force of 95.0 N is applied to a 60.0-kg crate on a rough‚ level surface. If the crate accelerates at 1.20 m/s2‚ what is the magnitude of the force of kinetic friction acting on the crate? (a) 23.0 N (b) 45.0 N (c) 16.0 N (d) 33.0 N (e) 8.80 N Newton’s second law gives the net force acting on the crate as This gives the kinetic friction force as ‚ so choice (a) is correct. 2. A 70.0-kg man stands on a pedestal of mass 27.0 kg‚ which rests on a level surface. What is the normal force exerted by
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A hockey stick exerts an average force of 39N on a 0.2kg hockey puck over a displacement of 0.22m. if the hockey puck started from rest‚ what is the final velocity of the puck? Assume no friction. Your physics teacher walking with the aid of a cane approaches a skateboard of 3.5 kg lying on the side walk. Pushing with an angle of 60 degree down from the horizontal with his cane‚ he applies a force of 115N‚ which is enough to toll the skateboard out of his way. Calculate the initial acceleration
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second lab performed on 2/1/12 involved two investigations concerning Maxwell’s wheel. Maxwell’s wheel is an apparatus that consists of a large disk with a long axle. The disk then bound to a support hanging from above with strings attached to each end of the axle. Maxwell’s wheel is considered to be an important apparatus to investigate physical phenomenon’s because it its ability to combine straight line motion and rotation of a rigid body. The two investigations performed during the lab included
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......8 3.4.1.1 Load Varying (or Constant Speed) Method ...............................8 3.4.1.2 Constant Loading Method ..........8 3.4.1.3 Mixed Loading Method...............9 3.4.2 General Remarks ...........................9 3.4.3 Skin friction correction force ........9 3.4.4 Measured Quantities....................10 3.4.5 Shaft Tare Test ............................10 3.4.6 Correction to Measured Forces ..........................................11 3 3.1.2 Installation......
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1) Net velocity = 3 cos 55°= 1.72‚ 3 sin 55°-2 = .46 1.72 + .46 = 1.78 m/s Direction traveled will be 14.97° NW. 2) Total distance from you: √50km2 + 2.1km2 + 1.5km2 = 50.067 km from you. Relation to x-axis: = 90° + arccos(2.1/50.067) = 177.59° Relation to y-axis: = arccos(1.5/50.067) = 88.28° 3) Horozontal distance covered is 0+ 1100 * 0.32 = 352m traveled horizontally Vertical distance traveled is -1/2(-9.8)(.32)2 = .502 m traveled vertically 4) If he is 8m from the pool edge and 20m
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certain gap between the kiln shell‚ can neither too big nor too small.If the clearance is too small‚ the expansion of the kiln shell is limited by a foetus wreath‚ kiln brick is easy to damage.If the clearance is too large‚ kiln relative movement‚ the friction between shell and tyre ring more interested‚ also can make the kiln shell elliptical deformation is more serious. Add the lubricating oil in between. Door can I through the kiln shell and relative motion between the foetus wreath with to estimate
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change in energy. Here‚ the object starts with kinetic energy [KE = (mv^2)/2] and friction brings it to rest by doing an equivalent amount of work. Work is‚ in this case‚ the product of the force of friction and the distance the object slides while coming to rest. Therefore‚ the distance the object slides is directly proportional to the kinetic energy of the object‚ and inversely proportional to the force of friction‚ i.e. d = (mv^2)/(2F). The 2-kg mass is half the mass of the 4-kg mass‚ and from
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Introduction to sheet metal forming processes INTRODUCTION TO SHEET METAL FORMING PROCESSES The documents and related know-how herein provided by SIMTECH subject to contractual conditions are to remain confidential. This documentation and related know-how shall not be disclosed‚ copied or reproduced by any means‚ in whole or in part‚ without the prior written permission of SIMTECH. © 1999 SIMTECH. All rights reserved Product names are mentioned for identification only and may be registered
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The First and Second Conditions for Equilibrium The first condition for equilibrium: The second condition for equilibrium: • • ΣF = 0 ΣΓ = 0 • In when both of these conditions are satisfied in static systems all forces and torques sum to zero. In problems where the first and second conditions of equilibrium are satisfied‚ the best strategy is to create FBD’s for both the first and second conditions‚ derive equations based on these FBD’s and then see what useful information may be gleaned from
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doing 5000J of work in the process. During this time‚ the car moves 25.0m. Neglecting friction between the car and road‚ find (a) v and (b) the horizontal force exerted on the car. 25) A daredevil on a motorcycle leaves the end of a ramp with a speed of 35.0 m/s as in the figure P5.25. If his speed is 33.0 m/s when he reaches the peak of the path‚ what is the maximum height that he reaches? Ignore friction and air resistance. 31) A horizontal spring attached to a wall has a force constant
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