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Natali [406]
4 years ago
10

A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic

friction between the block and the slope is 0.200. what is the acceleration of the block

Physics
1 answer:
Drupady [299]4 years ago
8 0
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
           Fn = (50)(9.81)(cos 16) = 471.5 N

When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:

Fnet = F - μk*Fn - mgsin30° = ma
          250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
          a = 2.84 m/s²

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Vertical distance h isΔg=\frac{-2GM}{ RE^{3} }

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Assuming h is small in comparison to the radius of the Earth, show that the difference in free-fall acceleration between two points separated by vertical distance h isΔg = (2GMeH) / RE³:

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Separated between two points=h

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Now, \frac{dg}{dr} =\frac{-2GM}{ RE^{3} }

dg =\frac{-2GM}{ RE^{3} } dr

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Two small plastic spheres are given positive electrical charges. When they are 15.0 cm apart, the repulsive force between them h
Tems11 [23]

Answer:

a) q_1=q_2= 7.42*10^-7 C

b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C

Explanation:

Given:

F_e = 0.220 N

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part a

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Using coulomb's law:

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Plug in the values and evaluate:

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part b

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Using coulomb's law:

F_e = k*q_1*q_2 / r^2

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