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goldfiish [28.3K]
3 years ago
9

Why do antoms move at different speeds depending pn weather they are in liqids or solids?

Physics
1 answer:
Salsk061 [2.6K]3 years ago
3 0
Liquid matter has atoms that move moderately freely; this is shown that whenever you pour water into a cup, it fills the shape.
Solids on the other hand, have atoms that basically don't move; thus is because the item is purely compact, and, for contrast, putting a rock in a cup wouldn't fill the shape of the cup, it would just lie there as it's original form.

Hope this helps!
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suppose you press a basketball against a wall. what happpens to the basketball when you are pressing it?
AURORKA [14]
It hardens because you are pressing it against something.
5 0
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________ is an indication of how near the air is to saturation rather than the actual quantity of water vapor in the air. mixing
Degger [83]
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6 0
2 years ago
A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at
yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)

Therefore, the work done by the worker in lifting the bucket is given as:

W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)

Now, plug in the values given and solve for 'W'. This gives,

W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0
3 years ago
You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice
Galina-37 [17]

The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

\sum F =ma

where

\sum F = F_f = -0.14 N is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)

m = 0.12 kg is the mass of the puck

a is the acceleration

Solving for a,

a=\frac{\sum F}{m}=\frac{-0.14}{0.12}=-1.17 m/s^2

The motion of the puck is a uniformly accelerated motion, therefore we can use the following suvat equation:

v^2-u^2=2as

where:

v = 0 is the final velocity (the puck comes to a stop)

u = 18.3 m/s is the initial velocity

a=-1.17 m/s^2 is the acceleration

s is the stopping distance

And solving for s, we find

s=\frac{v^2-u^2}{2a}=\frac{0-(18.3)^2}{2(-1.17)}=143.1 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

6 0
3 years ago
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