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2 years ago

What is the Principle of electric generator?

2 answers:

6 0

An electric generator works on the principle of Faraday's law of electromagnetic induction. Faraday's law of electromagnetic induction states that “whenever a current carrying conductor are placed in the magnetic field, then a flux is induces in the circuit due to which a current start to flow in the rotor also.

TiliK225 [7]2 years ago

5 0

An electric generator is a machine used for producing electricity. They convert mechanical energy into electrical energy.

electric generator is an application electro magnetic induction.

Its principle states that :

electric generator works on the principle that when a straight conductor is moved in a magnetic field then the current is induced in the conductor...

You might be interested in

A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

2 years ago

What fraction of the water must evaporate to remove precisely enough energy to keep the temperature constant? water at 37°c has

mart [117]

The fraction of the water must evaporate to remove precisely enough energy to keep the temperature constant when water at 37°c has a latent heat of vaporization of lv = 580 kcal/kg is 2.58 times 10 to the minus 3.

Vaporization is the process by which a substance is transformed from its liquid or solid state into its gaseous (vapour) state. Boiling is the term for the vaporization process when conditions permit the creation of vapour bubbles within a liquid. Sublimation is the process of directly converting a solid to a liquid.

Boiling and evaporation are the two processes that cause vaporization. Evaporation is the process by which a liquid body's surface changes from a liquid to a gas, as in the case of a drop of water on hot concrete evaporating into a gas. A liquid is said to be boiling when it is heated to the point at which it begins to give off steam, as when you boil water on a stove. The process of converting a substance from its liquid or solid state into its gaseous (vapour) state is known as vaporization.

To learn more about vaporization please visit - brainly.com/question/12625048
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5 0

1 year ago

Kelli weighs 425 N, and she is sitting on a playground swing that hangs 0.36 m above the ground. Her mom pulls the swing back an

kodGreya [7K]

Answer:

V = 3.54 m/s

Explanation:

Using the conservation of energy:

$E_i = E_f$

so:

$wh = \frac{1}{2}mV^2$

where w is te weigh of kelly, h the distance that kelly decends, m is the mass of kelly and V the velocity in the lowest position.

So, the mass of kelly is:

m = 425N/9.8 = 43.36 Kg

and h is:

h = 1m-0.36m =0.64m

then, replacing values, we get:

$(425N)(0.64m) = \frac{1}{2}(43.36kg)v^2$

Solving for v:

V = 3.54 m/s

7 0

3 years ago

The Moon's center is 3.9x10 m from Earth's center. The Moon is 1.5x10^8 km from the Sun's center. If the mass of the Moon is 7.3

nika2105 [10]

Explanation:

It is given that The Moon's center is 3.9x10⁸ m from Earth's center. The moon 1.5x10⁸ km from the Sun's center. We need to find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon.

The gravitational force is given by :

$F=\dfrac{Gm_em_m}{r^2}$