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PHY–302 K. Solutions for Problem set # 8. Textbook problem 7.16
PHY–302 K. Solutions for Problem set # 8. Textbook problem 7.16

... Thus, while the man walks 12 feet from the stern to the prow, the boat moves back 8 feet (i.e., 8 feet away from the pier). And by the way, the man’s displacement relatively to the pier is only 12 ft − 8 ft = +4 ft. Textbook problem 8.12: (a) For constant angular acceleration α, the angular velocity ...
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BLACKBOARD COURSE PHYSICS 1.2. PHYS 1433

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Conservative Forces and Potential Energy

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... These can be used with vmax = w A = ...
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... 27. A block of mass 'm' is connected to one end of a spring of 'spring constant' k. The other end of the spring is fixed to a rigid support. The mass is released slowly so that the total energy of the system is then constituted by only the potential energy, then’d’ is the maximum extension of the sp ...
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Newton`s Laws of Motion

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... • What does it take to stay on the road around a curve? – using s = 0.8 as average for tires on road, Ffriction = 0.8mg • (Normal force is just mg on level surface) ...
Friction and Gravity
Friction and Gravity

... An object in free fall accelerates as it falls. In a free fall, the force of gravity is an unbalanced force, and an unbalanced force cause an object to accelerate. Even though it seems hard to believe, ALL objects in free fall accelerate at the same rate regardless of mass!!! ...
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... The force probe should be mounted on the bracket attached to the cart. Open the file ForceVelAcc. Zero the force probe by clicking on the Zero button. The fan motor should be off. Your hand will be the only horizontal force on the cart. Click the Collect button. Hold on to the hook attached to the f ...
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... (b) Calculate the speed of the glider in m.s –1 when it is 3.0 cm from equilibrium. (c) Derive an expression for the acceleration a of the glider in terms of its instantaneous position x. (d) What is the acceleration of the glider in m.s–2 when it passes through x = 0? Reminder: The effective k-valu ...
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... Ignoring friction y = v0yt – 1/2gt2 t = x/v0x , v0y/v0x = h/d at x = d y = h – 1/2gt2 In the same time the monkey falls 1/2gt2 So the bullet always hits the monkey no matter what the value of v0 ...
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... shifting focus because of the different orbital periods. In reality, most planetary orbits follow very close to a circle with the two foci very near each other. For most problems, we can assume the two elliptical foci are close enough to be identical, and that planetary orbits approximate a circular ...
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... Assuming the frictional forces to be unchanged, calculate: (i) the new engine force (ii) the force exerted by the tow bar on the caravan. The car brakes and decelerates at 5·0 m s 2 . Calculate the force exerted by the brakes (assume the other frictional forces remain constant). ...
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... Example: Object rotating on a string of changing length. A small mass m attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v1 = 2.4 m/s in a circle of radius R1 = ...
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... Use Excel to plot Fk as a function of Fn and fit the data to a straight line to determine the slope. Open Excel, on column A insert the value of Fn and on column B insert Fk (Excel plots the first column on the x-axis and the second column on the y-axis). Using left-click on the mouse, select all yo ...
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Centripetal force

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