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Wewaii [24]
3 years ago
5

Now select the light bulb as the current indicator in the circuit. Move the magnet back and forth through the coil at different

rates and observe the variation of the brightness of the bulb. According to Faraday’s law of induction, the rate of change of magnetic flux through the coil is proportional to the voltage induced voltage in the coil.Why does the brightness of the bulb change when you move the magnet through the coil at different rates? Explain your answer with Faraday’s law and electrical
Physics
1 answer:
nataly862011 [7]3 years ago
6 0

Answer:

The magnitude of induced voltage varies with rate of change of magnetic field.

Explanation:

The experiment aims to demonstrate the principle of electromagnetic induction as shown by Michael Faraday. The intensity of the bulb is a physical indicator of the magnitude of induced voltage. A simple statement of Faraday's law of electromagnetic induction is that the induced voltage in a circuit is proportional to the rate of change over time of the magnetic flux through that circuit. This implies that, the faster the magnetic field changes, the greater will be the voltage in the circuit.

Moving the magnet at different rates implies changing the magnetic field hence the magnitude of induced voltage also changes accordingly. This change is indicated by changes in the brightness of the light bulb.

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A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete
HACTEHA [7]

Explanation :

Speed of car, v_1=50\ km/h=13.8\ m/s

Kinetic energy of the car, KE_1=400\ J

Speed of car, v_2=100\ km/h=27.7\ m/s

Let KE_2 is the kinetic energy of the car when it is moving with 100 km/h.

\dfrac{KE_1}{KE_2}=\dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}

\dfrac{KE_1}{KE_2}=\dfrac{v_1^2}{v_2^2}

KE_2=KE_1(\dfrac{v_2}{v_1})^2

KE_2=400\ J\times (\dfrac{27.7\ m/s}{13.8\ m/s})^2

KE_2=1611.6\ J  

Initial velocity of both cars are 0. Using third equation of motion :

So, S_1=\dfrac{v_1^2}{2a}=\dfrac{v_1t}{2}=\dfrac{v_1t}{2}

and S_2=\dfrac{v_2^2}{2a}=\dfrac{v_2t}{2}=\dfrac{v_2t}{2}

t is same. So,

\dfrac{S_1}{S_2}=\dfrac{50\ m/s}{100\ m/s}

\dfrac{S_1}{S_2}=\dfrac{1}{2}

S_2=2\ S_1

So, the braking distance at the faster speed is twice the braking distance at the slower speed.

4 0
4 years ago
Read 2 more answers
What happen to tbe brightness when one of the bulb removed from patallel circiut​
lawyer [7]
The current decreases as the overall resistance increases. In addition, if one bulb is removed from the “chain” the other bulbs go out. ... If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the “ladder”.
5 0
3 years ago
Wat is the volume of a marble in mL?
rewona [7]
4/3 pi (radius)^3
If the radius is in cm the volume will be in mL
6 0
3 years ago
What is the average kinetic energy of helium atoms in a region of the solar corona where the temperature is 5.90 x 10^5 K?
OLEGan [10]

<u>Answer:</u> The average kinetic energy of helium atoms is 1.222\times 10^{-17}J

<u>Explanation:</u>

To calculate the average kinetic energy of the atom, we use the equation:

K=\frac{3}{2}kT

where,

K = average kinetic energy = ?

k = Boltzmann constant = 1.3807\times 10^{-23}J/K

T = temperature = 5.9\times 10^5K

Putting values in above equation, we get:

K=\frac{3}{2}\times 1.3807\times 10^{-23}J/K\times 5.9\times 10^5K\\\\K=1.222\times 10^{-17}J

Hence, the average kinetic energy of helium atoms is 1.222\times 10^{-17}J

4 0
3 years ago
Please help answer this
ddd [48]
I believe the answer is the white dwarf.

The sun will cool then heat up again is my understanding. Hope this helps
3 0
3 years ago
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