In article "John Knight" <johnknight at usa.com> writes:
<
<
<"Jet" <thatjetnospam at yahoo.com> wrote in message
<news:3D3A5859.B7212C46 at yahoo.com...
<>
{...}
<>
<>
<> What do you have to say about the fact you can't answer any of the
<> questions you took girls to task for not being able to answer?
<>
<> Here's a really simple one:
<> http://christianparty.net/timssh04.htm
<>
<> Brian got it wrong, he thought the objects started out at the same
<> height and you don't have a clue.
<>
<> J
<
<What kind of a moron could you possibly be to presume that the person who
<supplied the test scores and the original problems and the test answers in
<the first place "can't answer any of the questions"?
<
<This is called "negative knowledge". You have all the data and evidence you
<need to answer the question correctly, but you instead get the *wrong*
<answer, even when you've quoted some of the data and evidence yourself.
<
<For example, you also presumed that .25x = .30, you acknowledged that this
<would make x = 1.2, you admitted that you knew that x = .92 and not 1.2, and
<you STILL couldn't get the correct answer, and instead wrote: "Sorry, dumb
<ass, I scored in the top 2% on the GRE math test. And I can answer H04".
<
<Please tell us how this statement is going to help you solve this simple
<problem, which STILL isn't even as complicated as an algebra, addition or
<subtraction, or multiplication or division problem? You haven't even
<advanced to that stage yet, and here you are LYING about your test scores,
<as if though you "think" this has something to do with solving the problem!
<It has utterly nothing to do with it.
<
<Is it your debate tactic to LIE about your test scores so that your flock of
<feminazis will just take your word for it, hope that they'll just go along
<with your program--and keep them ignorant for the rest of their lives?
<
<The CORRECT answer is that x = 93 1/3%. Why couldn't you even figure this
<out? Why do you keep insisting that .25x = .30, when you were handed the
<answer on a silver platter?
<
<What is it about your arrogance that, after a week of being unable to solve
<the problem, you were GIVEN all the proper details on a silver platter, yet
<you STILL can't solve for "x"?
<
<People need to know. They need to understand what it is about feminazis
<that they never seem to comprehend some of the simplest points. This
<emotional grab bag of false feminazi claims has just got to come to an end.
<
<John Knight
Heh heh heh...
Jet, when you've seen John in action as long as I, you'll realize
you just won -- when he wanders about rambling about averages and
feminazis and LYING and test scores and and putting hypothetical
arguments orthoginal to anything you said into you mouth -- in other
words, eight paragraphs of bush-beating which contains not a word
of direct response to your two straightforward observations:
What do you have to say about the fact you can't answer any of the
questions you took girls to task for not being able to answer?
Here's a really simple one:
http://christianparty.net/timssh04.htm
Brian got it wrong, he thought the objects started out at the same
height and you don't have a clue.
that he's evaded your challenge a dozen times now, and that Brian's
reasoning was so far afield that it wasn't just "not right", it
wasn't even wrong -- well then, you just took game, set, and match.
By the way, my posts are getting dropped with dreary regularity, so
allow me to append my critique of Brian's "proof", which did not
make it to Google:
[Brian:]
<
<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).
Well, not only did you do it incomprehensibly, you did it incorrectly,
since the correct answer would be t = sqrt (2S/a). Hint: ds/dt = v = a * t,
the integral of a * t * dt is 1/2 * a * t^2
<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.
The "resulting velocity would be the same" if both masses were experiencing
the same acceleration the instant of release, but they were not. The
bottom mass was experiencing -2mg downards due to gravity and +2mg upwards
due to the tension in the spring. The upper mass is experiencing a
now unopposed -mg downwards due to gravity and a -2mg downwards due
to the same spring tension. You figure it out.
-- cary
