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It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles. So assuming the spins of the electron-positron pair is averaged, and that you don't care about the photon ...

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You have scrambled across one of Zeno's many paradoxes - the so-called Dichotomy paradox. The resolution lies in the fact that sum of terms in an infinite series do not necessarily add up to produce an infinity, so the basic premise is flawed. In particular, the series $$S = \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = \frac{1}{2} + \frac{1}{4} + ... 12 http://wordpress.mrreid.org/2013/12/11/jerk-jounce-snap-crackle-and-pop/ Speaking derivatives to time: First position x, then velocity v=x'=\frac{dx}{dt}, then acceleration a=x''=\frac{d^2x}{dt^2}, then jerk x'''=\frac{d^3x}{dt^3}, then jounce/snap x''''=\frac{d^4x}{dt^4}, then crackle x'''''=\frac{d^5x}{dt^5}, then pop ... 8 The train does accelerate as it goes around a curve. Velocity is a vector, with magnitude and direction. Speed is the magnitude. The train changes direction. Acceleration is caused by a force. If the force causes a change in direction with no change in speed, it must be perpendicular to the direction of velocity. For example, A planet in a circular orbit ... 7 Yes. Rate of change of acceleration is called jerk. Yes its dimensional formula is [M^0, L^1, T^{-3}]. Similarly one could also define higher time derivatives of acceleration if required for a particular problem. 4 If the object was released from rest (u=0\,{\rm m/s}), what is its speed after 1\,{\rm s} if a=10\,{\rm m/s}^2? Using:$$v=u+at$$you will find that the object was not released from rest... 4 Who says all derivatives have to change continuosly? With a suitable setup, acceleration can jump from zero to something. Location and velocity can not change discontinuously, but acceleration can. No, there are no "infinitely many derivatives of velocity". In your question, you write "acceleration has to increase from zero and therefore the third ... 4 Well, if you have 2{\pi}r and you know the time in which you traveled it, then you could find speed. Nothing prevents this. 3 Yes. usually we name them a' . and there can be even a speed of my a' that I can call that a'' and it goes on like that. it is only used it real life calculation that the calculation should be very precise like rocket science. and the equation of displacement (with a constant a') will be :$$x =\frac16 a't^3 + \frac12 a_0t^2 + v_0t+x_0$$(EDIT: ... 3 The only way to purely rotate a rigid body about its center of mass is to apply a pure torque (no net force). If the net force applied is zero then the center of mass is not accelerating. However and combination of translation and rotation of the center of mass can be viewed as a pure rotation about the instant center of rotation. So to effectively answer ... 3 First of all, you need to be careful when saying the train does not accelerate. In Physics, acceleration is the rate of change of velocity. As velocity is vector, if the velocity of the train changes direction, as it does on the curved portion of the track, then the velocity is changing over time (changing in direction). Therefore, there must be an ... 2 As always, a diagram clears things up quickly: We are going to assume that there is no lift on the arrow (which is wrong - arrows do not fly like regular projectiles but that is not the point of this question). From thie diagram we can see the approach we need to take (I am not going to show the details of the steps, just give you some direction). the ... 2 I will try to explain this in simple english. Since the number of steps are infinite, then the distance you travel at each step is infinitly short. As such, the time teaken in each step is close to zero but not zero. Therefore the total time in taking this infinite number of steps does not necessary equal to infinite. Mathematically the total time take is ... 2 No. Automobile tires to not expand radially to any great extent - the steel belts will keep that from happening. So, the tire radius still determines how far the car travels per rotation. Now, if your tires are slipping on the road, or are slipping with respect to the rims, than yes you have speedometer problems, but you have lots of other problems as ... 2 It's a very strange question. In theory a ball is thrown in the shape of a parabola. You can get the graph of the thrown ball with this equation:$$y=\frac{-gx^2}{2V^2 \cos^2 (a)} + x\tan a$$where a is the angle at which the ball is thrown, g is the gravitational acceleration, and V is the initial velocity. If you want to calculate the initial ... 2 You must solve for the objects initial velocity first:$$ v(t)=u+at\\ v(0)=u\\ v(1)=u+10\text{m/s}\\ =20\text{m/s}\\ u=10\text{m/s} $$With this adjustment you should find the correct answer. 1 Your question is "Is there an infinite series of higher derivatives of position for this to work?" Answer: No. Acceleration can jump from zero to something. When it does, its derivative is not defined, so the series of position derivatives stops after the second one. From the question: "A change is velocity is acceleration, so the value of the ... 1 As the wheels try to roll they are prevented from rolling by the frictional force acting in the opp. direction.As the traction force exceeds the limiting frictional force the wheel starts rolling forward w.r.t rails. The force tries to induce relative motion between the wheels and rails.As the rails cannot move backwards (due to friction) the wheels have to ... 1 No, the baseplate will not move if the two motors are applying torques in such a way that the rotors remain at rest. To analyse this problem, it is best to consider the forces on the three parts (base disk and two rotors) individually, using free body diagrams. There are two motors, each connecting a rotor to the base plate, and the effect that each motor ... 1 Here is an extremely simple explanation: Force = Mass x Acceleration Force / Mass = Acceleration Mass x Acceleration due to Gravity / Mass = Acceleration Acceleration due to Gravity = Acceleration For further intuition, consider this: The greater the mass, the greater the inertia. The greater the inertia, the greater the difficulty to accelerate the ... 1 If only the forces of gravity are present, all objects fall at the same rate. This is what one calls equivalence principle. In classical mechanics it shows up in the force law for two particles of gravitating mass m_G and M_G, where M_G shall denote the earth's mass.$$ m_i \cdot \vec{a} = -G \cdot \frac{m_G \cdot M_G}{|\vec{r} - \vec{r} '|^2 } \cdot ...

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A nice way to compare both is to invoke the definitions: $${\vec a}_{\rm avg} = \frac{\Delta {\vec v}}{\Delta t}$$ and $${\vec a}_{\rm inst} = \lim_{\Delta t \rightarrow 0}\frac{\Delta {\vec v}}{\Delta t} = \frac{d{\vec{v}}}{dt}$$ Graphically, and if you consider change over an infinitesimal time period $\Delta t \rightarrow 0$, the same definition ...

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Yes, absolutely. This is in some sense parallel to the distance v displacement question in linear motion. It's true that, at the end of it all, the angle difference between the final and initial configuration can't be more than 360 degrees or $2 \pi$ radians or whatever. But the angle that was traveled through--which is really the important thing for your ...

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Yes it would be greater than 360°. Because what you basically do is you calculate $\frac{TraveledDistance}{circumference}$ . In your example this quotient is 0.5, but it can be also larger than 1.

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Complex mass means gravitational mass + i.lambda.higgs mass. The weak coupling constants are all proportional to (higgs) mass with Higgs vacuum value (=246 GeV) as the proportionality constant. Thus mass necessarily becomes a complex number. The real part produces attractive gravity forces and the imaginary part produces repulsive "weak" forces. The factor ...

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An actual example in which there is a non-zero change in acceleration, that is, jerk, occurs is a spring. A spring's motion is described by a sinusoidal function. The derivative of a sinusoidal function is just another sinusoidal function. As a result, you can differentiate such a function infinitely many times, and will never have a derivative that's 0/a ...

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If a object weighs 6 lbs and there are 5 1 lb forces pushing up, then it will not move. The ground pushes up, causing a zero net force.

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Sure, easily. We don't have to invoke any specific force like friction or gravity--just imagine a block at rest on the ground, and forces act either to the right or to the left. I could have four forces each of $F = 1N$ on the right, and one force on the left of $F = 4N$, and that would balance. The question is akin to asking "can we add up five different ...

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No its SPEED will remain same, velocity has a direction, it will change, both while entering and leaving. On the curve, frictional force will balance the centrifugal force. Assuming constant coefficient of friction throughout the track and constant force provided to engine.

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