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2 years ago

Input Voltage = 220V

Physics

1 answer:

muminat2 years ago

4 0

Answer:

Correct option is

450 V and 15 A

⇒T

200N

3000W

=15A

Efficciencyofthetransformer

η=

out

⇒

100

3000

100×6

90×5000

=450V

Hence,

option (A) is correct answer.

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According to Coulomb's law, if the separation between two particles of the same charge increases four times, the potential energ

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Answer:

c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R

V2 / V1 = (R1 / R2) = 1/4

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3 years ago

If the element gallium has an atomic number of 31 and an atomic mass of 70, how many neutrons does it have?

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The answer is 39.

Explanation:

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2 years ago

The electromagnetic waves that have the lowest frequencies are called

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A. Radio waves
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3 years ago

Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s

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Explanation:

In first case we are interested in one time 6 in six rolls

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3 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

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(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$