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3 years ago

When light waves passes straight through an object, it is called?

Physics

2 answers:

Julli [10]3 years ago

6 0

<span>When light waves passes straight through an object, it is called transmission.</span>

amm18123 years ago

4 0

<h3><u>Answer;</u></h3>

Transmission

<h3><u>Explanation;</u></h3>

<em><u>Light waves are types of waves that are electromagnetic waves, meaning they do not require material medium for transmission. </u></em>Light is transmitted as a transverse wave such that the vibration of particles is perpendicular to the direction of wave motion.
<em><u>A material medium enables the transmission of a wave by the vibration of particles, atoms or molecules. </u></em>The vibration of particles in a medium helps in the transmission of a wave such that energy is transferred from one point to another due to the disturbance caused by the wave.

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frez [133]

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Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

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4 years ago

What is the speed vfinal of the electron when it is 10.0 cm from charge 1?

fgiga [73]

Answer:

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