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2 years ago

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A custodian pushes a 119 kg box across a level floor. The box moves at a constant velocity. The custodian pushes horizontally on

the box with a force of 724 Newtons, what must be the friction force experienced by the box in Newtons?

1 answer:

bonufazy [111]2 years ago

8 0

The friction force experienced by the box in Newtons is 724N

The frictional force is the force that opposes the moving force acting on a body on a horizontal plane.

For a static body, the acceleration will be zero,

According to Newton's second law of motion:

∑Fx = ma
+Fm - Ff = ma

Given the following paramters

Moving force Fm = 724N

acceleration a = 0m/s²

mass m = 119 kg

Substitute the given parameters into the formula to get the frictional force:

724 - Ff = 119(0)

724 - Ff = 0

Ff = 0 + 724

Ff = 724N

Hence the friction force experienced by the box in Newtons is 724N

Learn more here: brainly.com/question/8859573

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Answer:

B) $\frac{2}{3}$

Explanation:

The electric force between charges can be determined by;

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Let the original charge be represented by q, so that;

$q_{1}$ = 2q

$q_{2}$ = $\frac{q}{3}$

So that,

F = $q_{1}$ $q_{2}$ x $\frac{k}{r^{2} }$

= 2q x $\frac{q}{3}$ x $\frac{k}{r^{2} }$

= $\frac{2q^{2} }{3}$ x $\frac{k}{r^{2} }$

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F = $\frac{2}{3}$ x $\frac{kq}{r^{2} }$

The electric force between the given charges would change by $\frac{2}{3}$ .

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3 years ago

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3 years ago

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Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

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where $r$ is the radius of curvature.

Now

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Thus, the electric field at the center of the curvature of the arc is:

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$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

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and this is

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