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So you have proper time and proper length. Use rapidity, $\eta$. For proper length $x$ and proper time $\tau$- $\eta=\sinh^{-1}{(\frac{1}{c}{\frac{dx}{d\tau}})}=\tanh^{-1}{(\frac{1}{c}{\frac{dx}{dt}})}$ From this you can calculate the velocity $\equiv dx/dt$

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And that tells us if the velocity speeds up the force will be stronger and the radius well be smaller. Only if the (linear) velocity remains the same between the two cases. This can be very hard to arrange if you are swinging a yo-yo on a string. If may be more helpful in the case of your yo-yo to consider the problem in terms of radius and frequency of ...

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Let's assume that the wheels aren't able to turn. That means the frictional coefficient is kinetic in nature because the car is skidding across the surface...assuming you already overcame static friction. The frictional force is as follows: $$f=u_{k}*m*g$$ You apply a force that opposes and overcomes the kinetic frictional force that is resisting your ...

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The mass of an object in a physics problem doesn't matter to that objects behavior when all the forces that act on it are fractions of the objects weight so that the acceleration has the from $$\vec{a} = \frac{\vec{F}}{m} = \frac{\vec{k}m}{m} = \vec{k} \,,$$ for some vector $k$. This is true of idealized projectiles and idealized objects siding down ramps ...

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Consider what you stated: $$a=v\frac{dv}{dx}$$ Now rewrite it: $$\frac{dv}{dx}=\frac{a}{v}$$ If $v$ is tiny, then you know that $\frac{dv}{dx}$ must be enormous in order to produce the acceleration that you darn well know exists. Accleration at the top of that trajectory is surely $9.8$ $m/s^2$. As $v$ gets smaller and smaller as it gets to the apex, $\... 0 Your question is as follows: Why centripetal force does not increase the value of tangential velocity? Answer: Assume you have circular motion as in the case where a person, with her hand, twirls a ball on a string. The string connects her fingers to the ball as the ball travels in a circle around her hand. In this case, the force in the string is ... 2 I think the OP has made a mistake in applying second law of Newton. The law (about a particle) says: $$\Sigma \vec F=m\vec a$$ As it is seen, this is a vector equation. This means that corresponding components of both side of the equation must be equal. Although it is not said in the law's body, but it is obvious that we must write the equation above with ... 0 The question tells you that the speed is decreasing. It is NOT constant. For the speed to decrease, there has to be a force, and hence an acceleration, acting opposite to the initial direction of motion. This decrease in speed required a -ve acceleration. The question also says that the train turns round a circular bend. The centripetal force required to ... 0 Problem statement says "the train slows down at constant rate", not "at constant velocity" - in fact, if velocity were constant the train wouldn't be slowing. What the statement means is that the train is slowing with a constant acceleration - at least, the tangential component of acceleration is constant. Since velocity is decreasing, the sign of the ... 0 The problem doesn't say the train "is slowing down at a constant velocity." That doesn't even make sense --- either it's slowing down, and the velocity is not constant, or it's not slowing down, and the velocity is constant. It can't be both accelerating and at a constant velocity, since the very definition of acceleration is change in velocity with respect ... 3 Centripetal force is the name of the force that points towards the centre. This is in the radial direction. Tangential velocity is, as the name suggests, a velocity direction tangent to the circle. The radial and tangential directions are by definition always perpendicular - in the same way that the x and y axes are. You are of course right that if any ... 1 1) Usually special relativity is taught at the end of the semester, after the class got through rotations, which they, on average, don't understand; torques, which are pseudo vectors and for this reason blow their minds up... By the time they get to relativity they are done! If you teach the same kind of course, it is unavoidable, that they will get confused.... 13 Your error is simply that you are assuming that$v(x)$is differentiable with respect to$x$at$v=0$. The chain rule needs that all derivatives involved exist before you can apply it. In the case of just letting go of something, the function$v(x) = \sqrt{2gx}$is not differentiable at$x=0$, which is where$v=0$, so you are not allowed to apply the chain ... 3 When an object starts at rest, the change in velocity when it has made an infinitesimal displacement is infinite - in other words,$\frac{dv}{dx}$is undefined. You can see this most easily by plotting the curve of$v$as a function of$x$for an object starting at rest: $$x = \frac12 a t^2\\ v = at\\ x = \frac12 a \left(\frac{v}{a}\right)^2\\ v = \sqrt{... -1 This equation is best understood in integral form$$ \left. {\rm d}x = \frac{v}{a}\,{\rm d}v \right\} x_2-x_1 = \int \limits_{v_1}^{v_2} \frac{v}{a}\,{\rm d}v $$It gives you the distance traveled by a varying acceleration between two speeds. "A car accelerating from 0 to 60 mph needs X distance". By stating that v=0 always not only it implies that a=0... 1 The kinematic equations are derived from differential equations. This means that you can start with the acceleration, then move to velocity, and then position given that the necessary information is available (like initial velocity or position). I find that understanding this helps me visualize the connection between the three. You can not rely on the ... 2 Juno probe's speed is in relation to what frame of reference? The escape speed of Jupiter is ~59.5 km/s and 150,000 kph ~ 67.1 km/s, so this speed must be in reference to the sun otherwise the spacecraft would not stay in orbit. Jupiter's orbital speed about the sun is ~13.1 km/s, which subtracted from the 67.1 km/s would result in ~54 km/s, thus more ... 1 Without seeing the particular example you have in mind it's impossible to be definitive, but without any context, for a probe in the Solar System, I'd assume heliocentric coordinates. Unless it's an orbital velocity around some other body, or an escape velocity from some other body, or of course if it's specifically stated to be in some other frame. 2 I would call this quantity "first moment of mass" or just "moment of mass". Have a look at this wikipedia article to read about the general concept of moments in physics. As pointed out in the comments to your question, the first moment of mass is closely related to the center of mass (CoM). For a collection of particles with masses m_i and positions \... 0 No. Acceleration and velocity have different units, so their magnitudes cannot be compared. Whether one is larger than the other numerically depends on what units are used. Acceleration and velocity can have the same direction, but this is not necessary - eg a ball thrown upwards has upward velocity but downward acceleration, until it reaches maximum ... 1 Gravity acceleration is... acceleration, measured in \mathrm{m/s^2}. It is the rate of change of the velocity.. Velocity is measured in \rm m/s. It is the rate of change of the position. The vertical velocity and the acceleration due to gravity of a body are collinear but they can have different magnitudes as well different orientations. Think about a ... 2 It is true that the weight will pull the machine to the north when it points north because the machine is pulling the weight to the south; by Newton's Third Law, that means the weight is pulling the machine northward. However, you have to think about what happens during the rest of the rotation. To have a faster velocity while the weight points north, the ... 1 Proper time is the reference time all observers can agree on. Makes for calculating things much much easier. Since tau is attached to the particle the space differentials are zero Imagine we used some other reference time (which you are more than welcome to do). Then the spacetime line element would have dx, dy, and dz not equal to zero. Furthermore ... 1 4-vectors have invariant length as defined by$$\vec{v}\cdot \vec{v} = g_{ab}v^a v^b.$$Coordinate velocity \frac{\mathrm{d}\vec{x}}{\mathrm{d}t} does not have this property. Proper velocity \frac{\mathrm{d}\vec{x}}{\mathrm{d}\tau} does. Coordinate velocity is defined. It's not a 4-vector, so it's not that useful. 0 Given a path x(s) on a manifold, the velocity with respect to that path is defined taking derivatives with respect to the invariant parameter the path is described with, on the manifold (the arc length). In general relativity such parameter is the proper length s (or proper time \tau = \gamma s). 1 The thing about the relativity principle is just that: all non-accelerating frames are equal, in the sense that no inertial frame is more "real" or "accurate" than an other inertial frame. If two frames are moving in a Minkowskian manifold with a constant velocity relative to each other, it doesn't matter which one you choose. No matter what, all the ... 0 Hint: Group velocity is a property of a linear superposition of plane wave solutions of the form e^{ik\cdot x-i\omega t}, with dispersion relation \omega(k) that relates frequency to wave number. From this, you can assume \psi(x,t)=\int\frac{d^3k}{(2\pi)^3}g(k)e^{ik\cdot x-i\omega(k)t} for some initial (normalized) wave function g(k). For your ... 1 As the force F is constant, and we know that the force of weight mg is constant too; we can say the net force acting on the ball is constant whole the motion. But, from t=0 to t=t_1 the net force is F-mg and after t=t_1 the net force is -mg. So, velocity graph will be linear but it has a breaking at t=t_1. For 0\le t\lt t_1 the gradient of ... 0 You have a few questions here. They mostly concern the energy loss due to friction. In this problem there is no energy loss because of friction. Friction is present between the rope and the pulley. This is static friction. It makes the pulley rotate. Energy is lost due to friction (dissipated as heat) only when there is relative motion between the ... 3 The equations of motion for the position determine the accelerations: they are second-order differential equations in time:$$\vec F = m\vec a = m\ddot{\vec x }$$So the acceleration, the second derivative of the location in time, has to be determined from the state of the physical system in some way. Typically, it is determined using the$F=ma$formula ... 0 Imagine a body moving with velocity$\vec{v}$in the$\hat{x}$direction and now imagine having it losing speed (decelerating, so$\vec{a} \propto -\vec{v}$) the friction force should be opposite the$\vec{a}$(acceleration) or the$\vec{v}\$ (velocity)? if you think carefully about it, you'll convince yourself that it should oppose velocity and not ...

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