kavlas

When i say "never fall" ,i mean the following: Suppose a body is so highly charged that after 1/10 sec there is no forces acting on the body ,then the speed with which the body will hit the ground will be 3,528km/hour. Suppose now the body is charged more so after 1/100sec there is no forces acting on the body ,then it will hit the ground with 0,3528 km/hour So we can highly charge the body up to the point that it will hit the ground with an infinitesimal speed. Probably with a speed : 3,528x 10^-1000Km/h

Then highly charged airplanes will never fall ?? Or ,in general ,highly charged bodies never fall. Suggest a practical experiment ,if this is possible.

Then the rule : all bodies fall in vacuum with the same acceleration 9,8m/s^2 ,should be altered to: All not charged bodies fall in vacuum with the same acceleration

You mean that the charge will hit the ground with an acceleration less than 9,8 meters/sec^2 ??

Then this work must of nearly infinite amount because you can let fall that charge from the top of a Cliff as many times as you want . Every time it will produce the same energy ,will it not ??

Where??

Sorry by "here" i meant the forum in concern. Certainly the discussion must be here. Where did the charge acquired the energy to form the electric field??

I would like to hear your opinion, if this possible, in the following paradox: here

[math] log_2 9 * log_3 4 * log_4 8 [/math]= [math]log_2 3^2*log_3 2^2*log_4 2^3=2log_2 3*2log_3 2*3\frac{1}{log_2 4}[/math]= [math] 12*\frac{1}{2log_2 2} = 6[/math] by using the formula : ................................[math]log_a b*log_b a = 1[/math]

According to the above 48/2 could be 24 ,or (48/1).2 = 96 ??, depending on what the author means ?? Since 2 = 1.2 ?? That is why i asked you how much is 48/2 Further more [math]\frac{48(x-a)^2}{(x-a)}[/math] could be 24(x-a) ,or: [math]\frac{48(x-a)^2}{2(x-a)}[/math] = [[math]\frac{48(x-a)^2}{2}[/math]].(x-a) = [math]24(x-a)^3[/math] Of course it is up to you to answer or not. I just wanted to show what i ment by asking those questions

What was my first question of the thread .....?

The question now is: Does Cap'n Refsmmat agree with you ??

How much is ,48/2 ,[math]\frac{48(x-a)^2}{2(x-a)}??[/math]

Well ,what did the OP asked?? Is there any mathematical book in the whole history of mathematics in which 48÷2y can be equal to [math](\frac{48}{2})y[/math]???

well ,the proof is high school level and it is the following: [math]\frac{48}{2y} = 48.\frac{1}{2y}[/math]=...........................................by using the definition of division: [math]\frac{a}{b}= a\frac{1}{b}[/math] =[math]48(\frac{1}{2}\frac{1}{y})[/math]=..................................................by using the theorem :[math]\frac{1}{a}\frac{1}{b}=\frac{1}{ab}[/math] and the axiom : 1a =a =[math](48.\frac{1}{2})\frac{1}{y}[/math]=...................................................by using the axiom : (ab)c =a(bc) =[math][(24.2)\frac{1}{2}]\frac{1}{y}[/math]=..................................................by using the theorem in Natural Nos 48 =24.2 =[math][24(2\frac{1}{2})]\frac{1}{y}[/math]=....................................................by using again the axiom : (ab)c = a(bc) =[math]24\frac{1}{y}[/math]=............................................................................by using the axiom : [math]a\neq 0\Longrightarrow a\frac{1}{a}=1[/math] and also the axiom :1a = a =[math]\frac{24}{y}[/math]................................................................................by using again the definition of division

the arc length in polar coordinates is : S = [math]\int_{0\to\theta}(\sqrt{(\frac{dr}{d\theta})^2+ r^2})d\theta[/math] So if you plug in the function r = Cr'+ [math]\frac{\theta}{2\pi}[/math] and do the appropriate calculation you will get the desired result

Well . Do we not define division in rational and in real Nos a÷b = [math]\frac{a}{b}= a\frac{1}{b}[/math]??

nouseforaname do you know that you can actually prove that 48÷2y= 24/y and not 24y??

Can the above solution be found in any known mathematical book??

The problem appeared in Gazeta Matematica and the difficulty of the problem is in proving that f is onto . Once this has been proved then the solution is easy since f is one to one . This i have dried so far but in vain

Suppose for the real valued funtions f,g,h ,where f is strictly monotone and the other two monotone the following relation holds: g(f(x)) =h(f(x)) =x for all ,x, belonging to the real Nos. Can we rove that : g=h for all ,x, belonging to the real Nos?? The domain of all 3 being the set of real Nos

Then we will have : t' =t Besides v is the speed of his reference frame

So to the standing observer a distance x from where he stands an event has taken place ( a cannonball has hit a wall 300000kilometers (c klm) away and come back to the cannon) after t time . That time t =[math]\frac{2c}{C/2 +C}[/math] = 4/3 sec. Now to the observer on the cannon the same event has taken place with new coordinates of time t' and x'=0. And from the Lorentz transformations: [math]x' =\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}[/math] we have: x = vt since x'=0.But v=c/2 and t=4/3 sec and so x=[math]\frac{4c}{6}[/math]klm. And from the trasformation: [math]t'=\frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}[/math] we can get t' by putting v=c/2klm/sec and x=4c/6 klm. That time will be shorter than t. Now is trere a checking procedure for the above??