dumbass wrote:
>> "Jet" <thatjetnospam at yahoo.com> skrev i melding
> news:3D3A3FCE.269F3BE9 at yahoo.com...> > > > You can see the original problem at
>http://christianparty.net/timssh04.htm> > > ,
> > > > Brian, or download the scores from
>http://christianparty.net/cisp1295.pdf> > > >
> > > > This is one of the 13 multiple choice questions (ONE THIRD of that
> test)
> > > > where American girls scored lower than if they had just guessed.
> > > >
> > > > But of course the feminazis can't even figure out what this means, so
> > > > they're claiming that we're lying. Maybe you can straighten them out,
> > > > because I sure can't.
> > >
> > > Ok, the problem isn't difficult.
> > > F=ma
> > > S=vt
> > >
> > > F=force - downward
> > > m=mass
> > > a=acceleration
> > > v=velocity
> > > t=time
> > > S=Hight of fall
> > >
> > > I did it this way, first I said that F=2m(v/t).
> > >
> > > Then I converted the equation to t=sqr(mS/ma), moving the t to the left
> > > side, and the F (ma) to the right side.
> > > I used both equations for the object separatly, and ended up with the
> same
> > > equation t=sqr(S/a).
> > > Sqr means square-root of the equation in the parenthesis ().
> > > So, the resulting velocity would be the same, as the same time is spent
> on
> > > the fall, and the tension would be zero.
> > >
> > > Brian
> >
> > Mensa, my big black ass. The time spent in the fall would not be the
> > same, because the higher object started out...higher.
> >
> > Now go ahead, call me "nigger", post some silly ass links, but it won't
> > change the fact you don't know how to solve the problem...and I do.
> >
> > Here's a hint: What is the tension on the string before the fall?
>> Pure Nigger BS.
The tension on the string before the fall was Pure Nigger BS? Why can't
you answer the question, monkey boy? Oh, yeah you are too stupid!
>> Brian
>> 1.F=m(v/t)
> 2.t=mv/F
> 3.t=mv/ma
> 4.t=v/a
> 5.t=(S/t)/a
> 6.t=(S/ta)
> 7.t²=(S/a)
> 8.t=sqr(S/a)
>> do you understand the equation ?
Yes, I do, monkey boy. But you don't understand that S is different for
each object.
> It says that the time it takes for ANY object to fall is only dependent on
> two things, acceleration and height.
That's nice. But the heights were not the same.
http://christianparty.net/timssh04.htm
I'm really laughing at you, monkey boy.
> And it is regardless of mass.
> Need I say Copernicus and Gallileo ?
> Stupid Nigger, never comprehended the factoid that I have slam-dunked her
> into the scarp-yard of history.....
>
You can't figure out the answer! Here's another clue: "frame of
reference".
LOL, you stupid fucking monkey boy! You got the answer WRONG! How could
S be the same for both objects when one is higher than the other?
It could be you are just a moron who couldn't figure that out! ROTFLMAO!
J
> Brian
