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julsineya [31]
2 years ago
11

PLEASE HELP ME!!!!! You draw an arrow from left to right on a piece of paper and place the paper well behind a glass of water. Y

ou observe that the direction of the arrow appears reversed behind the water glass!
What conclusions can you draw from this experiment?


(1 point)


The arrow appears reversed because light is bent as it enters the water, and again as it exits. The two light paths cross, making the direction of the arrow appear crossed.

The arrow appears reversed because light is bent as it enters the water, and again as it exits. The two light paths cross, making the direction of the arrow appear crossed.


The arrow appears reversed because light is reflected by the glass of water. Light is reflected so much that the arrow appears to face the other direction.

The arrow appears reversed because light is reflected by the glass of water. Light is reflected so much that the arrow appears to face the other direction.


It is not possible for the arrow to appear reversed behind a glass of water. The most likely explanation is that a mistake was made during the experiment.

It is not possible for the arrow to appear reversed behind a glass of water. The most likely explanation is that a mistake was made during the experiment.


The arrow appears reversed because light is split as it enters the water, and combines again as it exits the water. When the light re-combines, the arrow appears to be flipped.

Physics
1 answer:
julia-pushkina [17]2 years ago
7 0

Answer:

I think it is the first one: <u>The arrow appears reversed because light is bent as it enters the water, and again as it exits. The two light paths cross, making the direction of the arrow appear crossed.</u>

Explanation:

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h is the horizontal distance between the center of mass and the rotational axis of the rod

I = (\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2})^{2}[/tex]

I = (\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2})^{2}[/tex])

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   at the maximum height velocity = 0 therefore final kinetic energy = 0

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where (H-h) = rise in the center of mass

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   (H-h) = 0.15 m

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Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

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x is the stretching of the spring from the equilibrium position

In this problem, we have

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x = 0.2 m (stretching)

Therefore, the spring constant is

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