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Alex
3 years ago
13

Calculate the work done by an applied force of 71.5 N on a crate for the following. (Include the sign of the value in your answe

r.) (a) The force is exerted horizontally while pushing the crate 4.30 m.
Physics
1 answer:
Nastasia [14]3 years ago
8 0

Answer:

W = 307.45 joules

Explanation:

the work done is defined as the product of force applied in the direction of displacement and the displacement.

Use the formula for work directly

(work) = (force) x (displacement)

Work is a scalar quantity and its SI unit is Joule.

W = F x d

F = 71.5 N

d = 4.30 m

W = 71.5 × 4.30

W = 307.45 joules

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The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

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where

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Since the block is in equilibrium, we can write

mg sin \theta - F_f = 0

And substituting, we find the force of friction:

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3 years ago
A stationary charge is located between the poles of a horseshoe magnet. Is a magnetic force excerted of the charge? why?
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A stationary charge is located between the poles of a horseshoe magnet. The magnetic force exerted by the charge is zero.

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You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball
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Explanation:

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In this case, the ball landed at the coordinates \left(x,y\right)=\left(61,0\right), so

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Answer:

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