# Tag Info

4

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 ...

4

If you balance the forces due to the electric and the magnetic fields on the charged particle in such a way that there is no resultant force on the charged particle, then that is called a velocity selector. It means that the Lorentz force on the particle is 0. $$F =Q(E + v \times B) = 0$$ This allows you to measure the velocity of the charged particles ...

2

Reusing your car example: use the fact that acceleration is "change in velocity". This can be positive (acceleration in the usual/common sense), but everyone knows that velocity can also be decreasing. This is what phsicists call negative acceleration.

2

There are two things here that aren't quite right. The speed of light is the same for all observers. The speed that light travels is invariant for any observers in any reference frames observing the same beam of light. I can move at 100 meters per second (as measured in one reference frame) in one direction, while you move 256 meters per second (as ...

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.

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...

1

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, $\frac12 ... 1 It is because the ball was traveling with you when you threw it. Imagine the following question: I am in a car traveling at 5m/s holding a ball. Where will the ball be relative to me in 10 seconds? Answer in my hand. To be obtuse: ball has velocity 5m/s. After 10s it will have moved forward 10 x 5 meters = 50. I have velocity 5m/s. After 10s it will have ... 1 The value of velocity depends on the observer. Therefore the value of$\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. ...

1

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

1

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

1

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 ...

1

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 ...

1

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 ...

1

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 ...

1

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 ...

1

Positive means speeding up, negative means slowing down. Now this is assuming you are traveling in the positive direction but through an axial change you could always guarantee this. I think this would be a good starting point for a child.

1

I would explain that we can feel acceleration in a car. As our speed increases we are pushed back into the car seats. This is positive acceleration. As we slow down we are pulled back by the edge of the seat/seat bell. If the child is happy that acceleration is a change of speed then they should be able to tell with their eyes closed if the ...

Only top voted, non community-wiki answers of a minimum length are eligible