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KengaRu [80]
3 years ago
8

The mcb of rupa's room is tripped and keeps on tripping again and again . if it is a domestic circuit, what could be the reason

of phenomenon?
Physics
1 answer:
kap26 [50]3 years ago
4 0

Overloading

Explanation:

The reason reason why the mcb of rupa's room keeps tripping is due to the fact that excessive current has being supplied to his room.

MCB stands for a miniature circuit breaker.

A miniature circuit breaker opens up the electrical circuit by switching it off when there is overloading or faulty connections.

  • A MCB helps to control the flow of current and it is designed to hand certain limits of electrical voltage.
  • If the MCB keeps tripping, it suggests a surge in current supplied, overloading or probably a faulty connection.

learn more:

Circuit brainly.com/question/12472944

#learnwithBrainly

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Explanation:

Principle focus is the point on the axis of a convex lens, where the parallel rays of light from one side of the lens. meet on other side after refraction. Distance between optical centre to principle focus point is the focal length.

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If you run 12 m/s for 15 minutes, how far will you go?
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A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence
Anestetic [448]

Answer:

The linear velocity is  v=4.08m/s

Explanation:

According to the law of conservation of energy

   The potential energy possessed by the  hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

        The kinetic energy can be mathematically represented as

                    KE = \frac{mv^2}{2} + \frac{Iw}{2}

Where I is the moment of inertia possessed by the hoop  which is mathematically represented as

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Here R is the radius of the hoop

         w is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

                          w = \frac{v}{r}

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

      Now expressing the above statement mathematically

            potential \ energy = \frac{mv^2}{y} + \frac{Iw^2}{2}

               mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}

=>            mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}  

=>          mgh = \frac{mv^2}{2} + \frac{mv^2}{2}

=>           mgh = mv^2

=>              v = \sqrt{gh}

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                v = \sqrt{9.81 * 1.7}

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Answer:

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Work done = 1.098 + 2.529

Work done = 3.63 KJ

Therefore, the non conservative workdone is 3.6 KJ approximately

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