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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 ...

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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 ...

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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 ...

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If you mean force proportional to velocity, that restricts the second law to only the specific case when the force is proportional to the velocity (the object will feel a drag or will accelerate exponentialy with time, depending on the sign of the proportionality constant). Such a law will not decribe the dynamics of any other objects.

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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 $$v(t) = v(0) + \sum a \times \delta t$$ ...

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In the case of the atmosphere, we usually assume that the stationary atmosphere is adiabatic, that is, there is very little transferral of heat between regions of different height . Combining hydrostatic equilibrium with the adiabatic law of an ideal gas gives you a variation of temperature (and pressure) in the direction of acceleration. Check out Richard ...

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

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In physics, magnitude is the size of a phusical object, a property by which the object can be compared as larger or smaller than other objects of the same kind. More formally, an object's magnitude is an ordering (or ranking) of the class of objects to which it belongs.

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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. ...

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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? ...

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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: $$... 1 You may be surprised to learn that you have done the right thing and have got the right answer. It's just that you've used non-standard units. Because you've put the distance in as millimetres and the time as milliseconds the value you calculate for g/2 is in units of millimetres per millisecond squared. You need to multiply it by a thousand to convert it ... 1 Let's start out by understanding what's going on when we take logaithms. We have the relationship$$ y=kx^n $$There's no obvious method for working out either k or n from the graph of y against x; instead, we plot \log(y) against \log(x). Why do we do that? Well, if we take logarithms of both sides of the above equation, we get:$$ ...

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I believe the Earths Inertia would prevent any of the other 3 objects from moving it. Their Gravity can not overcome the Earths inertia. Therefore The Earth stay's where it is and the 3 land at exactly the same moment.

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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: $2 g R = v_{esc}^2$.

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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.

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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 ...

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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 ...

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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 ...

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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.

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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.

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You are actually completely right, and then at the last minute you look the wrong way!!! Lets examine the last bit. If you were to continue applying a net force on an object for an infinite time it would indeed accelerate infinitely. There is nothing wrong here. The key point is the NET force, that is, the net amount of force, and its direction, that is ...

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This is an extended comment on Valter's answer, so please upvote his answer not this one. In Relativity (General and Special) there is no unique way to divide spacetime into space and time. Different observers, using different coordinate systems, will disagree about whether a four vector is just a displacement in time or just a displacement in space. So to ...

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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: ...

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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 ...

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The second law states that the net force on an object (F) is equal to the rate of change (that is, the derivative with respect to time) of its linear momentum p (p = m*v, where v is the velocity) in an inertial reference frame. When F is zero, due to the above equality, dp/dt is also zero. This means that d(mv)/dt = 0 --> mv = const --> v = const. ...

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Here's what you do: You know $v_0$, $v_f$, and $x$. This means you can use the equation $$v_f^2=v_0^2+2ax$$ Re-arrange it to get $$a=\frac{v_f^2-v_0^2}{2x}$$ You also know the mass, $m$. You can use Newton' second law, $F=ma$, here: $$F=m\frac{v_f^2-v_0^2}{2x}$$ This should lead you to the answer.

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Here's a simple demonstration: Consider flat space (i.e. Minkowski), viewed in a rotating frame (in e.g. cylindrical coordinates one just replaces $\phi$ by $\phi'=\phi+\omega t$). One can calculate (without too much trouble) that, in these coordinates, a spatial line element can be expressed in terms of the canonical cylindrical coordinates as $$... 3 But what about the case where the velocity only changes its direction keeping its magnitude constant? The first equation you give is a vector equation which, in this case, means there are three equations:$$\vec a = \frac{d\vec v}{dt}\Rightarrow a_x = \frac{dv_x}{dt}; a_y = \frac{dv_y}{dt}; a_z = \frac{dv_z}{dt} If we have the additional ...

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There’s a slight issue with some of your wording. Measuring and calculating a quantity are two very distinct tasks - one is experimental in nature, and one lies upon mathematical formalism and theory. That is not to say that they are entirely separate, but it’s worth noting from a pedagogical standpoint. Fundamentally, the answer to your question relies on ...

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