small block with mass 0.0400 kg is moving in the y-plane. The net force on the block is described by the otential-energy function (x, y) = (5.70 J/m²)z² – (3.55 J/m³)y³.
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A: Potential energy function is given by : U(x,y) = 3x6y - 8x To find : Force at the point (x,y)
Q: A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described…
A: Given data: Mass of the block, m=0.04 kg Potential energy function, Ux,y=5.8 J/m2x2-3.6 J/m2y3…
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Q: A particle has a function for Potential Energy as shown below. U = (7.30 J/m1) x1 + (9.40 J/m4) x4…
A: Given: U=7.30 Jmx+9.40 Jm4x4 The expression for force (F) on the particle is: F=-dUdx Substitute all…
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A: It is given that, F=167 Nx=7.2 cm=0.072 m
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A: as per the question:hereforce between m and M,F=kr4
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Q: A potential energy function for a system in which a two- dimensional force acts is of the form U=…
A: x-component of the force Fx=-∂U∂x=-∂3x3y-7x∂x=-∂3x3y∂x+∂7x∂x…
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A: Given that The figure here shows a plot of potential energy U versus position x of a 0.898 kg…
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A: The vector force acting on system kept in a potential field U is given by- F→=-∂U∂xi^-∂U∂yj^
Q: 20 - Bolum-7+A2B1C3uUxiD5ViBM7s27Vi20FXF231G A2.00 kg particle has a speed of 4.00 m/s at point A…
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A: Given data as per question U = 5.50x2 - 3.60y3 x = 0.22 y = 0.62
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Q: 9- Bolum-7+A2B1C3uUxiD5VIBM7s27V120FXF231G A 2.00 kg particle has a speed of 4.00 m/s at point A and…
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A: Given- Mass of the particle m= 0.888 kg UA=15 JUB=35 JUC=45 JThe particle is released at x=4.50…
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Q: The figure here shows a plot of potential energy U versus position x of a 0.876 kg particle that can…
A: a) At x=4.50, UA=15.0 JAt x=1.00 mUB=35.0 JInitial velocity at x=4.50 mvi=7.96 m/smass m=0.876…
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