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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 ...


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It seems to me that the key to this trick is to build up enough elastic energy in the rope - which requires you to "build tension" by riding the edge hard, as explained in this video. If you do the trick too close to point A, there is limited lateral motion needed to build tension - but as you take off, the force you are looking for will disappear as the ...


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\begin{equation} T-mg\cos\theta=m\,a_\textrm{centripetal}=m\,\dfrac{v^{2}}{r} \tag{01} \end{equation} At maximum $\:\theta\:$ since $\:v=0\:$ \begin{equation} T-mg\cos\theta=0 \tag{02} \end{equation}


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The concept of "mediation" depending on a photon comes from the use of Feynman diagrams. Feynman diagrams are the calculational tool of quantum electrodynamics because they give a prescription of how to calculate the integrals in each order of the pertubation series expansion for proton proton or proton electron scattering. Lets make it simple, because the ...


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The drag force on an arrow is proportional to the square of the velocity. The equation of motion is simply $\vec F=m\vec a$, or $$\frac{dv}{dt}=-\frac{b}{m}v^2,$$ for an arrow of mass $m$ in the $Ox$ axis. We integrate this equation $$\int_{v_0}^v\frac{dv}{v^2}=-\frac{b}{m}\int_0^t dt,$$ to obtain $$v(t)=\frac{mv_0}{m+v_0bt}.$$ Notice that the velocity goes ...


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That question refers to apparent weight, as opposed to the true weight. Your true weight, $W=mg$, is the force of gravity pulling on you. Your apparent weight is how hard the ground has to push on you to keep you from falling. This is the weight you "feel". When you are accelerating up or down in an elevator while standing on a scale, the scale reads ...


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There are two reasons. One is that the Earth is not perfectly spherically symmetric and so at any point on the Earth's surface, the net gravitational force does not point directly towards the centre of the Earth. Secondly, because the Earth is rotating, there is a pseudo-force, the centrifugal force, that appears to act directly away from the Earth's ...


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First off, there would have to be a force transfer between the road (blue) into the asphalt (black) beneath it for the sensor to trigger. We need to view the road as not infinitely rigid, it's somewhat "squishy", which allows force transfer at and near to the point(s) of contact. The ability for the force at one point to communicate to the other point ...


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Velocity is a vector quantity. It has magnitude as well as direction, so if suddenly all gravitational force disappeared then all planets will move tangentially. It can be proved by a simple experiment: Tie a weight to a string and rotate it and suddenly cut that string without disturbing the system. It will move tangentially. In this experiment tension in ...


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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 ...


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I think that the components must be done in the perpendicular directions $T\cos\theta=mg$ as the bob can't go any further. It is wrong. Because $\Sigma \vec F=\vec 0$ doesn't mean that the bob will be fixed. A body can move because of its inertia even if $\Sigma \vec F=\vec 0$ (first law of Newton).


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The sign of the force direction is just given by convention as $q \vec{v} \times \vec{B}$. Whilst the direction of the force on the particles is clearly something that can be measured, the sign of the charge and the direction of the magnetic field lines are man-made constructs. For example you would get the same direction for the force if we decided that ...


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For a pulley that has mass and moment of inertia, there must be a net torque on the pulley for the pulley to demonstrate an angular acceleration. Assuming that mass "M" is greater than mass "m", a net torque necessarily requires that the counterclockwise torque from mass "M", given by the equation Torque1 = T1(R), is larger than the clockwise torque from ...


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Yes. Tension can vary if external forces are acting between the ends of the string - such as gravity (if the string has mass) and friction where the string makes contact with other objects (such as the pulley). For example, suppose you attach one end A of a uniform massless string to a support and the other end C to a vertically hanging mass M. This ...



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