Roger Dynamic Motion

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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 onlyPhotons 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
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I offer my sympathy and support.Why would the terrorists want a lessnegotiationminded adversary in control? Surely they don't really think they can win a fullon showdown. If I were in their position I'd want the most conciliatory, discussionminded 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."
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thank you.I gave examples of weak force/electroweak interactions.
Read above links.
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Can you elaborate please.There are also electron capture, beta decay minus, double beta decay minus, interaction with (anti)neutrinos..
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is the questionA photon is charge neutral so there is no attraction due to charge. However all forms of energy/density including photons can produce mass if of sufficient value. For that you will need to apply the energy momentum tensor.
(string theory already takes the above under consideration)
Are electrons affected by a gravitational feels.
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A photon is charge neutral so there is no attraction due to charge. However all forms of energy/density including photons can produce mass if of sufficient value. For that you will need to apply the energy momentum tensor.
(string theory already takes the above under consideration)
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yes thanksAs far as I'm aware electrons are subject only to the Coulomb force and gravity. Given that they are both inverse square forces and the Coulomb force is much stronger regardless of radii, electrons would always repel each other.
(I'm guessing that was your question  the title sort of chopped it off)
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Hello Every one. I'm working on a model that would demonstrate the veracity of the string Theory .
Please your answer will be very much appreciated, if Yes why and if No Why Thanks .
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Hello Every one. I'm working on a model that would demonstrate the veracity of the string Theory .
Please your answer will be very much appreciated, if Yes why and if No Why Thanks .
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Please.an answer would be much appreciated .
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got to go lunch time
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I thought that electrically charged particle would emits photons and died.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 xray 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.
thanks you SenseiThat'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..
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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 .
is in it the other way around ?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 xray 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.
I always thought the kinetic energy of electron was dependent of the abortion of photons
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Hello Every one. I'm working on a model that would demonstrate the veracity of the string Theory .
Please your answer will be very much appreciated, if Yes why and if No Why Thanks .
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Uff.. what is that supposed to mean. A spinning top or ball has moment of momentum but zero momentum.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?
ok I see it now
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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 .
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I have my Modern Physic Book for scientists and engineers and it said 186000 sI've always heard 183,000, but neither is exact.
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This is only an exercise to prove that the velocity of the man with the wagon must be deduced to the total velocity over allThe 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.
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you mention here that ,,*The speed of light is 182,000 miles per second . Where did you get that ?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.
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just a question Kipingram isn't that dynamic acceleration?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.
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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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I agree Swansont I monopolize.You could have mentioned that earlier rather than waiting until the 29th post in the conversation.
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Can a bundle of photons attract an electron?
in Classical Physics
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