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Nady [450]
3 years ago
15

Find the equivalent resistance in the circuit between A and B ​

Physics
1 answer:
Monica [59]3 years ago
5 0

Answer:

6Ω

Explanation:

The 6Ω resistor and the 3Ω resistor are in parallel.

R = 1 / (1/6 + 1/3)

R = 1 / (1/6 + 2/6)

R = 1 / (3/6)

R = 6/3

R = 2

The 8Ω resistor and the other 8Ω resistor are also in parallel.

R = 1 / (1/8 + 1/8)

R = 1 / (2/8)

R = 8/2

R = 4

This 4Ω resistance is in series with the 4Ω resistor.

R = 4 + 4

R = 8

The 2Ω resistor and the 6Ω resistor are also in series.

R = 2 + 6

R = 8

These two 8Ω resistances are in parallel.

R = 1 / (1/8 + 1/8)

R = 1 / (2/8)

R = 8/2

R = 4

Finally, this 4Ω resistance is in series with the 2Ω resistance we found earlier.

R = 4 + 2

R = 6

The total equivalent resistance is 6Ω.

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An ice cream maker has a refrigeration unit which can remove heat at 120 Js'. Liquid ice
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Answer:

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C is 45,360 J

Explanation:

The given parameters for the refrigeration unit and the ice cream are;

The power of the refrigeration unit = 120 J/s

The mass of the liquid ice cream, m = 0.6 kg

The initial temperature of the liquid ice cream, T₁ = 20°C

The freezing point temperature of the ice cream, T₂ = -16°C

The specific heat capacity of the ice cream, c = 2,100 J/kg⁻¹·°C⁻¹

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, ΔQ, is given as follows;

ΔQ = m × c × ΔT

Where;

ΔT = T₁ - T₂

∴ ΔQ = m × c × (T₁ - T₂)

Therefore, by substituting the known values, we have;

ΔQ = 0.6 × 2,100 × (20 - (-16)) = 45,360

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C = ΔQ = 45,360 J.

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Newtons first law 1 to 5. <br>What is each of the net force for all of the 5 questions? ​
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Answer:

1. 65 N.

2. 160 N.

3. 0 N.

4. 210 N.

5. 90 N.

Explanation:

1. Determination of the net force.

Force applied to the right (Fᵣ) = 80 N

Force applied to the left (Fₗ) = 145 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 145 – 80

Fₙ = 65 N

Thus, the net force is 65 N

2. Determination of the net force.

Force 1 applied to the left (F₁) = 35 N

Force 2 applied to the left (F₂) = 125 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 35 + 125

Fₙ = 160 N

Thus, the net force is 160 N.

3. Determination of the net force.

Force applied to the right (Fᵣ) = 75 N

Force applied to the left (Fₗ) = 75 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 75 – 75

Fₙ = 0

Thus, the net force is 0 N

4. Determination of the net force.

Force 1 applied to the right (F₁) = 150 N

Force 2 applied to the right (F₂) = 60 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 150 + 60

Fₙ = 210 N

Thus, the net force is 210 N.

5. Determination of the net force.

Force applied to the right (Fᵣ) = 115 N

Force applied to the left (Fₗ) = 25 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 115 – 25

Fₙ = 90 N

Thus, the net force is 90 N

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