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2 years ago

A man pushes a box across a floor. as he increases the force he applies horizontally to the box

2 answers:

6 0

When a man pushes a box across a floor, as he increases the force he applies horizontally to the box, then the kinetic friction remains the same.

To find the answer, we need to know about friction.

<h3>What is friction?</h3>

Whenever a body moves, over the surface of another body, a force comes into play which acts parallel to the surface of contact and opposes the relative motion. This opposing force is called friction.
There are two frictions acting on a surface.
Static friction: It come into play between two bodies before when it starts moving. Self-adjusting force and the max value of it is called limiting static friction.
Kinetic friction: It come into play between two bodies when it starts moving. It is not a self-adjusting force.
As we increase applied force, kinetic friction will remain same.

Thus, we can conclude that, kinetic friction remains the same when we increase the applied force.

Learn more about the friction here:

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marshall27 [118]2 years ago

3 0

As a guy pushes a box across a floor, the kinetic friction stays constant no matter how much force is applied on the box horizontally.

We must understand friction if we are to find the solution.

<h3>Describe friction?</h3>

A force that works parallel to the surface of contact and opposes the relative motion is always present whenever a body moves over the surface of another body. Friction is the name for this opposing force.

On a surface, there are two frictions operating.

Static friction: It occurs between two bodies before motion is initiated. Limiting static friction is the term used to describe the maximum amount of self-adjusting force.

As soon as two bodies begin to move, kinetic friction is triggered. It is not a force that adjusts itself.

The amount of kinetic friction will not change when we apply more force.

We can therefore draw the conclusion that kinetic friction does not change as the applied force is increased.

Learn more about the friction here:

brainly.com/question/28048527

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2 years ago

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3 years ago

Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two

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Answer:

197.5072.

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force of interaction between two charges $\rm q_1$ and $\rm q_2$ which are separated by the distance $\rm d$ is given by

$\rm F = \dfrac{kq_1q_2}{d^2}.$

<em>where,</em> k is the Coulomb's constant.

For the case, when,

$\rm q_1 = Q.$
$\rm q_2 = Q.$
$\rm d=r.$
$\rm F=12.3442.$

Then, using Coulomb's law,

$\rm 12.3442 = \dfrac{kQQ}{r^2}=\dfrac{kQ^2}{r^2}\ \ \ \ .......\ (1).$

For the case, when,

$\rm q_1 = 2Q.$
$\rm q_2 = 2Q.$
$\rm d=\dfrac r2.$

Then, using Coulomb's law, the new electric force between the charges is given by,

$\rm F' = \dfrac{k(2Q)(2Q)}{\left (\dfrac r2\right )^2}\\=\dfrac{k\ 4Q^2}{\dfrac{r^2}{4}}\\=4\times 4 \times \dfrac{kQ^2}{r^2}\\=16\ \dfrac{kQ^2}{r^2}\\=16\times 12.3442\ \ \ \ \ \ \ \ (Using\ (1))\\=197.5072.$

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3 years ago

A person jogs at an average velocity of 3 m/s West. How long will it take this person to run a 30-meter city block?

grigory [225]

Answer:

It will take the person 10 seconds to run a 30-meter city block.

Explanation:

To solve this problem, we must remember the formula for velocity:

velocity = displacement/time

If we plug in the values that we are given, we get:

3 m/s = 30 m/time

We can see from the above equation that our answer should be in seconds, so we can get rid of the units to simplify the equation slightly.

3 = 30/t

Next, we can multiply both sides by t:

3t = 30

Then, we should divide both sides of the equation by 3 to get the variable t alone on the left side of the equation.

t = 10

Therefore, the correct answer is 10 s.

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3 years ago

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