A 12.0 Kg object is dragged across a 5.0 m long horizontal?
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http://answers.yahoo.com/question/index;_ylt=AkhKDjAz8JvL7oBmULlsYTW7zKIX?qid=1006020111393
824936
a) You have to draw gravity as a straight line down. Then you have to realize that gravity does no work if the block was on a horizontal surface. Work done by gravity is alway M*G*H.  The mass is 4.8 kg and the gravity is 9.81 m/s^2 .  The height will be sin (30) = H/2.  the sin (30 degrees) is 0.5 so .5=H/2.  the height will be .5/2 = .25 m.\n\nsoo you will have 4.8*9.81*0.25\n\nb) the work done by the frictional force will always be the frictional force times the distance.  The frictional force is going to be the normal force times Uk.  The normal force has components to it, so its going to be sin(30)*g*m.  The you need to multiply that by the friction coefficient and then by the distance traveled.  You will have 2*0.436*sin(30)*4.8*9.8\n\nc.  The work done by the normal force is just going to be cos(30) (i think) times 2 ( the distance traveled) times 9.81 (gravity) times 4.8 (the mass)
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T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
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5 meters from any side
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50 Km is the answer\n\nIf the car moved 50 km north and then 50 south its displacement would be 0\n\nif it go in a direction like sw or ne or nw then you have to use a triangle to solve its displacement
274972
a foot is equal to 12 inches.
511118
gravity draws everything towards the centre
643953
depends on the direction of the force
783086
Gravity is pulling down on the object, air is resisting it (pushing up, if you will) and causing friction, and if there's any wind it's pushing sideways on the object. I doubt you'd have wind in your problem, though. Wind is complicated :)
222948
Just drag it upwards.
430468
First of all the distance is 20 METERS and not 20 METERS/SECOND. Also the unit of acceleration is m/sec*sec and not m/sec.\n\nNow, there are both horizontal and vertical speeds. Also the position is a vector that has both a horizontal factor and a vertical factor.\n\nTo find the horizontal speed:\nInitial speed: 0 m/sec.\nAcceleration: 5m/s^2\n=> Speed2 = Speed1 + Acc.* Time\n=> After 1 sec. speed= 0 + 1*5=5m/s\n=>After 2 secinds speed = 0 + 2*5= 10m/sec.\n\nTo find the vertical speed:\nInitial speed: 0 m/sec.\nAcceleration: -9.8m/s^2 [acceleration due to gravity]\n=> Speed2 = Speed1 + Acc.* Time\n=> After 1 sec. speed= 0 - 1*9.8=-9.8m/s\n=>After 2 seconds speed = 0 - 2*9.8= -19.6m/sec.\n\nTo find the horizontal position:\nDisplacement = (initial speed) * time + 0.5*acc.*time^2\n=> after 1 sec. displacement = 0 + 0.5*5 * 1 = 2.5 m\n=> position = 2.5 meters to the right of the initial horizontal position of the balloon\n\n=> after 2 seconds displacement = 0 + 0.5*5* 4 = 10 m\n=> position = 10 meters to the right of the initial horizontal position of the balloon\n\n\nTo find the vertical position:\nDisplacement = (initial speed) * time + 0.5*acc.*time^2\n=> after 1 sec. displacement = 0 + 0.5*-9.8 * 1 = -4.9 m\n=> position = 20[initial position] - 4.9 = 15.1 meters above the ground\n\n=> after 2 seconds displacement = 0 + 0.5*-9.8 * 4 = -19.6 m\n=> position = 20[initial position] - 19.6 = 0.4 meters above the ground
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Final Examination Physics 104 19 May 1977 Attention: in this ...\nThe pipe turns vertically upwards as shown in the sketch; the vertical part has a smaller ... Dick, together with his suit, has a mass of 175 kg, and Jane, ...\nwww.iit.edu/~johnsonp/exams104/final190577.html - 31k - Cached - Similar pages\nNewton's Laws Applications Review - Q#13-Q#21 Answers\nSince there are two forces pulling upwards and since the sign is hanging ... (0.250 kg)•ax = Ftens - 0.25 N. Analyzing the Fy forces on the 50-g mass yields ...\ngbs.glenbrook.k12.il.us/academics/gbssci/phys/chemphys/reviews/u8breview/u8ans2.html - 28k - Cached - Similar pages\nNewton's Laws Review\nIf a football is moving upwards and rightwards towards the peak of its ... A 0.250-kg rightward moving air track glider decreases its speed from 0.872 m/s ...\ngbs.glenbrook.k12.il.us/Academics/gbssci/phys/chemphys/reviews/u8areview/u8aprint.html - 32k - Cached - Similar pages\n[ More results from gbs.glenbrook.k12.il.us ]\n[PDF]\nPHYS 1002 Physics 1 (Fundamentals) – Semester 1, 2006\nFile Format: PDF/Adobe Acrobat - View as HTML\nof mass, torque and static equilibrium. You will also learn the concept of momentum, energy, ... 0.250 kg, m. 2. = 0.375 kg, g = 9.81 ms ...\nwww.physics.usyd.edu.au/ugrad/jphys/1002_mod2_mech.pdf - Similar pages\nThe Physics Classroom\nNote that the vertical forces balance but the horizontal, forces do not. The net force is 129.9 N, right - 60 N, left. The mass is 20 kg (Fgrav/g); ...\nwww.physicsclassroom.com/Class/vectors/U3L3d.html - 31k - Cached - Similar pages\n[PDF]\n∑ ∑\nFile Format: PDF/Adobe Acrobat - View as HTML\nvalue in the upwards direction. (b) For no vertical acceleration we have: ... 6 A ball having a mass of 0.250 Kg is held at point A against a spring which ...\nwww.physics.villanova.edu/faculty/shaw/2400/tests/s97all4.pdf - Similar pages\n[PDF]\nPhysics 2400 Test #2 (8:30 Class) Spring 2002\nFile Format: PDF/Adobe Acrobat - View as HTML\n3 (a) A boy's toy cannon fires a 0.250 (kg) rubber ball vertically. ... When the cannon is fired, compressed gas below the ball exerts an upwards force, F ...\nwww.physics.villanova.edu/faculty/shaw/2400/tests/s02all2.pdf
572487
Distribution of force over area.
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the amount of pressure exerted on an object
611971
Mass is a constant; weight varies by gravitational pull.
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Lift, thrust, gravity, drag.
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Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons