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Alinara [238K]
2 years ago
10

Particles q1, 92, and q3 are in a straight line.

Physics
1 answer:
NNADVOKAT [17]2 years ago
8 0

The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

<h3 /><h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

\rm F_{31}} = \frac{Kq_1q_3}{r^2} \\\\ \rm F_{31}}  = \frac{9 \times 10^9 \times -28.1 \times 10^{-6}\times \times -47.9 \times 10^{-6}}{(0.600)^2} \\\\ F_{31}} =33.64 \ N

The force,by the charge q₂ on the q₃;

\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}}  = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09  \ N

The net force is the sum of the two forces;

\rm F_{net}=F_{32}+F_{31}\\\\\ \rm F_{net}=36.6-19.9 \\\\ \rm F_{net}=17.51 \ N

Hence, the net force on q₃ will be 17.51 N.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

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1. 12 V

2a. R₁ = 4 Ω

2b. V₁ = 4 V

3a. A = 1.5 A

3b. R₂ = 4 Ω

4. Diagram is not complete

Explanation:

1. Determination of V

Current (I) = 2 A

Resistor (R) = 6 Ω

Voltage (V) =?

V = IR

V = 2 × 6

V = 12 V

2. We'll begin by calculating the equivalent resistance. This can be obtained as follow:

Voltage (V) = 12 V

Current (I) = 1 A

Equivalent resistance (R) =?

V = IR

12 = 1 × R

R = 12 Ω

a. Determination of R₁

Equivalent resistance (R) = 12 Ω

Resistor 2 (R₂) = 8 Ω

Resistor 1 (R₁) =?

R = R₁ + R₂ (series arrangement)

12 = R₁ + 8

Collect like terms

12 – 8 =

4 = R₁

R₁ = 4 Ω

b. Determination of V₁

Current (I) = 1 A

Resistor 1 (R₁) = 4 Ω

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 1 × 4

V₁ = 4 V

3a. Determination of the current.

Since the connections are in series arrangement, the same current will flow through each resistor. Thus, the ammeter reading can be obtained as follow:

Resistor 1 (R₁) = 4 Ω

Voltage 1 (V₁) = 6 V

Current (I) =?

V₁ = IR₁

6 = 4 × I

Divide both side by 4

I = 6 / 4

I = 1.5 A

Thus, the ammeter (A) reading is 1.5 A

b. Determination of R₂

We'll begin by calculating the voltage cross R₂. This can be obtained as follow:

Total voltage (V) = 12 V

Voltage 1 (V₁) = 6 V

Voltage 2 (V₂) =?

V = V₁ + V₂ (series arrangement)

12 = 6 + V₂

Collect like terms

12 – 6 = V₂

6 = V₂

V₂ = 6 V

Finally, we shall determine R₂. This can be obtained as follow:

Voltage 2 (V₂) = 6 V

Current (I) = 1.5 A

Resistor 2 (R₂) =?

V₂ = IR₂

6 = 1.5 × R₂

Divide both side by 1.5

R₂ = 6 / 1.5

R₂ = 4 Ω

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