Say D⊂ℝ. Then a ∈ ℝ is called accumulation point of D, when there is a sequence (an) in D "without" {a} and an→a (→ means goes towards)
thats how the prof defined it
So we just recently did accumulation points in my maths class for chemists. I understood everything that was taught but ever since I was trying to find a reasonable explanation if the sequence an = (-1)n has 2 accumulation points (-1,1) or if it doesnt have any at all. I mean it's clear that its...
Homework Statement
So we have been doing complex numbers for about 2 weeks and there is this one equation I just can't solve.
It's about showing the set of solutions in graphical form (on "coordinate" system with the imaginary and the real axis). So here is the equation:
Homework Equations...
Hey there!
I just have a quick question regarding the graphical adaptation of the solution for the following complex "equation":
|(z+i)/z| < 1
What is the solution for this and how do I convert it into an image (with the imaginary and the reality-axis)
Thank you!
I thin they are more considered like "monkey see, monkey do" which is maddening. The blue block formula comes from the elastic equation ( where no KE is lost) (combined of momentum conservation and energy conservation)
Yeah well I get it ;) I am just gonna solve it my way and then blame the professor or his students who made these examples...
And you have to consider that I have to translate the whole thing from German to English
Thx guys, really appreciate your help:kiss::biggrin:
Ok, I just gotta make a statement: I am not actually bad at physics, but some of the problems to me pose further problems on how to solve the question, just as you said, the incompleteness of the questions shows that they are not thought out well in my eyes. The example before this on my...
Yeah for c) they are loosely coupled, so we have inelastic collsions between all pairs. But how will the velocity change there?
and for a) and b), I assume, regarding other examples we got for exercise, that they would have mentioned it if the first waggon would "stick" to the other 10 so that...
Ok so could you explain a little further why b) is wrong? would the formula: v1' = ((m1-m2)/(m1+m2)*v1 be correct for the one waggon? and v2'= (2*m1/(m1+m2))*v1 for the rest of the waggons be correct?
to c) well I thought of it like this: The first waggon hits in an elastic collision waggon...