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Ahat [919]
3 years ago
12

Which is the correct order from least amount of atomic (particle) movement to the most amount of atomic particle movement?

Physics
1 answer:
babymother [125]3 years ago
8 0

Answer: particles movement in solid< particles movement in liquids< particles movement in gases.

Explanation:

Atoms are very small, it is not easily seen even with the help of light microscopes. However, We use multiple models of atoms toexplain describe particles of an atom behaviour.

In solids, the particles are packed together tightly in an ordered arrangement. The particles only vibrate about their position in the structure because the particles are held together too strongly to allow movement. Thereby,making the particles MOVE THE LEAST

In liquids, the particles are close together and they move with random motion in the container. The particles move rapidly in all directions but there is more colision between itself even more than particles in gases. This means that the particles here are MORE FASTER THAN THAT OF THE SOLID.

Particles in gases move the FASTEST, more than the particles in solids and liquids. Although, the average speed of the particles depends on their mass and the temperature.

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Compare and contrast infrasonic and ultrasonic vibrations
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Answer:

Explanation:

Ultrasonic waves are acoustic waves that are so high in frequency that humans can't hear them; however, infrasonic waves are sound waves that are lower in frequency than what humans can hear. A subsonic wave is a wave that is traveling slower than the speed of sound and a supersonic waves travels faster

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3 years ago
A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery
GuDViN [60]

Answer:

at the beginning: 2.3\cdot 10^{-10} F

when the plates are pulled apart: 1.1\cdot 10^{-10} F

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=k \epsilon_0 \frac{A}{d}

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

A=0.210 m^2 is the area of the plates

d=0.815 cm=8.15\cdot 10^{-3} m is the separation between the plates at the beginning

Substituting into the formula, we find

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F

Later, the plates are pulled apart to d=1.63 cm=0.0163 m, so the capacitance becomes

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F

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