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The Equivalence Principle of General Relativity holds that acceleration and gravity can be described identically. With an accelerometer, you can tell whether or not you are accelerating in empty space, regardless of whether another object is available to act as a reference point. Under acceleration, your weight will change just as though you were ...

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You have not specified which clock you pass first. When you pass over the X-clock, both it and your own clock read zero. The Y-clock (according to you) reads some time earlier than 0 if you've already passed over it, or some time later than 0 if you haven't passed over it yet. According to you, both clocks run (equally) slow, and the Y-clock is either ...

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The speed at that both X and Y run only corresponds to your relative velocity, not to your distance. So from your frame of reference both X and Y run slower than your clock, at the same, constant speed. What Y and your clock show when you pass it is easiest answered from the earths frame of reference: You are moving at high velocity, so your clock runs ...

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The reason for the "apparent" confusion, is that you ere, inadvertently, changing the frame of reference! In addition, the formula you are using is not correct for the cases in question. The energy being calculated, is the energy required to make a change in velocity $$E = m(\Delta v)^2 \ were\ \Delta v = v - v_o$$ For the case "walking on ground" $v_o ... -6 Because gravity works on you based on your distance from other masses. That changes little when you are at 30K', but measurably slightly less. This still forces your feet against the floor and friction coef. allows for traction. a Smart Academic answer would be because the "Captain has turned off the seatbelt sign". and you have legs. the energy came from ... 52 Due to momentum being conserved, when you accelerate yourself forwards relative to the plane, the tangential force you're applying to the floor will accelerate the rest of the plane backwards. Since the plane has a lot more mass than you, its velocity will not change by very much. Thus, an inertial observer who was initially at rest with respect to the ... -6 You, and the air and everything inside the aircraft, are travelling at the speed of the aircraft, and your motion is relative to that. Lest there be turbulance, you would certainly no longer be moving relative to the speed of the aircraft, and you would be accelerated by the difference. That's why they have seat belts. 35 Kinetic energy is not invariant under Galilean transformations. To see this consider the following: In the rest frame of the plane you apply a force$F$of 100N for one second to accelerate yourself to 1 m/s. During this time you move a distance$d\$ of 0.5m so the work done is: $$W = Fd = 100 \times 0.5 = 50\,\text{J}$$ This of course is equal to your ...

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"Virtual" is not a property of a particle at all. And it is not true that pure electric and magnetic fields are "made out of virtual photons" or that a combined electromagnetic field is "made out of real photons". A "virtual particle" is not a particle. It is an internal line in a Feynman diagram, which is in turn a graphic notation for a certain integral. ...

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Assuming that the twin star system is isolated, there is no external force acting on it so the center of mass should not move if the system was at rest initially. Centre of mass doesnot move Centre of mass moves although no external force is applied and the system was at rest initially

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