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3 years ago

How does a plasma differ from a gas

1 answer:

3 0

<span>They are different and unique from the other states of matter. Plasma is different from a gas, because it is made up of groups of positively and negatively charged particles. In neon gas, the electrons are all bound to the nucleus. In neon plasma, the electrons are free to move around the system.

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The first electric generators used direct current, so they needed to be reversed manually on a regular basis in order to work pr

djyliett [7]

Could I see the questions

4 0

3 years ago

[TRUE/FALSE] If an object is moving at constant velocity it must be in equilibrium.

DENIUS [597]

Answer:

yes

Explanation:

objects with constant velocity also have zero net external force. this means the forces on the object are balanced. this mean they are in equilibrium

4 0

3 years ago

An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

$V_x'=0.8c\\V=0.6c\\m=4*10^5kg$

Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

$KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

So, to $V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016$ find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
We have an expression from Lorents transformation for relativistic law of addition of velocities as,

$V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V$

Substituting values, we get,

$V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c$

Thus, the KE will be,

$KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J$

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

3 0

2 years ago

Solve this questions please

Rudik [331]

Answer:

10.37×10³j

Explanation:

hope this helps you !!!

4 0

4 years ago

write aparagraph explaining the difference bln the heat energy in a substances temperature in physics

iVinArrow [24]

The core difference is that heat deals with thermal energy, whereas temperature is more concerned with molecular kinetic energy. Heat is the transfer of thermal energy, whereas temperature is a property the object exhibits.

5 0

3 years ago