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Tcecarenko [31]
3 years ago
11

The setting sun is actually visible after it has dropped below the horizon due to the bending of light waves entering the atmosp

here at a non-perpendicular angle to the surface, a process known as?
Physics
2 answers:
Yuki888 [10]3 years ago
4 0
That's one result of "refraction".
mamaluj [8]3 years ago
3 0

Refraction is the process in which light  will bend away or towards the normal at the point of incidence when light comes from one optical medium to another optical medium.

During sunset, sun is still visible after it has dropped below the horizon.It is so because the light coming from sun is refracted when it passes through the atmosphere.

Our atmosphere has not a constant density.It varies depending its position from earth surface. Hence, refractive index of each layer of atmosphere varies which results in the refraction of sun light for which sun is visible though it is geometrically below the horizon.


Hence, the correct answer to the question is refraction.

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The water in this pot is in the process of changing from a liquid to a gas. Thermal energy is still being added to the water as
olga nikolaevna [1]

The extra energy allows the molecules to convert states of molecules.

<h3>What is heating?</h3>

When the temperature is increased , the body or particle is said to be heating.

The water in this pot is in the process of changing from a liquid to a gas. Thermal energy is still being added to the water as it vaporizes, but its temperature and the motion of its particles are not increasing.

When the liquid is converted to gas, the temperature reaches the boiling point of water. After boiling, the heat added up will only make the particles to convert its phase and go away in then atmosphere.

Thus, extra energy added helps molecules of water to change its state from liquid to solid.

Learn more about heating.

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6 0
2 years ago
Cardiovascular Conditioning is __________________________________________________.
shutvik [7]
B . Because cardiovascular has everything to do with your heart how the blood flows and pumps and all that good stuff .
5 0
3 years ago
Read 2 more answers
a cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5m/s2 for 8.0
Andreas93 [3]

The final velocity is +15.0 m/s

Explanation:

The motion of the cart is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

t is the time

For the cart in this problem, we have:

u = +3.0 m/s (initial velocity)

a=1.5 m/s^2 (acceleration)

t = 8.0 s (time)

Substituting, we find the final velocity:

v=3.0+(1.5)(8.0)=+15.0 m/s

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4 0
3 years ago
Write the factors of potential energy
IgorLugansk [536]
It depends on two factors height and mass

6 0
3 years ago
A 2.40 kg snowball is fired from a cliff 7.69 m high. The snowball's initial velocity is 13.0 m/s, directed 49.0° above the hori
Crazy boy [7]

Answer:

a) W = 180.87 J , b)  ΔU = -180.87 J , c)  ΔU = -180.87 J

Explanation:

a) Work is defined as

         W = F .ds

Where bold indicates vectors, we can write the scalar product

         W = F s cos θ

Where the angle is between force and displacement.

The force of gravity is the weight of the body, which is directed downwards and the displacement thickens the tip of the cliff at the bottom, so that it is directed downwards, therefore the angle is zero degrees

        W = F_{g} y

        W = m g y

For this problem we must fix a reference system, from the statement it is established that the system is placed at the base of the cliff, so that final height is zero and the initial height (y₀ = 7.69m)

    W = 2.40 9.8 (7.69-0)

    W = 180.87 J

b) The potential energy is

            U = mg y

The change in potential energy,

        ΔU = U_{f}- U₀

       ΔU = mg (y_{f}- y₀)

      ΔU = 2.4 9.8 (0 -7.69)

      ΔU = -180.87 J

 

c) in this case we change the reference system to the height of the cliffs, for this configuration

          y₀ = 0

          y_{f} = -7.69 m

          ΔU = 2.4 9.8 (-7.69 -0)

          ΔU = -180.87 J

3 0
3 years ago
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