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Two questions and two errors: The (hypothetical) point of view of the photon is not "frozen" because you need time to perceive something frozen. But the photon has proper time zero, everything is reduced to one instant, thus nothing can be frozen. "Dodge" a photon: Information is transmitted with light speed. As the photon is moving with light speed it is ...


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A photon cannot be said to have its own inertial reference frame, because inertial reference are defined to be a family of coordinate systems that satisfy the two fundamental postulates of SR, one of which is that light moves at c in all frames. You could construct a coordinate system where the photon was at rest, but since this coordinate system wouldn't be ...


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It's a little bit hard to follow your reasoning. Let me try to give the method I would use - simple balance of energy. If the car reaches a velocity $v$ after time $t$, the power of the engine will have been used mostly for five components: Kinetic energy of car: $\frac12 m v^2$ Rotational kinetic energy of tires: $4 \times \frac12 I_t \omega_t^2$ ...


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There is no simple intuitive answer to your question - it follows from various group theoretical theorems: the Euclidean group and Poincaré group decompose to so called semi-direct products of (1) a group of translations and (2) the rest of the isometric transformations. In the Euclidean group, the "rest" turn out to be rotations, and in the Poincaré group ...


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A chaotic pendulum, or for the matter, any pendulum, would go on forever if it were only under the influence of the gravitational force and not influenced by any non-conservative forces. (i.e. ones which dissipate the energy of the system, examples would be air drag and friction in common cases). Also, any pendulum which is left initially at its lowest ...


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When a force is applied to a body initially at rest for a finite amount of time, the velocity of that body is given by conservation of momentum: $$m\Delta v = F\Delta t$$ When initial velocity is zero, $\Delta v = v$ - the final velocity is equal to the change in velocity. So we can write $$v = \frac{F\Delta t}{m}$$ From this it can be seen that ...


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In a system, the total sum of forces when added together equals mass times acceleration: $$ \sum F = \frac{\mathrm{d}p}{\mathrm{d}t} = \frac{\mathrm{d}mv}{\mathrm{d}t} = m\frac{\mathrm{d}v}{\mathrm{d}t} = ma $$ Since the sum of the forces on the robots is zero, there is no acceleration. However, the tension of the string is not contingent on the movement. I ...



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