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Pavlova-9 [17]
3 years ago
12

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

roton and an electron are situated 997 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed . You find that forces are:a. Attractiveb. Repulsiveand the magnitude of force is :_______
Physics
1 answer:
Oksana_A [137]3 years ago
8 0

Answer:

a.Attractive

2.31531\times 10^{-16}\ N

Explanation:

When it comes to charges, the charges which are alike repel each other and the charges which are different will attract each other.

Here, there is a proton and electron which are different particles hence, they will attract each other.

q_1=q_2 = Charge of electron and proton = 1.6\times 10^{-19}\ C

r = Distance between them = 997 nm

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(997\times 10^{-9})^2}\\\Rightarrow F=2.31531\times 10^{-16}\ N

The force of attraction between the particles will be 2.31531\times 10^{-16}\ N

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Ammonia gas occupies a volume of 0.450 L at a pressure of 96 kPa. What
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0.426 L

Explanation:

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6 0
3 years ago
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

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.

8 0
3 years ago
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