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attashe74 [19]
2 years ago
12

Light hits a clear plastic bottle and a granite rock. Which choice most accurately describes the effect of visible light on the

clear plastic and a grey rock?
1. Visible light is absorbed by both the rock and the plastic, making both objects warmer. Visible light is also reemitted by both objects, allowing visible light through the rock and the plastic.

2. Visible light is absorbed by both the rock and the plastic, making both objects warmer. Some of the light hitting the rock is reemitted, allowing visible light through the rock. None of the light hitting the plastic is reemitted.

3. Visible light is absorbed by the rock, making it warmer. Light is also absorbed by the plastic making it warmer. None of the light hitting either object is reemitted.

4.Visible light is absorbed by the rock and the plastic making both objects warmer. Some of the light hitting the plastic is reemitted, allowing visible light through the plastic. None of the light hitting the rock is reemitted.
Physics
2 answers:
balu736 [363]2 years ago
8 0

Answer:

#4 is the accurate answer.

LenKa [72]2 years ago
5 0

Answer:

4

Explanation:

When the light hits an object, it is absorbed, and the energy of it is transferred by the object by heat, thus it warms. When the object is dark the light is not reemitted because the absorption factor of dark color is almost 100%. But, for clear objects, the absorption factor is small, does, some of the light must be reemitted through it.

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Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov
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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

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Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

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2 years ago
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