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Practice test Midterm 2-1_Chapter 7
Practice test Midterm 2-1_Chapter 7

... The rigid body shown rotates about an axis through its center of mass and perpendicular to the paper. If M = 2.0 kg and L = 80 cm, what is the kinetic energy of this object when its angular speed about this axis is equal to 5.0 rad/s? Neglect the mass of the connecting rod and treat the masses as pa ...
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... When things change momentum, most of the time the object is changing its Velocity, and not its mass. Think about it… while walking down the hallway, if Wanted to change your momentum, would you change your mass or your speed? So… If you change your velocity, you are ACCELERATING – if you are acceler ...
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... Mechanical energy = KE + PE = mv2/2 + mgy In the absence of applied forces and friction: (that is, non-conservative forces) (change in KE) + (change in PE) = 0 so mechanical energy is conserved. KE + PE = constant When an applied force does work, the work-energy theorem becomes: W = ΔKE + ΔPE Monday ...
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Physics@Brock - Brock University

... 42. Inertia is (a) the tendency of a body to remain at rest or moving in a straight line at constant speed. (b) the tendency of a body not to move. (c) the tendency of gases such as neon and argon not to enter into chemical reactions. (d) the tendency of a body to remain moving in a circle at a cons ...
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PHYSICS 12 Centripetal Acceleration/Centripetal Force

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... carrying the 10 kg brick, at a constant speed. How much work did they do? What is the equation for the change in PE of the brick? Change in PE = PE2 – PE1 = (mgh2 – mgh1) But h2 = h1, so no PE is added or change in PE = 0. So no work is done because no potential energy is added to the package. (Agai ...
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... Draw the weight vector also! Now, what is the net Force? Net force = +19 N – 5 N – 20 N = Net Force = - 6 N What is the acceleration? a = Fnet / m = a=-6/2 a = -3 m/s2 It will accelerate downward. ...
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... Now that you have deduced the force function F(x) from the motion data x(t), you can check your result by directly measuring F(x) using a spring scale. This may seem like an easy method to find the force, but remember, someone had to calibrate the force meter to convert a length measurement (stretch ...
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PHY 105 (Module 2) March 30 * April 13, 5 hours

... Examples for Discussion A 921-kg sports car is moving rightward with a speed of 29.0 m/s. The driver suddenly slams on the brakes and the car skids to a stop over the course of 3.20 seconds with the wheels locked. Determine the average resistive force acting upon the car. [F = d(mv)/dt] A 1250-kg s ...
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Classical central-force problem



In classical mechanics, the central-force problem is to determine the motion of a particle under the influence of a single central force. A central force is a force that points from the particle directly towards (or directly away from) a fixed point in space, the center, and whose magnitude only depends on the distance of the object to the center. In many important cases, the problem can be solved analytically, i.e., in terms of well-studied functions such as trigonometric functions.The solution of this problem is important to classical physics, since many naturally occurring forces are central. Examples include gravity and electromagnetism as described by Newton's law of universal gravitation and Coulomb's law, respectively. The problem is also important because some more complicated problems in classical physics (such as the two-body problem with forces along the line connecting the two bodies) can be reduced to a central-force problem. Finally, the solution to the central-force problem often makes a good initial approximation of the true motion, as in calculating the motion of the planets in the Solar System.
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