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3 years ago

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:

"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"

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:

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Answer:

B. Any object that has motion has potential energy, wow any object not in motion light with the potential to do work and kinetic.

Explanation:

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3 years ago

Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,

klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble= $2\hat{i}-3\hat{k}$ N

Position vector= $r=0.5\hat{j}-2\hat{k}$ m

a.We have to find the resulting torque on the pebble about origin.

Torque= $r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque= $-k-1.5i-4j$ N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

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3 years ago

If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star

dangina [55]

1) In the first case, the correct answer is
A.Wavelengths measured would match the actual wavelengths emitted.
In fact, the stars are not moving relative to Earth, so there is no shift in the measured wavelength.

2) In this second case, the correct answer is
A.Wavelengths measured would be shorter than the actual wavelengths emitted.
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