | \ a -------|------> a \ | / x \|/...