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Jet001 [13]
3 years ago
12

A circuit is supplied with a constant voltage. As the resistance of the circuit decreases, the power given to the circuit

Physics
1 answer:
Yakvenalex [24]3 years ago
7 0

Here, we are required to determine the reaction of the power given to a circuit supplied with a constant voltage as the distance of the circuit decreases.

  • In a circuit supposed with a <em>constant voltage</em>, as the resistance of the circuit decreases, the power given to the circuit Increases as evident in the formular: P = V²/R

The power given to a circuit is given by the formula;

Power, P = V × I

where, V = voltage supplied

and I = Current through the circuit.

However, from Ohm's law;

V = I × R

In essence, I = V/R

By substituting, I = V/R into the equation P = V × I

Therefore, we have;

P = V²/R.

Ultimately, if the circuit is supplied with a constant voltage, As the resistance of the circuit decreases, the power given to the circuit Increases.

Read more:

brainly.com/question/16488641

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A steel ball with mass m=5.21 g is moving horizontally with speed ????=412 m/s when it strikes a block of hardened steel with ma
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Answer:

a) The speed of the block immediately after the collision is v_{2f}=(0.289\±0.002)m/s.

b) The impulse exerted on the block is p_{2}=(4.2772\±0.0296)kg*m/s.

Explanation:

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a) As this is a perfectly elastic collision, we can use the formula  v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}+ (\frac{m_{2}-m_{1}}{m_{1}+m_{2}} )v_{2i}, due v_{2i}=0m/s, we obtain v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}. Then with the data that we know m_{1}=5.21g=0.00521kg, m_{2}=14.8kg and v_{1i}=412m/s, therefore v_{2f}=(\frac{2(0.00521kg) }{0.00521kg+14.8kg} ) 412m/s=0.289m/s or v_{2f}=(0.289\±0.002)m/s adding uncertainty.

b) Now that we know the speed we can use p_{2}=m_{2}*v_{f2} =14.8kg*(0.289\±0.002)m/s=(4.2772\±0.0296)kg*m/s.

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Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta

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B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

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E = (3)(230\ m/s)(0.61\ T)Sin90^o

<u>E = 420.9 N/C</u>

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