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Katena32 [7]
2 years ago
10

The word “virtual” refers to something that exists in effect but not in actual fact. How does this definition relate to the virt

ual image you see of yourself in a plane mirror?
Physics
2 answers:
atroni [7]2 years ago
8 0
Because you see yourself the opposite way in a mirror. So yes your “seeing” yourself but not how everyone else sees you.
givi [52]2 years ago
5 0

Answer:

The image seems to be behind the mirror, but nothing is really there.

Explanation:

Sample response edge 2021

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A 10kg box is sliding at 50m/s. Find the momentum
MAXImum [283]

Answer:

The momentum of the ball is 500 kg·m/s

Explanation:

The momentum is given by Mass × Velocity

The given parameters are;

The mass of the box = 10 kg

The velocity by which the box is sliding = 50 m/s

Therefore, the momentum of the ball is given as follows;

The momentum of the ball = 10 kg × 50 m/s = 500 kg·m/s

The momentum of the ball = 500 kg·m/s

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What happens when light passes through a prism or diffraction grating?
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The light scatters in all directions.

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A forest fire wipes out all of the plant life in an area, causing the animals to leave. The ground becomes a mixture of soil and
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What is an electric motor? Explain its operation. 2-3 sentence
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Answer:

An electric motor is a device that changes electrical energy into mechanical energy. This change occurs due to the interaction between the magnetic field of magnets and the magnetic field due to the  electric current in the  loop. The interaction between the two produces a torque that makes the loop rotate on a shaft.

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A 2 kg ball is moving 3 m/s when it starts rolling up a hill.
AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]

Replacing the values on the equation we have:

Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\

Now we have:

h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

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3 years ago
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