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zysi [14]
2 years ago
15

What is the equivalent resistance of a circuit that contains three 10.0 Ω

Physics
2 answers:
yanalaym [24]2 years ago
8 0

Answer:

O D.30.0 Ω

Explanation:

this <em>is </em><em>the </em><em>correct </em><em>answer</em><em>!</em>

Cerrena [4.2K]2 years ago
8 0

<u>Question</u><u> </u><u>:</u>

  • What is the equivalent resistance of a circuit that contains three 10.0 Ω resistors connected in series to a 6.0 V battery?

Options given :

  • A. 0.600 Ω
  • Β. 60.0 Ω
  • C. 16.0 Ω
  • D. 30.0 Ω

<u>Solution</u><u> </u><u>:</u>

As we know that,

When a circuit is connected in series resistance is calculated by :

  • Rs = R1 + R2 + R3 .... Rn

Here,

  • R is resistance

As a circuit contains three resistors. So R1 , R2 and R3 would be :

  • R1 = 10 Ω
  • R2 = 10 Ω
  • R3 = 10 Ω

Now,

>> Rs = 10 + 10 + 10

>> Rs = 30 Ω

Therefore,

  • Option D is correct.

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Nikolay [14]

The original kinetic energy will be 0 J and the final kinetic energy will be 7500 J and the amount of work utilized will be similar to the final kinetic energy i.e., 7500 J.

<u>Explanation:</u>

As it is known that the kinetic energy is defined as the energy exhibited by the moving objects. So the kinetic energy is equal to the product of mass and square of the velocity attained by the car. Thus,

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Thus, the initial kinetic energy of the car is zero, the final kinetic energy is 7500 J and the work utilized by the car is also 7500 J.

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