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dividebyzero
post Nov 29 2013, 10:47 AM

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QUOTE(Critical_Fallacy @ Nov 28 2013, 10:55 PM)
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Thumbs up for both of you. I kept trying x = y/(y-2) and even (x-a)(x-b)(x-c)=0 to solve and failed superbly.

Now, another math question appears.
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So far I have reached here. How can I execute these highlighted areas?
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This post has been edited by dividebyzero: Nov 29 2013, 10:06 PM
dividebyzero
post Dec 1 2013, 12:01 AM

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QUOTE(Critical_Fallacy @ Nov 30 2013, 02:58 AM)
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ohmy.gif How can I repay my gratitude for such detailed explanation? You sir, deserve unlimited medals of honour.

I'm still on my way to solve cos 4θ. Wish me luck.
dividebyzero
post Dec 1 2013, 12:14 AM

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QUOTE(delsoo @ Nov 30 2013, 04:38 PM)
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Perhaps you may consider this: Work done to increase kinetic energy is not considered since no change of speed is observed (keyphrase: constant speed). If an external force other than electrostatic force, the equation W = q(V - V\inf) is not applicable since V is the electric potential / driving force at a certain location.
Is there any other type of forces which brings charged particles around? IMO there should be none other than electrostatic force.

This post has been edited by dividebyzero: Dec 1 2013, 12:17 AM
dividebyzero
post Dec 1 2013, 09:17 PM

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QUOTE(delsoo @ Dec 1 2013, 08:39 AM)
do u mean if the external force is not stated in the question, then the charged paticle should be driven by the elcstrostatic force alone? shakehead.gif
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dividebyzero
post Dec 29 2013, 12:54 PM

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Quick question once again:

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First part was cleared. The solution of the second part was blunt and depressing.

user posted image

What logic is this? Please, enlighten me.

In the meantime I'm off to study Abelian groups etc etc.
dividebyzero
post Dec 29 2013, 06:18 PM

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QUOTE(Critical_Fallacy @ Dec 29 2013, 05:40 PM)
ZERO,

This is purely algebraic manipulations. laugh.gif It is pretty obvious that

user posted image

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I understood every part of your explanation except this. Why would Sn be equivalent to the original roots? To my understanding, when the sum of two numbers is raised to any power, they give rise to intermediate terms like those in binomial theorem. Correct me if I'm wrong.
dividebyzero
post Dec 29 2013, 11:03 PM

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QUOTE(Critical_Fallacy @ Dec 29 2013, 06:42 PM)
ZERO,

It was originated from here:

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Therefore, when we add up the three equations below:

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Ahh, that's why -3 appears. Thanks much, my brain has been aloof.
dividebyzero
post May 9 2014, 11:44 PM

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Such treacherous fraction in my moment of inertia!
dividebyzero
post May 26 2014, 10:33 PM

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QUOTE(Vincent_Lam @ May 12 2014, 04:21 PM)
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Pulling my hair trying this question, anybody able to assist me in this?  icon_question.gif  tongue.gif
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I don't have enough time to explain, but try taking moments about point G. Since it is in equilibrium, both total anticlockwise and total clockwise moments are equal. Make sure the moments generated by the force is perpendicular to point G.

 

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