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That's a novel way of looking at it.But does it show that the third option I suggested takes longer? It would suggest that, as the height is lost quicker, it would get there faster so the inequality wouldn't always always apply.So what is the shape of fastest path from top to bottom?

From a height R falls (from zero speed) a mass point which follows a curved trajectory of 1/4 circumference (so the total lenght is π/2*R).From the same height R falls, at the same instant of time, from zero speed, a mass point along a stright inclined plane which is long π/2*R as well.

I had posted a response, but then realized my math was off, so here goes again. The constraints on the problem are that total length traveled = R*Pi/2, and that the mass drop a distance of R total. I think that your numbers are off in the last case you posted. Here's what I get:Step 1: The mass falls vertically the distance R. The time for this istv=Sqrt[2]*Sqrt[R/g],and the final velocity isv=tv*g=Sqrt[2gR]Step 2: The mass maintains its velocity and finishes the total distance horizontally. The remaining distance is R*Pi/2-R=R*(Pi-2)/2. The time to travel this distance is justth=R*(Pi-2)/(2*tv)=R*(Pi-2)/(2*Sqrt[2*g*R])=(Pi-2)/[2*Sqrt(2)]*Sqrt[R/g]The total time is thent=tv+th={(Pi-2)/[2*Sqrt(2)]+Sqrt(2)}*Sqrt[R/g]~1.82*Sqrt[R/g].This is the fastest of the three times, as it should be.

You are right, of course and it solves the problem, as presented.I read it wrong but the fastest journey from A to B (i.e. displacement) is a more likely scenario to deal with than how quickly you can travel a certain distance.I guess, if you had to drop a ball down a fixed length of tubing from a given height above ground, the original question would deal with that scenario.

Does no one want to calculate the phase angle or the "Q" ?.

How can we include √(L/C) in our calculations when we are told nothing about them ?.

We are told nothing about the value of R we can only assume a convinient value.In my opinion the only information we can derive from the data given is the Q factor (1.25) and the phase angle of the current (53.13°).

Why does E_{k} = 0.5mv^{2} but E = mc^{2}

How are the two equations related? What happened to the 'm'?

Quote from: syhprum on 07/07/2009 11:54:45How can we include √(L/C) in our calculations when we are told nothing about them ?.Not understood this one.QuoteWe are told nothing about the value of R we can only assume a convinient value.In my opinion the only information we can derive from the data given is the Q factor (1.25) and the phase angle of the current (53.13°).Exactly. We have 3 equations and 4 unknown. Solved them we have the parameters I wrote, from which we can only find what you said.

E_{k} = 0.5mv^{2} is valid only for bodies with non zero mass and at non relativistic speeds; E = mc^{2} is valid only at zero speed.The correct one, valid for all speeds and with massive or non-massive bodies (in special relativity) is:E^{2} = (cp)^{2} + (mc^{2})^{2}where p = momentum. If m = 0 (photons, gluons, ecc.) then E = cp. At zero velocity p = 0 so E = mc^{2}.Kinetic energy is total energy minus rest energy:E_{k} = E - mc^{2}