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In order to move through a concave path, an agent has to impart force to otherwise a linearly-moving object. The object , by virtue of its motion, under the absence of any external force, always travels or tends to travel in the direction of the velocity vector at the concerned instant. So, when the object has to transverse a curve trajectory, the main ...


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Intuitively, You need a tension towards the center of curvature (if you image a weigtless rope connected to the particle instead of following a path). The same reason when you turn in a bike you mostly lean in the curve. Otherwise you would be flung out and not be turning the way you want. It actually helps to think about it the other way. When a moving ...


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Cicero gave you the math; let me give you the picture. Looking at this: The blue blob is an object initially moving horizontally to the right. At point B I give it a push upwards; it will then follow a new trajectory. You probably have no difficulty determining that the mass had to accelerate vertically to get its new direction. Now make the steps ...


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Because the non-tangential component of the acceleration always points toward the concave side. This is a mathematical result, with the proofs given, but to provide physical intuiton consider the non-tangential component of acceleration. This component doesn't affect the magntitude of the velocity vector, but changes its direction in a circular fashion. ...


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2 points: first, $F=ma$ describes the acceleration of an object due to the sum of all forces acting on the object. If these forces are in different directions, they may partly or fully cancel each other out. In the case where the object is not accelerating (so it's moving with constant speed in a constant direction, or it's not moving at all), the sum of ...


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Although an object that moves with constant velocity has no acceleration, it has kinetic energy and it has momentum. Acceleration is not a conserved quantity. It is not passed from one object to another. Momentum and energy, however, are conserved quantities that pass from one object to another. If a moving object hits a target, kinetic energy will be ...


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$\vec{F} = m\vec{a}$ means that an object with a force $\vec{F}$ exerted upon it accelerates by an amount $\vec{a}$, not that an object accelerating with $\vec{a}$ exerts a force $\vec{F}$ on something else. Typically, the force exerted by an object has nothing to do with Newton's second law, but is given by other laws (like Coulomb's law in electrostatics). ...


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The definition of acceleration, $a = \frac{dv}{dt}$ is always correct, and any other equations of acceleration are derived from it, depending on the conditions. To use this equation, you need to have a basic understanding of calculus. For example: condition: constant acceleration. This means $\frac{dv}{dt}$ is constant. This means that $v = at + constant$ ...


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When the slope is constant (left part of the drawing with the skier, flatness in the sled example) any possible acceleration is tangent to the surface. Thus, when the sled is speeding along perfectly level and flat ground, it experiences no acceleration. When the skier is going downhill, he experiences only acceleration along the slope. But, as already ...


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In your context, the second interpretation is correct. The fact is that falling objects accelerate both on Earth and on the Moon. The sentence is saying that the amount of this acceleration, regardless the source, is six times greater on Earth than on the Moon. In other words, things accelerate towards the surface of Earth six times faster than they ...


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The explanation comes from earlier in that paragraph: If all the co-ordinates and velocities are simultaneously specified, it is known from experience that the state of the system is completely determined and that its subsequent motion can, in principle, be calculated. This is just saying the familiar thing that if you know the laws of physics for the ...


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The formula you've referenced, $\vec a_c = -\omega^2\vec r$, must define $\vec r$ as a function of the angular position, which depends on time (uniform motion and all). Something like $\vec r(\omega t) = r(\hat x \sin \omega t + \hat y \cos \omega t)$. $\omega$ essentially represents the angular speed, which means for constant $\omega$ you have an orbital ...


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The net force on the 5kg block will be 15N in the direction of the hand pushing. the subsequent acceleration or the 5kg block will be 3m per second per second. the net force on the 10kg block will be 30N giving an acceleration 3m per second per second. The accelleration will continue until the the hand is removed resulting in the blocks continuing at a ...


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First We'll see the FBD (Free Body Diagram)and we get: Where Fc is normal force acting (Force of contact). From FBD of 5 Kg block (By newton's 2nd law) $$F-F_c = ma$$ (1) From FBD of 10 Kg block $$F_c=Ma$$ Solving above equations we will get: $$F=ma+Ma$$ $$a=F/(m+M)$$ Putting the values you may get your result and your resultant force.


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If you're talking about a real car, the answer generally is that it takes longer to accelerate at higher speeds... but that's because of the gears, not the forces at work. If you're just talking about acceleration, at least at speeds low enough to ignore relativistic effects, then any acceleration that produces a certain change of speed will take the same ...


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You can't extrapolate like this and the function may be neither linear nor logarithmic. The car is subject to frictional forces from air-resistance, rolling-resistance, gear-train resistance, resistive forces inside the engine and other resistive forces. Each of these depends not only on speed but also on other factors like shape, selected gear, engine RPM, ...



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