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Let us not forget that the interactions in two of the machines violating the energy conservation law which were constructed by me (the machine ADAM - TWT-II, i.e., the cemented Faraday disk, or the so-called N-machine of Bruce de Raima, and the machine MAMIN COLIU - TWT-III) were magnetic.The latest information which reached me from Bill MLiller from the Canadian British Columbia (I met Bill and his wife in Hannover in 1987 and his charming daughter Carmen in Einsiedeln in 1989, where she reported on the work of her father) is that Mliller is on the threshold of closing the energetic circle in his over-unity converter and thus to run it as a perpetuum mobile.Then I think that the pushing force will depend neither on the size of the bridge. (20) The force acting on the circuit at change of its characteristic size, a, will be f = 8W/3a = (l/2)I^(8L/aa).*- The pushing force acting on the Ampere bridge can be found also proceeding from the formula for the inductance of the loop: If L is the inductance of the circuit, then the magnetic energy of the circuit when current I flows through it is' ' (p. (21) The inductance of a square loop with side a and radius r of the wire is^ ' k«..,v.« = 8xl0'^a = 8xl0"^a, square (22) while the inductance of a circle with diameter a and radius of the wire r is^ ' (p.' '^ - 20 - ^* * Marinov neither Thus I am inclined to conclude that, the pushing force acting on a II- form bridge, nor the force acting on a U-form bridge can be calculated mathematically.And taking into account the negative result of my experiment presented in figs.3 and 4, I make the general conclusion: the pushing force acting on the Ampere bridge does not depend on its form.This conclusion, of course, has to be confirmed by careful expe- riments with U-form and 11- form bridges with very long legs, to see whether, indeed, the pushing forces will be equal.

But he does not know about Faraday's experiments on unipolar induction in 1831, either.

361) "-circle " 27Txlo"^a = 6.28xl0"^ax = - (Mo Vl/47ry)a(y^ a^)-^/^ x (7) will be in parallel to the x-axis and thus will not generate induced tension (I dislike the term electromotive force ) in the wire DA or AE. The rest-transformer electric intensity E,est-tr = - ^«(t)/»t.

(8) where vt 1 A(t) = (y^, I/4TT)/(y-y')'Vy = - (u^, I/4i T)ln(l - vt/y)y (9) is the magnetic potential generated by the additional vertical current appearing in the rail DA at time t as a result of the motioii of the wire AB with a velocity v, and y is the ordinate of the reference point in the wire AE (for a reference point in the wire DA one must write in formula (9) y' -y instead of y-y'), so that ^rest-tr = " (V^/^^^Y " vt)"^y. For the motional electric tension induced in the wire AB we shall have by - 30 - Marinov Integrating formula (4) and taking Into account also the tension induced as a result of the magnetic action of the current wire CB (^^ the tensions act on linear wires, we ^ can consider them as vectors) "mot = ■ {%^l/^^)n^/^)i^^*^^)'^^^'^ = - (v I/27T)/(dx/x))( = - (y^vl/2ii)ln(a/a^)5!

(10) The calculation of the rest-transformer electric intensities induced in the wires CB and BF is similar and thus we see that the polarities of the electric tensions in- duced in the wires AB, DA, AE, CB, BF are as indicated in fig. (11) where a is a small length, as for a^ = the integral on the right becomes illimited.

The mathematical determination of the small quantity a is impossible at this level of mathematical calculations (in Ref. 140), for AB/AD = Z^' (3 so that q^ and iqr/c)'\ are the electric and magnetic energies in whid charge q takes part. (2) to its usua form (known as the Lorentz equation) : dpjdt = -q{V a A/ca/) -»- {qy/c) x rot A.

Familiarity with either would quickly tell him what is going on and what the answer to the "conundrum" is.

அவள் போடா அதுக்கெல்லாம் வேற ஆள பாருனு சொல்லி ரூமுக்கு வெளியே சென்றாள். அவள் சரி போதும் நான் கிழம்புறேன் என்று சொல்லி நகர்ந்தாள். அவள் இரு முலைகளையும் பிடித்து சப் சப்பென்று சப்பினேன். அவள் ஆம், அவள் பாய் பிரண்ட் அவளை 3 முறை ஓத்திருக்கிறான்.