In article  "John Knight" <[email protected]> writes:
<
<
<"Jet" <[email protected]> wrote in message
<news:[email protected]...
<>
{...}
<>
<>
<> 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://fathersmanifesto.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".
<
<problem, which STILL isn't even as complicated as an algebra, addition or
<subtraction, or multiplication or division problem?  You haven't even
<as if though you "think" this has something to do with solving the problem!
<It has utterly nothing to do with it.
<
<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
<
<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://fathersmanifesto.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

[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

In article  "Parse Tree" <[email protected]> writes:
<
<"Cary Kittrell" <[email protected]> wrote in message
<news:[email protected]...
<> In article  "Parse Tree" <[email protected]> writes:
<> <
<> <"Cary Kittrell" <[email protected]> wrote in message
<> <news:[email protected]...
<> <> In article  "John Knight" <[email protected]> writes:
<> <>
<> <> <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.
<> <
<> <But they really are experiencing the same acceleration at the instant of
<> <release.
<>
<> That is right as far is acceleration due to gravity is concerned,
<> as is implied in my statement above.  But each body is experiencing
<> additional forces due to the spring, so they will no be subject
<> to the same accelerations.  If you mentally switch off gravity,
<> the two bodies will move towards one another with an acceleration
<> proportional to the 2mg tension in the spring.  If you now switch
<> gravity back on, the whole system will accelerate downards at
<> 1 g, but this acts equally on the whole system, so you're back
<> to considering things in the frame of referrence of the system
<> itself -- as Jet implied.
<
<This is not true.  If you switch off gravity, then each sphere will stay at
<rest.  Firstly, you're assuming tension again, and secondly, the tension you
<assume exists only because of gravity.
<

This suddenly grows more interesting.  I've been reading "spring"; it
in fact says "string".

To be pedantic, in the real world, my argument would still apply --
a string is over short ranges in fact a spring, one with an extremely large
spring constant.  But that's obviously not the intent of the question.

<Regardless, you can simulate this using two balls and a string.  Just put
<them on a table and attach them with some string.  Then pull on them and
<release.  They don't move together with a force proprotionational to how
<much you pulled them apart.
<

If you assume an infinitely strong string, then you are correct. Otherwise
they will indeed move, unless you've stretched the string inelastically.
However, I'm being picky, and you're on to the intention of the question,
I think.

<> <Also, you're assuming the value of the unknown.
<> <
<>
<> Um, beg pardon?  Assuming the value of what unknown?  If you mean
<> the spring tension, I simply said that the /initial/ spring
<> tension is 2mg, because the lower mass is being pulled downwards
<> by a force of 2mg due to gravity.  Since it isn't moving initially,
<> there must be an equal and opposite force: 2mg of tension in the spring.
<
<The initial spring tension is unknown.  You're assuming that the bottom
<sphere is suspended from the top one.  It simply says that it's suspended at
<rest.  Which could simply mean that the system is suspended at rest.  Who
<knows?  Actually, I find many of these questions to be very imprecise.
<
<Regardless, the acceleration of the system is g.  And the acceleration of
<all of the parts are g.  Thus the string's tension should be 0.

Assuming an infinitely strong string -- one whose relaxation is zero -- then
you are correct.

-- cary

"Cary Kittrell" <[email protected]> wrote in message
news:[email protected]...
>
> In article  "Parse Tree" <[email protected]> writes:
> <
> <"Cary Kittrell" <[email protected]> wrote in message
> <news:[email protected]...
> <> In article  "Parse Tree" <[email protected]> writes:
> <> <
> <> <"Cary Kittrell" <[email protected]> wrote in message
> <> <news:[email protected]...
> <> <> In article  "John Knight" <[email protected]> writes:
> <> <>
> <> <> <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.
> <> <
> <> <But they really are experiencing the same acceleration at the instant
of
> <> <release.
> <>
> <> That is right as far is acceleration due to gravity is concerned,
> <> as is implied in my statement above.  But each body is experiencing
> <> additional forces due to the spring, so they will no be subject
> <> to the same accelerations.  If you mentally switch off gravity,
> <> the two bodies will move towards one another with an acceleration
> <> proportional to the 2mg tension in the spring.  If you now switch
> <> gravity back on, the whole system will accelerate downards at
> <> 1 g, but this acts equally on the whole system, so you're back
> <> to considering things in the frame of referrence of the system
> <> itself -- as Jet implied.
> <
> <This is not true.  If you switch off gravity, then each sphere will stay
at
> <rest.  Firstly, you're assuming tension again, and secondly, the tension
you
> <assume exists only because of gravity.
> <
>
> This suddenly grows more interesting.  I've been reading "spring"; it
> in fact says "string".

Yes, I noticed that.  But it's true even with a real spring.

> <Regardless, you can simulate this using two balls and a string.  Just put
> <them on a table and attach them with some string.  Then pull on them and
> <release.  They don't move together with a force proprotionational to how
> <much you pulled them apart.
> <
>
> If you assume an infinitely strong string, then you are correct. Otherwise
> they will indeed move, unless you've stretched the string inelastically.
> However, I'm being picky, and you're on to the intention of the question,
> I think.

They will move, but they won't move the same amount that they are pulled on.
Even a spring won't, unless it is perfectly elastic.

> <> <Also, you're assuming the value of the unknown.
> <> <
> <>
> <> Um, beg pardon?  Assuming the value of what unknown?  If you mean
> <> the spring tension, I simply said that the /initial/ spring
> <> tension is 2mg, because the lower mass is being pulled downwards
> <> by a force of 2mg due to gravity.  Since it isn't moving initially,
> <> there must be an equal and opposite force: 2mg of tension in the
spring.
> <
> <The initial spring tension is unknown.  You're assuming that the bottom
> <sphere is suspended from the top one.  It simply says that it's suspended
at
> <rest.  Which could simply mean that the system is suspended at rest.   Who
> <knows?  Actually, I find many of these questions to be very imprecise.
> <
> <Regardless, the acceleration of the system is g.  And the acceleration of
> <all of the parts are g.  Thus the string's tension should be 0.
>
> Assuming an infinitely strong string -- one whose relaxation is zero --
then
> you are correct.

Yes.  There are too many assumptions in these questions though.  I can see
why they're difficult.  There was another question about probability which
didn't even seem to specify if the two values involved were independent or
not.