Prev: Nano catalysts, and the death of Coal ;)
Next: It's a heatwave
From: BURT on 28 May 2010 19:43
On May 28, 6:46 am, PD <thedraperfam...(a)gmail.com> wrote:
> On May 28, 4:02 am, boltar2...(a)boltar.world wrote:
>
>
>
>
>
> > On Thu, 27 May 2010 12:14:39 -0700 (PDT)
>
> > Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> > >Maybe it's time to sit down and do the calculation? Try it, and you'll
> > >see that this isn't a problem.
>
> > >The faster the rocket goes, the more KE the unexhausted exhaust starts
> > >with. When the rocket goes faster than the exhaust, the change in the
> > >KE of the exhaust is larger than when the rocket goes at the speed of
> > >the exhaust, even if the final KE of the exhaust in not-zero.
>
> > >If this troubles you, do some examples yourself. Many examples. Try
> > >the completely inelastic collision problem (i.e., where the two
> > >colliding bodies stick together). Start with the same relative speed
> > >between the two bodies, but with different speeds for the bodies. For
> > >the different speeds, the KE of each body, and the total KE, will be
> > >different. The final KE will be different. But in all cases,
> > >KE_initial - KE_final =3D the same value, the amount of energy lost in
> > >the collision. (Where "lost" means converted to other forms of
> > >energy.)
>
> > Ok , I understand now. A good explanation , thank you.
>
> > I still can't figure out how it works when what provides the force doesn't
> > have any mass itself such as some sort of field however since then the field
> > won't have any KE to start with to donate to the rocket.
>
> This is a misconception. Fields are deemed to be physical entities
> partly because they do exhibit important physical properties like
> stored energy and momentum. You might ask how they do that if they
> don't have mass. You may have been misled that the momentum of
> ANYTHING is defined as mass times velocity, or that the kinetic energy
> of ANYTHING is half the mass times the velocity squared. If you got
> that impression, then it was an unfortunate mistake. That rule for
> momentum works only for massive objects and even then only for those
> moving slowly (compared to the speed of light). There are different
> rules for finding momentum for other kinds of physical entities.
>
>
>
>
>
> > >This isn't at all new. The equivalence of coordinate systems was
> > >discussed by Newton and Galileo. But it has been confusing students
> > >for hundreds of years. You don't have to take it on faith; you can try
> > >it for yourself, and caculate examples, solve problems, until you see
> > >how it works. That's one purpose of those long lists of end-of-chapter
> > >problems.
>
> > I know its true. I'm not one of these nutters trying to prove it to be false, I
> > just didn't understand why it was true :)
>
> > B2003- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -

Fields are forces all with a finite range. No force has an infinite
range. Strong is the smallest field followed by the electric then
magnetic field aether then gravity with the longest range.

Mitch Raemsch
From: John Polasek on 2 Jun 2010 17:47
On Wed, 26 May 2010 14:16:03 +0000 (UTC), boltar2003(a)boltar.world
wrote:

>Before anyone asks , no this isn't some school project I'm working on. Its
>a simple problem I can't figure out for myself.
>
>If you have a rocket motor giving a constant thrust then the rocket - not
>accounting for weight change due to fuel use - will accelerate at a constant
>rate. However if say its accelerating at 1m/s/s then it will take twice as
>long to accelerate from 2 m/s to 4 m/s as it did to get from 1 m/s to 2 m/s.
>Yet going by 0.5 mv^2 the kinetic energy for a rocket weighing 1 kilo at the
>given speeds is:
>
>1 m/s = 0.5J
>2 m/s = 2J
>4 m/s = 8J
>
>So in the first second it gained 1.5J but in the next 2 seconds it gained
>6J. I don't get it? The motor wasn't putting out any extra energy so where
>did it come from?
>
>B2003
>
With constant thrust, the power output is proportional to the velocity
W = FV force times velocity
so the power output is definitely not a constant. At 100 m per second
the power is 100 times what it is at 1 m per second, relative to the
Observer.
W = FV = FAT (A = constant acceleration)
Energy equals FA* integral TdT = FAT^2/2 = FX equals force times
distance traversed.
Constant thrust is kind of tricky.
John Polasek
First  |  Prev  | 
Pages: 1 2 3 4 5
Prev: Nano catalysts, and the death of Coal ;)
Next: It's a heatwave