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kkurt [141]
3 years ago
11

What happens to the molecules of water when it moves from a liquid to a gas?

Physics
2 answers:
mr_godi [17]3 years ago
5 0

its A or D but im not sure which one ik it moves fast

harina [27]3 years ago
3 0
It’s D because the water molecules are turning into a gas wich makes them go faster
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Consider the concepts of kinetic energy (KE) and gravitational potential energy (GPE) as you complete these questions. A ball is
Lena [83]

Answer:

When the ball is held motionless above the floor, the ball possesses only GPE  energy.If the ball is dropped, its GPE energy decreases as it falls.If the ball is dropped, its KE energy increases as it falls.

Explanation:

If the ball is held motionless, then its kinetic energy is equal to zero, since kinetic energy depends on the velocity. And the ball is held above the ground, which means it possesses gravitational potential energy.

If the ball is dropped, its height will decrease, therefore its gravitational potential energy will decrease. Along the way, the ball will be in free fall, and therefore its velocity will increase, hence its kinetic energy.

K = \frac{1}{2}mv^2\\U = mgh

3 0
3 years ago
An a.c. supply is connected to a wire stretched between the poles of a magnet. Which way does the wire move?​
snow_lady [41]
The wire vibrates back and forth between the poles of the magnet.
The frequency of the vibration is the frequency of the AC supply.

7 0
2 years ago
Two hockey players , big Jim and little Tim collide head on and get tangled while going for the puck 65kg time was traveling at
12345 [234]
I will assume that big Joe is big Jim. The equation for the momentum is p=m*v, where m is the mass of the body and v is the velocity. Big Joe has a mass m=105 kg and speed v=5.2 m/s. When we input the numbers:

p=105*5.2=546 kg*(m/s).

So big Joe's momentum before the collision is p=546 kg*(m/s).

7 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
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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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A young owl forgets how to fly for a moment and is free falling towards the ground. Before crashing, he remembers how to fly and
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did you figure it out

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