"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.
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 ?
It says that the time it takes for ANY object to fall is only dependent on
two things, acceleration and height.
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.....
Brian
