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worty [1.4K]
2 years ago
7

A water wave approaches a wall with a small hole in it. The wave is blocked by the wall, except for the part that goes through t

he hole. The waves coming out of the hole begin to spread out as the new source of wavefronts. This is an example of what property?
a Interference
b Amplitude
c Collision
d Diffraction
Physics
1 answer:
Eduardwww [97]2 years ago
7 0

Answer:

diffraction

Explanation:

took da test

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Question 23
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If the gymnast mass were doubled, her height (h) from the top of the board would be as follows,

с  Stay the same

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  • The Mass of an object or body does not affect the acceleration due to gravity in any kind of way.
  • Light weight objects accelerate more slowly than the heavy objects because when the forces other than the gravity also plays a major role.
  • Mass increases of a body when an object has higher velocity or the speed.
  • The greater the force of gravity, it would give a direct impact on the object's acceleration; thus considering only a force, the heavier the object is, it would accelerate faster. But an acceleration depends upon the two factors which are  force and mass.
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3 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
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