Recent content by haruspex

There are various reasonably standard functions for this. See e.g. Iverson Bracket notation, Indicator Function and Heaviside Unit Step Function.

I don’t see that there is any connection with part a. Just treat part b on its own terms.

Fortunately, it also works for cards😉

Only if you add the condition that the next card is a different colour.

Looks wrong. You first have to decide how many ways you can pick two non overlapping 2x2 squares, then ask for the probability of choosing exactly the eight points corresponding to one of those pairs.

@NTesla, that does it! Look for help elsewhere.

Ok, but you wrote that you wanted the average number of clusters. But I am unsure how you are defining that number. With 6 particles in a 2x3 block, is that one 2x2 or two? I don't see any hope of an exact general solution, but maybe some bounds. Simplify it a bit to start with. Try s=2...

To explain it, we have to come up with a model for how more work is being conserved than we might guess. Consider a rod hinged to a fixed point at one end and subject to a perpendicular force ##\vec F## at the other. No gravity. The reaction from the hinge normal to the rod is ##\vec F/2##...

Ok, so what velocity does that imply, assuming mechanical work is conserved?

##\frac {v}{2}## is not a length. Do you mean it falls with velocity ##v##? When the top has fallen by ##y##, say, how much GPE has been lost?

In the r=0 limit, there are only two velocities and one of them is zero.

I believe this problem belongs to class that includes the chain fountain effect. In practice, it seems that when the link at the bottom gets rotated by the force from the static side its inertia leads to a significant pull down on the descending side. E.g. consider a rod lying EW on a smooth...

Based purely on the text description, I think you are supposed to treat O and A as the same point. Or was the diagram provided? The diagram is certainly inaccurate since it would start as a catenary.

Yes, very neat, but might not be clear to all. A bit of elaboration might help… After n tosses each, - A has more heads than B with probability p - B has more heads than A with probability p - equal counts with probability 1-2p A's extra toss doesn’t change the outcome in the first two cases...

Mostly, yes. An awkward exception is kinetic friction. When writing, say, ##\mu mg\sin(\theta)## in an equation, the sign matters and might not be apparent.