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Rainbow [258]
3 years ago
11

Explain what happens to a light wave that hits the surface of a pool.

Physics
1 answer:
brilliants [131]3 years ago
5 0

A light wave that hits the surface of a pool gets refracted and gives us an apparent image of the surface of the pool, following the concepts of refraction.

<u>Explanation:</u>

Let’s recall the concept of refraction when a light wave passes from medium of rarer to denser. There is a change in the speed of light while travelling from medium of rarer to denser.

There can be a change in the direction as well. This property is known as “Refraction” and the best example to see refraction is watching the surface of a clean pond, lake or pool.

When the light travels from a rarer medium (air) to a denser medium (water), it changes its angle of direction and gets refracted and hit to our eye lenses. With this, we see the surface of the pool at a changed angle and it seems to be a bit shallow than its original depth.

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Which diagram shows diffraction as light passes through an opening?
Bingel [31]
I don’t see the diagram
3 0
3 years ago
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Calculate the force exerted by a mental ball having a mass of 70kg moving with speed of 20m/s&gt;2
lord [1]

Answer:

F = 1400 N

Explanation:

It is given that,

Mass of the ball, m = 70 kg

It is moving with an acceleration of 20 m/s². We need to find the force exerted by the ball.

Force is given by the product of mass and acceleration. So,

F = ma

F=70\ kg\times \ 20m/s^2\\\\F=1400\ N

So, the force of 1400 N is exerted by a metal ball.

8 0
3 years ago
( Can someone help? )
Murrr4er [49]

Answer:

Answer would be 0.33

Explanation:

Calculations

8 0
3 years ago
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A shovel is the third class lever​
Nookie1986 [14]

Answer:

Yes

Explanation:

In a third-class lever, the effort force lies between the resistance force and the fulcrum. Some kinds of garden tools are examples of third-class levers. When you use a shovel, for example, you hold one end steady to act as the fulcrum, and you use your other hand to pull up on a load of dirt.

8 0
3 years ago
Un objeto se suelta desde determinada altura y emplea un tiempo t en caer al suelo. Si se cuadruplica la altura desde la cual se
blondinia [14]

When an object falls from a h height, you should work with the uniformly accelerated linear movement equations:

y=½*a*t²+Vo*t+yo

You should consider:

a=-g=-10m/s²

yo=h

If it’s a freefall, it means it starts from rest, which means it has no initial velocity:

Vo=0

Replacing that information in the equation:

y=½*(-10m/s²)*t²+0*t+h=-5m/s²*t²+0+h=-5m/s²*t²+h

So this is the

Besides, if you want to find out how long it takes for it to get to the floor, you should put the height of the floor as final height, which would be 0 (assuming the initial height has been measured from there):

y=0

0=-5m/s²*t²+h

5m/s²*t²=h

t²=h/(5m/s²)

t=√(h/(5m/s²))

t=√(hs²/(5m))

t=(√(h/(5m)))s

<span>If we <span>quadruple </span>h:</span>

t2=(√(h2/(5m)))s=(√(4*h1/(5m)))s=(√4)*(√h1/(5m)))s=2*(√h1/(5m)))s=2*t1

This 4 goes inside the square root, so then it converts to 2. So the new time is twice as much the previous time.

Concerning velocity, you have to use the other equation:

v=at+vo

As I said before, a is gravity and vo is zero.

v=-10m/s²*t+0=-10m/s²*t

Final velocity is directly related to time, so if time is doubled, so is velocity.

v2=-10m/s²*t2=-10m/s²*(2*t1)=2*(-10m/s²*t1)=2*v1

<span>So the correct answer is A, and the other ones are false.</span>

8 0
3 years ago
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