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Roger Dynamic Motion

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Posts posted by Roger Dynamic Motion

  1. An intelligible question would be appreciated! What do you mean by a "bundle of photons"? A single photon or a "bundle" would go past the electron "at the speed of light" so while there will be a slight gravitational attraction the photons would go past too fast for it to be noticeable.

    thank you .HallsofIvy

    An intelligible question would be appreciated! What do you mean by a "bundle of photons"? A single photon or a "bundle" would go past the electron "at the speed of light" so while there will be a slight gravitational attraction the photons would go past too fast for it to be noticeable.

    well they say the Earth is attracted by the Sun
  2. Photons by definition travel at c, since they have no rest mass. Once they exist, they move, regardless of what's going on at the source. Also, from a classical view, Maxwell's equations show that EM waves propagate, again without reference to their source.

     

    ]EM waves propagate, again without reference to their source.

    Quote meaning they have a momentum only

     

    Photons by definition travel at c, since they have no rest mass. Once they exist, they move, regardless of what's going on at the source. Also, from a classical view, Maxwell's equations show that EM waves propagate, again without reference to their source.

     

    ]EM waves propagate, again without reference to their source.

    Quote meaning they have a momentum only

     

    Quote meaning they have a momentum only
  3. Why would the terrorists want a less-negotiation-minded adversary in control? Surely they don't really think they can win a full-on showdown. If I were in their position I'd want the most conciliatory, discussion-minded adversary possible. Not that I think they really want to negotiate, but that would be the environment that would let them "get away with more."

    I offer my sympathy and support.
  4. Charged particles that are decelerating, are emitting photons to conserve energy.

    f.e. in vacuum tube electron gun is emitting electrons, they are passing through hole in positive charged electrode, and flying through tube (their path can be controlled by external electric field or magnetic field (typically electromagnet)).

    They hit something which you will put there, f.e. piece of metal,

    and x-ray photons are emitted after collision.

    Maximum energy of photon emitted depends on kinetic energy of electron prior collisi

     

    Charged particles that are decelerating, are emitting photons to conserve energy.

    I thought that electrically charged particle would emits photons and died.

    That's direct the case in photoelectric effect.

     

    In vacuum tube, you can accelerate electron to f.e. 1 keV kinetic energy, thus being able to have at max 1 keV photons, when you will provide 1000 Volts to electron gun electrodes..

    thanks you Sensei
  5. I think this depends on exactly what you mean by "can exist." Electrons can exist for periods of time without photon association. But the HUP makes that period of time variable, and I don't know to what extent you could "isolate" an electron such that you prevented such associations for any given period of time T. So in that sense I think such associations are allowed and expected, but in a probabilistic way.

    Thanks you very much for the answer .

    Charged particles that are decelerating, are emitting photons to conserve energy.

    f.e. in vacuum tube electron gun is emitting electrons, they are passing through hole in positive charged electrode, and flying through tube (their path can be controlled by external electric field or magnetic field (typically electromagnet)).

    They hit something which you will put there, f.e. piece of metal,

    and x-ray photons are emitted after collision.

    Maximum energy of photon emitted depends on kinetic energy of electron prior collision.

     

     

     

    Maximum energy of photon emitted depends on kinetic energy of electron prior collision.

    is in it the other way around ?

    I always thought the kinetic energy of electron was dependent of the abortion of photons

     

  6. If the p component perpendicular to radius is conserved and L and p cannot both be conserved, how is it possible to conserve L?

     

    Does the specific equation I am using not clarify which of your examples applies?

    Uff.. what is that supposed to mean. A spinning top or ball has moment of momentum but zero momentum.

    ok I see it now

  7. I think there must be some misunderstanding here (Nver suprising with Resnick and Halliday).

     

    Moment of momentum and Momentum may well refer to different things.

     

    A spinning top or ball has moment of momentum but zero momentum.

     

    A ball travelling in straight line, without rotating on its axis, has momentum but zero moment of momentum.

     

    A ball travelling around on a whirling string has both momentum and moment of momentum.

    Your quote] A spinning top or ball has moment of momentum but zero momentum. My quote But yes it has momentum when the top start warbling .

  8. The equations work. I won't do it relativistically because it's so much more involved, but for your example classical techniques would say the following:

     

    [ (wagon/man mass) + (ball mass) ] * (initial speed) = initial_momentum

     

    "tiny push": (some_force * some_time) = added_momentum

     

    (wagon/man mass) * (wagon final speed) + (ball mass) * (ball final speed) = final_momentum

     

    Then you'd have initial_momentum + added_momentum = final_momentum

     

    What about this is it that you think is wrong? You do need to know the ball's final speed in order to find the wagon's final speed, and all of the components of momentum are included.

    This is only an exercise to prove that the velocity of the man with the wagon must be deduced to the total velocity over all
  9. Then there is no way you can have a 3/4 sec time difference between the arrival of the light and the ship.

    Obviously not. If the measurements are being made by someone in the ship, and they turn on the light when they are 186,000 miles from the target, then by their reckoning it will take 186,000/(186,000+46500)*= 0.8 sec for the light to hit the target. They will meet up with the target in 186,000/46500 = 4 sec, or 3.2 sec after the light hits it.

     

    I don't know what you are doing to get 3/4 of a sec with the numbers you have given, but whatever it is, it's wrong.

     

    *The speed of light is 182,000 miles per second and 0.25 c is 46500 miles per sec. For someone in the ship the light is traveling away at 186,000 miles per sec, and the distance between themselves and the target is decreasing at a rate of 46500 miles per sec, and thus the light and target will meet when the target is 148,800 miles from the ship. The light will have traveled this distance at 186,000 miles per sec, which takes 0.8 sec. The distance between ship and target will have decreased by 37,200 miles, which at a relative velocity of 46500 miles per sec takes 0.8 sec to traverse.

    you mention here that ,,*The speed of light is 182,000 miles per second . Where did you get that ?
  10. In a central force system p changes to compensate for changes in r and L is conserved. And you are right - that force cannot change the components of p perpendicular to the radius, so the overall change to p necessary to conserve L is achieved by affecting the p component parallel to radius.

    just a question Kipingram isn't that dynamic acceleration?
  11. Referring to the rider man who tosses a ball from a motorcycle ahead of him the equation is no accurate because the inertia of the body the (ball) is not taking in consideration

    even if the ball has little inertia it should be included in the equation.

    For example . let the man the rider be sitting on a small wagon and they together make 100 pounds and let the ball be 100 pounds so they both have the same inertia; now lets give a tiny push let say with a velocity 5 feet second to the wagon and tell the man to trow the ball ahead of him

    should the reduced speed of the wagon , be consider in the equation ; no need to know the velocity of the ball in this case.

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