/sci/ - Science & Math

>>6734667
Well, firstly, not all forms of Extroversion are the same. There are four different cognitive functions that an Extraverted type may have.

>>Extroverted Feeling
>>Extroverted Thinking
>>Extroverted Intuition
>>Extroverted Sensing

The variance between the activities from the Extroverted dominant types are extremely high. I think that the function that you harbor the most disdain towards would be Extroverted Feeling.

Also, every person is complex, and by widening your gaze to the way that ALL people think, you can gain greater insights within yourself, your strengths, and your weaknesses.

In fact, the way you present your argument may have hidden motivations of jealousy or admiration laden with envy. That isn't nesseacarly a bad thing, but by opening yourself to other abilities you can learn to hone and adapt them to yourself, becoming a more fulfilled and well-rounded person.

>>5668683
dF=(dq*Q)/(4*pi*E0*r^2)

Q<- the particle
dq<- very small piece of the arc
pi<- 3.14
E0<- vacuum permitivity
r<- radius of your arc

let L=dq/dl be the liniar density of your arc

thus dq=dl*L;

rewrite F=(Q*dl*L)/(4*pi*E0*r^2)
One of the components of F will be 0; Fx=0 cause of the simmmetry
Fy=F*sin(c)
so you now have

Fy=[(Q*dl*L)/(4*pi*E0*r^2)]*sin(c)

lenght of a tiny piece of the arc dl=r*dc
Fy=(Q*r*dc*L)/(4*pi*E0*r^2)*sin(c)
Integrate over your angle : from 0 to bla bla
The only part you have to integrate is sin(c)*dc the rest of the terms are constant
You should end up with sth
{(Q*r*L)/(4*pi*E0*r^2)}*(-cos(0)+cos(whatever angle you have))
where L=the total amount of charge on the arc / the lenght of the arc

>>4648389
Nice sage. As a fellow anon said, either KoF or Street Fighter