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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 toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
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Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

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