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irinina [24]
3 years ago
6

A chair exerts a force of 20 N on a floor and is not moving. What force does

Physics
1 answer:
AysviL [449]3 years ago
8 0

The floor exerts 20 N of force on the chair

Explanation:

We can answer this question by using Newton's third law, which states that:

<em>"When an object A exerts a force (called action) on an object B, object B exerts an equal and opposite force (called reaction) on object A"</em>

In this problem, we can identify:

- Object A as the chair

- Object B as the floor

This means that the force of 20 N exerted by the chair on the floor is the action, and so the force exerted by the floor on the chair is the reaction. Newton's third law states that these two forces are equal and opposite: therefore, the force exerted by the floor on the chair is also 20 N, but in the opposite direction.

Learn more about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

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3 years ago
Please help. I don’t understand this
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The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>

and under constant acceleration,

<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2

According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2

∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2

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5 0
3 years ago
A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni
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Answer:

The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

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We need to calculate the electric field at the point x = 2 m

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})

Put the value into the formula

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Hence, The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

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