course phy 201
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23:56:53 The displacement between any two clock times is equal to the area of the trapezoid between these two clock times. This area is equal to the product of the average altitude of the trapezoid and its width, which gives us the first basic equation of uniformly accelerated motion: `ds = (v0 + vf) / 2 * `dt. Under the same conditions the slope of the graph will be the acceleration a, and the change in velocity between two clock times will be represented by the rise of a slope triangle who slope is a and whose run is the time interval `dt between the two clock times. The velocity change will therefore be `dv = a * `dt, and the final velocity will be equal to the initial velocity plus this change: vf = v0 + a `dt. This is the second basic equation of motion.
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RESPONSE --> the area between two interval is the average height and width which is really 'ds=(vf+v0)/2*'dt the slope a='dv/'dt 'dv=a*'dt
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23:57:41 When a wound-up friction car is released on the level surface, what do we see as a result of the potential energy conversion?
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RESPONSE --> the winding is PE when release become KE
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23:57:53 Most of the work done to wind the spring goes into the elastic potential energy of the spring. When it is released, most of this energy is then converted to kinetic energy. In each part of the process there is some friction loss in the mechanism.
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RESPONSE --> right and some if lost through friction
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23:59:10 When a wound-of friction car is released up a ramp, what to we see as a result of the potential energy conversion?
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RESPONSE --> KE increase and PE increases because gravity is acting on it.
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23:59:28 The car climbs the ramp, increasing its gravitational PE. The car also speeds up, increasing its KE. Some of the energy is dissipated in the form of thermal energy as a result of friction.
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RESPONSE --> there is thermal energy as a result of friction
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00:00:20 What forces act on an object sliding up or down an incline?
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RESPONSE --> Normal force, gravity, friction most of the time
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00:00:38 A normal force will be exerted by the ramp. Gravity acts on the object; usually the resulting weight is expressed as two components, one parallel and one perpendicular to the incline. There is also a frictional force acting in the direction opposite to the motion of the object.
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RESPONSE --> yes the gravity is in two parallel and perpendicular
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00:02:22 What energy changes take place as an object slides up or down an incline?
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RESPONSE --> `dW = w sin(`theta) * `ds `dW = w * `dy
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00:02:29 The parallel component of the weight is wParallel = w sin(`theta), and to slide the object through displacement `ds at a constant velocity requires a force equal and opposite to this component. As a result work `dW = w sin(`theta) * `ds, if the positive direction is chosen as up the incline. If we simply raise the object vertically through a distance equal to its vertical rise, we will have to exert a force equal in magnitude to its weight. The vertical distance through which we lift the object will be `dy = `ds sin(`theta), so the work done will be `dW = w * `dy = w * (`ds sin(`theta)). It should be clear that this is the same work contribution found before.
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RESPONSE --> right
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00:04:36 If we know the mass and length of a pendulum, how can we determine the force required to displaced pendulum a given small distance from equilibrium?
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RESPONSE --> not sure
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00:05:14 When a pendulum is displaced a certain distance from its equilibrium, the forces on the pendulum will be in equilibrium if the tension force directed along the pendulum string has a vertical component equal to the weight and a horizontal component equal to F. Therefore, we will have equilibrium if the ratio of F to the weight is the same as the ratio of the displacement x to the length L. That is, F / mg = x / L. It follows that F = mg * (x / L), which is the weight of the pendulum multiplied by the ratio x / L of displacement to length.
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RESPONSE --> so if you were to change the ration os F/mg=x/L then you would have a change in equilibrium
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