Recent content by RooftopDuvet

ahhh, i had completely forgotten about the perpendicular bisector approach, thanks for reminding me. I'll try it now. Although I'm still not sure about the equation of a circle approach. I disagree with some of your working for that way. In subsituting values in you've written your Q x...

I'm doing a question in my book on conics where there is a circle cutting through a parabola. There are three points S - the focus P - a point on the parabola and Q - the point where the tangent at P meets the directrix. The focus is at S[1,0] and the dirextrix is x=-1 point P is [t^2, 2t]...

My question was more... when the light reflects off the mirror it begins to travel perpendicular to the motion of the ether; these two motions are independant aren't they? It's not as though the light is being bent at a slight angle more, so that it's resultant motion is in the perpendicular...

Because of mixups and unwittingly bad decisions made about my subjects it means I'm coming out of three years of sixth form with 3 and a half A levels, which I am predicted high As in though. I desperately need some advice on physics related experience i could get which would put me in a...

I've just started reading the dustiest book on relativity i could find at the local library and like most others I've seen it includes the Michelson Morley experiment. I just need a bit of clarification on the formulae they use. The equations t1=2l/(c^2 - v^2)^1/2 and t2=2lc/c^2 - v^2...

damn it, i had a feeling it was going to be something that simple. Thanks a lot, you've put my mind at rest.

thanks for the heads up... btw... you wouldn't happen to be doing some kind of radical calculus crossword puzzle would you? edit: wait... i don't seem to be finding any LaTex. I may need a bigger heads up...

appreciated... p.s. "The adjective is calculous by the way.." - :-p i know

How would i go about writing an equation for the velocity of an object released at a high altitude above the Earth which takes into account the sqaure increase in acceleration with displacement (s^2). I posted it here because I'm aware it's more of a calculus question than physics.

p.s. I'm new here. Where is everyone getting their maths type?

I've not looked at these type of equations before either but i'll speculate an answer if it's alright with the sire. You're told the values of y' and y for when x is 1 so substitute them in with x as 1 to get: (1^2)y'' + 2(1x5) - (30x6) = 1 y'' = 171 Then integrate to get: y' = 171x...

I'm doing some research on space elevators and have found a site (http://www.zadar.net/space-elevator/#transverse") which gives some insight into the maths behind the elevator cable. If you click on the link and scroll down, up, whichever direction to equation (2) and a picture of a blue...