Re: Electrolysing Steam

From: charliew2 (charliew2_at_ev1.net)
Date: 10/27/04 

Date: Wed, 27 Oct 2004 12:12:45 -0500

Bill Ward <bwardREMOVE@ix.netcom.com> wrote in message
news:417de0cd.4659636@localhost...
> On Mon, 25 Oct 2004 23:11:43 GMT, Dan Bloomquist
> <EXTRApublic21@lakeweb.com> wrote:
>
> >
> >
> >Franz Heymann wrote:
> >>
> >> No, no and no again.
> >> The correct expression is
> >> F = dp/dt = m dv/dt + v dm/dt
> >> The first term is ma and the second term is an absolutely essential
> >> modification if the mass of the system varies.
> >
> >Hi Bill,
> >When you consider the instantaneous case it is natural to 'think' that
> >mass is not changing in a zero time interval. But the same would apply
> >to dv/dt, a, with that thinking! So, the second term does contribute
> >even if it contradicts your sense. Do you see it?
>
> Hi back atcha, Dan.
>
> I think I understand the equation, I just don't see the
> relevance. I'm looking for instantaneous acceleration given
> a constant force and variable mass. The equation provides
> the force as a function of acceleration, velocity, mass, and
> dm/dt.
>
> Why won't instantaneous values work in A = F/M?
>
> What am I missing here?

If F actually was defined by Newton as dp/dt, then it is apparent that for
most common problems, mass of an accelerating object either doesn't change
at all, or changes so slowly that the term dm/dt has been defined as
identically equal to zero. This simplification resulted in the common F=ma
equation that everyone uses. For problems in which mass does change
appreciably with time, F=ma is going to be incorrect because the starting
premise that dm/dt=0 is incorrect.

>
> More in the response to Franz.
>
>
> >
> >Here is another one that messes with ones senses.
> >Imagine the earth was just a shell, like a ping pong ball, say a foot
> >thick. This shell is some material so dense that when you stand on it
> >you weigh what you do on earth. Now, step through a hatch in the shell
> >so you stand on the inside of the shell. How much do you weigh? The
> >answer will surprise you.
>
> I think I've seen this one. IIRC, everything cancels and
> you're zero G.
>
> But I'll need to polish up my rusty calculus to see for
> sure.
>
>
> Thanks for your response.
>
> Regards,
>
> Bill Ward
>

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