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Fantom [35]
3 years ago
5

A series circuit has a voltage supply of 12 V and a resistor of 2.4 kΩ. How much power is dissipated by the resistor?

Physics
1 answer:
adell [148]3 years ago
4 0

Answer:

(B) 0.06W

Explanation:

power absorbed by the resistor is given by \frac{V^2}{R}=I^2R

Where V = voltage

           R= resistance

            I =  current through the circuit

we have given V =12 Volt and resistance =2.4 K\Omega

current I=\frac{V}{R}=\frac{12}{2.4\times 1000}=5\times 10^{-3}A=5mA

power using voltage and resistance equation

p=\frac{12^2}{2.4\times 1000}=60mW =0.06W

using current equation P=I^2R=5^2\times 12=60mW= 0.06W

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jarptica [38.1K]

Answer:

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tino4ka555 [31]

Answer:

 i = 0.477 10⁴ B

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Explanation:

For this exercise let's use the Ampere law

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Where the path is closed

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From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.

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Segment on the y axis

        L₀ = (y2-y1)

        L₀ = 3-0 = 3 cm

Segment on the point x = 2 cm

        L₁ = 3-0

        L₁ = 3cm

       B L = μ₀ I

       B 2L = μ₀ I

        i = 2 L B /μ₀

        i= 2 0.03 / 4π 10⁻⁷   B

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The current is perpendicular to the magnetic field whereby the current flows in the  counterclockwise

8 0
3 years ago
A car has uniformly accelerated from rest to a speed of 25m/s after traveling 75m. What is its acceleration in m/s^2
Roman55 [17]
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4 0
3 years ago
A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnit
gayaneshka [121]

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns

4 0
3 years ago
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