Recent content by Jorgen1224

I have conducted an experiment involving projecting a laser beam onto a surface and observing speckles that move relative to us. Those speckles become less visible (maybe smaller?) when the surface is smooth and that is especially the case with milk. I have dyed milk and water to see whether in...

And I'm still extremely confused. I do understand now the beginning of the first equation, but I don't see where did x2/2l1,2 come from. How was this calculated?

And where did that relationship come from? i mean l1l2(1-cos(θ1-θ2))=h(l2sin(θ2)-l1sin(θ1))

Okay, now i think i do understand 3rd equation, but I'm still stuck at the first one. In this equationl1l2(1-cos(θ1-θ2))=h(l2sin(θ2)-l1sin(θ1)) if we make an approximation l_2\sin(\theta_2)=l_1\sin(\theta_1) doesn't that make the right side equal to zero? And should i use here...

Homework Statement Two balls of mass m are attached to ends of two, weigthless metal rods (lengths l1 and l2). They are connected by another metal bar. Determine period of small oscillations of the system Homework Equations Ek=mv2/2 v=dx/dt Conversation of energy 2πsqrt(M/k) The Attempt at a...

So yeah, the answer i got is v > sqrt(g⋅h(2-d/2l) from similar triangles and their ratios. New height on which the center of mass is, is h((hl-d)/4l) so then we plug it as height in equation mv2/2 > mgh and rearrange it to get velocity.

Ah yes, I'm terribly sorry for not stating that at the beginning. It should be "what initial velocity does the train need to go up and down the hill without a drive"

Between them. There's no 90 angle in this triangle then i could separate it in two , but then anyway i know c=d/2, but i don't know a or b(which is some part of the height) so i can't use pythagoras theorem

When the train is divided into two parts each with length d/2

I honestly have no idea. Pythagoras theorem doesn't seem to be working here

That it is in between of both centres of mass meaning beneath the top of the hill?

This is the drawing supplied by my teacher. So this train bends and takes shape of the hill while it's on the top?

According to the image it would be hanging there unless the top is flat for a distance equal to the length of the train. I don't really see any other option of train's movement, so i'd say that the centre of mass is going to be on top of the hill.

oh, yeah, its' center of mass has to be at the top of the hill for it to reach maximum potential energy

Well when the front of the train touches the top of this hill then the center of mass is on the left side.