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GarryVolchara [31]
2 years ago
9

How is force useful as well as harmful.Comment with the statement.​

Physics
1 answer:
vladimir1956 [14]2 years ago
7 0

No, we can't say force is useful or harmful. All else being equal, force is either not harmful or useful depending on the body that is applying it and where it is applying it.

Given how it affects motion, force is a crucial idea. It is a relationship that, in the absence of an opposing force, modifies an object's motion. However, a push or a pull that any object feels is the simplest definition of force. Due to the fact that force is a vector quantity, it possesses both a magnitude and a direction.

How force is useful

A body at rest can move with enough force.A body in motion may be slowed down or stopped by it.It has the power to quicken the pace of an object in motion.Along with its shape and size, it can also alter the direction of a moving body.

How force is harmful

  • Force has the power to alter an object's state of motion.
  • Moving objects can shift direction due to force.
  • Moving things' speeds can be increased by force.
  • Moving items can become slower due to force.
  • Force has the power to alter an object's shape.

Learn more about force here

brainly.com/question/14362949

#SPJ9

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Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

This question is about refraction and total internal refraction.

Here I will take refractive index of air and water

n_{air}=1\\ n_{water}=1.33=4/3

Now let's look at the diagram I have attached here

At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.

According to snell's law

\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3

At critical angle B = 90°

\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)

Therefore

A = 48.6^\circ

With this, we can find the radius of the circle (refer to my diagram)

h* tan (A) = R\\R =11.3 m

And with that we can find the area

A = \pi R^2=404\ m^2

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from <em>any</em> source to be traveling at c. This causes light that originated under water, which has the speed of

v_{water} = \frac{c}{n_{water}} = 0.75c

to appear as if it has traveled with the same duration as light with speed c

In order for this to happen our brain perceive shortened length  which is the apparent depth.

To put it in mathematical term

t_{travel}=\frac{h_{apparent}}{v_{water}} =\frac{h}{c}

So we get apparent depth

h_{apparent}=0.75h = 7.5\ m

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