# Tag Info

## New answers tagged velocity

0

some context is missing here but it could be the difference between Eulerian and Lagragian quantities, i.e. the spatial derivatives vs the particles derivative. It is mostly used for continuous materials (e.g. fluids), but can extend to other cases (e.g. field or stream of objects). The spatial (i.e. Eulerian) velocity in a field is the one at a given ...

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In the particular setup that you show, where both the car with the photon source and the surface with the detector move in the same direction at the same speed, the result is the same regardless of the emitted object is a ball or a photon: it will hit the detector. This is best understood by a transformation of reference frames: Instead of looking at the ...

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Yes it will be affected by the car's movement (if you are outside the car), since momentum is conserved. If you are moving with the car, the source is not moving relative to you, so the photon is moving down straight. PS from the outside point of view, it will seem like the photon traveling diagonally covers more distance. But since the speed of light ...

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If the shore is really far away, you can't tell the water is moving. Then it is immediately obvious that the boat must have spent the same length of time moving away from the crate as moving towards the crate - 1 hour each way. If we look at just the crate, we see that it moved 3 km in 2 hours (it was found "5 km downstream from the turnaround point" which ...

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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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Think of a position axis $x$ starting at zero, positive to the right and negative to the left. It is like a number line. A positive velocity means that the value of $x$ is increasing eg going from $x=+3$ to $x=+5$ or $x=-7$ to $x=-4$ or $x = -3$ to $x=+4$. A negative velocity means that the value of $x$ is decreasing eg going from $x=-3$ to $x=-5$ or $x=+7$ ...

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well you have given answer in your own question! Velocity and acceleration are both vector quantities, meaning they have magnitude and 'direction'. The sign (+/-) will depend on the direction. To simplify, let me give you an easy example.. Case 1: An object is moving down from the top of a mountain. The acceleration (in this case 'g') will act in the ...

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The acceleration at the point of reflection is actually quite complicated. It is caused by the elastic forces of the surface and the ball and has a complicated time dependence. However, the timespan in which the ball touches the ground is very short (especially if the ground and the ball are very rigid), therefore we can simplify the actual acceleration ...

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Imagine that the wheel is stationary. A force is applied which accelerates the wheel horizontally. This add translational momentum to the wheel. Now, since the wheel does not slide, a frictional force is produced at point P which produces a torque on the wheel. The torque causes the wheel to start rolling, adding rotational momentum. Once the wheel is ...

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As the definition of "rest mass", when the item is relatively static in your frame, the mass you observed is the rest mass.

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When the particle is stationary in your frame, the mass you observe is the rest mass.

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You are totally correct! Yes, velocity is relative, and therefore "relativistic mass" $m = \gamma ~m_0$ is relative: different people see different values. However, here's the crucial part, anyone who sees you travelling at speed $v$ with mass $m$ agrees on this number $m_0 = m / \gamma$. It is what's called a "Lorentz scalar": every coordinate system ...

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The value of velocity depends on the observer. Therefore the value of $\gamma$ depends on the observer. Therefore (since the value of $m_0$ does not depend on the observer) the value of $m=\gamma m_0$ depends on the observer. It's not entirely clear what's confusing you, but you appear to be assuming that $m$ should be observer-independent. It's not. ...

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It is because the ball was traveling with you when you threw it. Imagine the following question: I am in a car traveling at 5m/s holding a ball. Where will the ball be relative to me in 10 seconds? Answer in my hand. To be obtuse: ball has velocity 5m/s. After 10s it will have moved forward 10 x 5 meters = 50. I have velocity 5m/s. After 10s it will have ...

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Have you learned about gravitational potential? The force of gravity follows an inverse square law, $F=\frac{GM_1M_2}{r^2}$; the potential, which is the integral $\int F dr$, goes as $-\frac{GM_1M_2}{r}$. As you go from "infinity" (where the potential is zero) to some smaller radius, some potential energy is released and converted to kinetic energy, $\frac12 ... 0 The derivative with respect to proper time is the derivative with respect to time of the instantaneously comoving inertial frame. This does not mean the particle is at rest. That's why I had to have the word instantaneously in there. As for four velocity, that's the unit tangent vector to the world line. It tells you what direction in spacetime something ... 1 Your approach is correct; your ability to read data from a graph is suspect (the divisions are 2 m/s each). The initial velocity is -12 m/s, and at time t=9 s it is up to 18 m/s That should change your answer... 2 The average velocity is given by $$\bar v=\frac{1}{T}\int_0^T v(t)\mathrm dt=\frac{1}{T}(v_1t_1+v_2t_2)$$ where$t_1$is the time spend on the first interval,$t_2$is the time spend on the second one, and$T=t_1+t_2$. Using $$v_1t_1=v_2t_2=d$$ you get $$\bar v=2\frac{v_1v_2}{v_1+v_2}$$ I believe you can take it from here. 0 If a step in the temperature of the entering air occurs, there will be a lag time before this temperature change appears at the air outlet, can be sensed by the control system, can result in a change in valve position and thus a change in thermal input to the exchanger, and can then change output air temperature. This control response time delay must be ... 1 If you take the two well-known formulae 1. E = hν (Einstein) 2. pλ = h (de Broglie) and write the usual formula for phase velocity (v'), and do a bit of algebra, you end up with v' = E/p Now p is the momentum, which is well known to be = mv, v the particle velocity, m the mass, but what is E? If you woodenly put in E = mc^2, you end up with v' = ... 6 She will feel motionless, seeing the other car approach with the combined velocity of the two cars: $$v_\mathrm{tot} = \frac{v_\mathrm{\scriptsize{1}} + v_\mathrm{\scriptsize{2}}}{1 + v_\mathrm{\scriptsize{1}} v_\mathrm{\scriptsize{2}}/c^2} = 0.8c,$$ with all the usual implications of observing something moving at$0.8c^\dagger$.$^\dagger\$E.g. an epic ...

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Actually, you can consider a transformation to a tachyonic frame where the relative velocity is greater than c. Then the solution becomes simple, and is no longer meaningless. In the Earth's frame, XYZ was launched before TY. In TY's frame, his spaceship was launched before XYZ. Hence, it is XYZ that needs to catch up to TY! Not the other way round. ...

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