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svetoff [14.1K]
3 years ago
11

s it possible for a system to have negative potential energy? Is it possible for a system to have negative potential energy? No,

because the kinetic energy of a system must equal its potential energy. Yes, as long as the total energy is positive. Yes, since the choice of the zero of potential energy is arbitrary. No, because this would have no physical meaning. Yes, as long as the kinetic energy is positive.
Physics
1 answer:
frez [133]3 years ago
8 0

Answer: Yes it possible for a system to have negative potential energy, since the choice of the zero of potential energy is arbitrary.

Explanation:

Potential energy is defined as the energy present in an object due to its position.

Mathematically,   P.E = mgh

A potential energy can be negative also. This is because zero potential energy is arbitrary. For example, at the top of a building the choice of potential energy can be zero.

Therefore, the potential energy below the building will be negative.

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Answer:

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Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

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