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serg [7]
2 years ago
5

A +26.3 uC charge qy is repelled by a force

Physics
1 answer:
Musya8 [376]2 years ago
5 0

Answer:

+1.46×10¯⁶ C

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C

Force (F) = 0.615 N

Distance apart (r) = 0.750 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Charge 2 (q₂) =?

The value of the second charge can be obtained as follow:

F = Kq₁q₂ / r²

0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²

0.615 = 236700 × q₂ / 0.5625

Cross multiply

236700 × q₂ = 0.615 × 0.5625

Divide both side by 236700

q₂ = (0.615 × 0.5625) / 236700

q₂ = +1.46×10¯⁶ C

NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C

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Option D is the correct answer.

<h3></h3><h3>Power </h3>

The work done by an object in a given time interval is called the power of that object.

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Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that t
ArbitrLikvidat [17]

Answer:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T_H in the cycle is twice the minimum absolute temperature T_L in the cycle

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now, we find the efficiency of the Carnot cycle engine

η_{th = 1 - T_L/T_H

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η_{th = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η_{th = 1 - W_{net/Q_H

where W_{net is net work done, Q_H is is the heat supplied

we substitute

0.5 = 60 / Q_H

Q_H = 60 / 0.5

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Now, we apply the first law of thermodynamics to the system

W_{net = Q_H - Q_L

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Q_L = 60 kJ

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we substitute

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q_L = 2400 kJ/kg

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T_L = 40 + (5) × (0.5)

T_L = 40 + 2.5

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Therefore, The temperature of the steam during the heat rejection process is 42.5°C  

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Answer:

a

Explanation:

they became stronger

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