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shepuryov [24]
3 years ago
13

The diagram shows changes of state between solid, liquid, and gas. The atoms of a substance gain energy during a change of state

. Before the change, the atoms are close together and cannot slide past one another. After the change, the substance can fill its container. A diagram has a triangle at center with gas on top, liquid on bottom right, solid on bottom left. An arrow from gas to liquid is labeled O, and an arrow from liquid to gas is labeled N. An arrow from solid to liquid is labeled P, and an arrow from liquid to solid is labeled Q. An arrow from solid to gas is labeled L, and an arrow from gas to solid is labeled M. Which arrow represents the change of state described above? L M N O
Physics
2 answers:
dolphi86 [110]3 years ago
3 0
  • L is sublimation ,M is deposition,N is evaporation, O is condensation .
pav-90 [236]3 years ago
3 0

Answer:

L

Explanation:

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denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

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