# Tag Info

6

What happens if the twin in the spaceship doesn't return? Would he still be younger than his other twin? It's really a moot point, because you can't compare clocks. There is no absolute time! You can't say, "What's each twin's age at this instant?" because "this instant" depends on the observer. Is the symmetry broken simply by accelerating out of earth? ...

5

You need to integrate acceleration to get the velocity. $$v(t) = \int_{t=0}^{t} a. dt$$ Ther are a number of ways of doing this numerically. I assume that you get these readings regularly with a spacing of $\delta t$, for example $\delta t = 100 ms$ or something like that. About the simplest way to do it is $$v(t) = v(0) + \sum a \times \delta t$$ ...

5

You mix the relations between the things. A force produces changes in the linear momentum. The acceleration $a = F/m$ produces changes in velocity $v = p/m$. So, your question should be either why don't we take the force proportional to the linear momentum, or why don't we take the acceleration proportional to the velocity. Now, the second thing is ...

4

As far as we know and can test, space is continuous, not discrete. Since space is continuous, then the labels we associate with it (i.e., positions) are also continuous. Calculus requires continuous functions to do the derivative and integral, so this implies that velocities and accelerations are also continuous because they are derivatives of positions: $$... 3 Well, technically we do:$$ F_{net}=m\frac{dv}{dt} $$The net force is proportional to the rate of change of velocity, which we call acceleration. Note also that it's not F\propto \Delta v (force proportional to the changed velocity because the changed velocity occurs over a period of time, \Delta t, that is also important--consider the difference in ... 2 Force was defined to be proportional to acceleration because that definition makes description of classical physics simple. For example on Earth we have a downward pointing constant force - gravity. If we define another quantity, one which is proportional to velocity, lets call it push, it would be quite useless in describing gravitational interactions. An ... 2 I'm hardly a GR expert, so if you want a more technical analysis I'm sure others will be able to give you one. However, the answer to your apparent questions is fairly straight forward. It is not the curvature of space or the curvature of time that causes accelerations, it is the curvature of space-time. We live in a four dimensional universe (ignoring ... 2 In a certain sense (regime) acceleration is caused by the curvature of time more than the curvature of space. Actually, the curvature is of the spacetime so that, making rigid distinctions has no much sense. However, if you consider the motion of a particle free falling in a region of spacetime, the equation of its story is the geodesical one: ... 2 When a charged particle comes into the vicinity of another, it's path is deflected. It decelerates in one direction, and accelerates in another. All charged particles that are accelerated/decelerated by another charged particle, or a magnetic field, emit radiation. See: Bremsstrahlung. Synchrotron radiation. Cyclotron radiation. 2 The equivalence principle tells us that we can evaluate \nabla_u u in a co-moving reference frame and that for geodesics we should find no acceleration (to the occoupants of an elevator in free-fall, the contents seem to be experiencing no acceleration). Therefore, if we evaluate this when we are not along a geodesic (elevator sitting on earth), we find ... 1 A central force does not perform work only if the motion is tangential. When the particle moves radially, it has a component of the speed that is not tangent. The cetripetal force and the velocity are no longer paralell so there the centripetal force actually does work on the system. The work is actually W=\int_{r_i}^{r_f}F_{centripeta}(r) dr NOTE: for ... 1 What do you mean you don't know what to do with epsilon(angular acceleration) ? Remember the crucial equation that: translational acceleration = angular acceleration * distance to rotational axis a = epsilon*r. Otherwise I am not sure what the problem is. 1 No, to translate escape velocity v_{esc} at the surface of a planet into gravitational acceleration g at the same location, you also need the radius R of the planet. The equation that applies is: 2 g R = v_{esc}^2. 1 Let's start out by understanding what's going on when we take logaithms. We have the relationship$$ y=kx^n $$There's no obvious method for working out either k or n from the graph of y against x; instead, we plot \log(y) against \log(x). Why do we do that? Well, if we take logarithms of both sides of the above equation, we get:$$ ...

1

You may be surprised to learn that you have done the right thing and have got the right answer. It's just that you've used non-standard units. Because you've put the distance in as millimetres and the time as milliseconds the value you calculate for $g/2$ is in units of millimetres per millisecond squared. You need to multiply it by a thousand to convert it ...

1

The change in velocity of the electron give rise to emission of X-rays. The electrons arrive at the anode with very high velocity and end up at thermal velocities - which must mean they slowed down. Both statements are therefore true.

1

The fast electrons slow down in the cathode, mostly due to interactions with atomic electrons. But hard X-rays are produced mostly due to deflection to large angles in the field of atomic nuclei. Roughly speaking, an atomic electron can stop the projectile electron in a head-on collision, but a nucleus can "reflect" the projectile back, so here the ...

1

When I square $v_x$ and $v_y$, I get \begin{align} v(t)&=\sqrt{\left(\omega-\omega\cos\omega t\right)^2+\omega\sin^2\omega t}\\ &=\left[\omega^2+\omega^2\cos^2\omega t-2\omega^2\cos\omega t +\omega\sin^2\omega t\right]^{1/2}\\ &=\omega\sqrt{2-2\cos\omega t} \end{align} which, due to the square root term, is slightly different than what you have. ...

1

Acceleration is simply a rate of change of velocity. So the magnitude tells you, how quickly velocity changes.

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