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adoni [48]
4 years ago
15

The three different states of matter are liquid, solid, and gas. A solid is something you can hold in your fingers, while a liqu

id is something you can hold in a cup. Gases are all around you, making up the air we all breathe. These three states are very different. However, liquids and gases do have some things in common. One of these is A) that liquids and gases are both less dense than solids. B) that liquids and gases are both more dense than solids. C) that liquids and gases both take the shape of their container. D) that liquids and gases do not take the shape of their container. Eliminate
Physics
2 answers:
prohojiy [21]4 years ago
5 0

A) Liquids and gases are both less dense than solids*.


*Except for the very special case of water.  Ice is less dense than liquid water.  That's why cubes float in soda, and bergs float in the ocean.

nika2105 [10]4 years ago
3 0

The answer is C. that liquids and gases both take the shape of their container.  

Think of it this way, if you take an ice cube and put it in your glass, it will stay in its shape and stay that way until it melts.  But if you put liquid or a gas into a glass, it will take the shape of the glass that it is put into.  

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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?
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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,

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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}.

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