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Marina86 [1]
3 years ago
12

A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at th

e effective radius of the flywheel).
(a) How much work is done to bring this wheel from rest to a speed of 120 rev/min in a time. interval of 30.0 s?
(b) What is the applied torque on the fly-wheel (assumed constant)?
Physics
1 answer:
musickatia [10]3 years ago
4 0

Answer:

A)5524J,

B) 29.2Nm

Explanation:

This question can be treated using work- energy theorem

Work= change in Kinectic energy

W= Δ KE

Work= difference between the final Kinectic energy and intial Kinectic energy.

We know that

Kinectic energy= 1/2 mv^2 .............eqn(1)

This can be written in term of angular velocity, as

KE= 1/2 I

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fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

<em>We have a formula for such condition,</em>

Q=m.c.\Delta T.....................................(1)

where:

  • \Delta T= temperature difference
  • Q= heat energy
  • m= mass of the body
  • c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

  • we have equal masses of two solutions A & B, i.e. m_A=m_B
  • equal heat is supplied to both the solutions, i.e. Q_A=Q_B
  • specific heat of solution A, c_{A}=2.0 J.g^{-1} .\degree C^{-1}
  • specific heat of solution B, c_{B}=3.8 J.g^{-1} .\degree C^{-1}
  • \Delta T_A & \Delta T_B are the change in temperatures of the respective solutions.

Now, putting the above values

Q_A=Q_B

m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1

Which proves that solution A attains a higher temperature than solution B.

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Answer:

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Harlamova29_29 [7]

The correct answer to the question is False i.e the tendency of an object in motion to remain in motion is not called the orbital speed.

EXPLANATION:

Before going to answer this question, first we have to understand Newton's first laws of motion.

As per Newton's first laws of motion, every body continues to be in state of rest or of uniform motion in a straight line unless and until it is compelled by some external unbalanced forces.

Hence, as long as no unbalanced force is acting on a moving object, it will be in motion. This tendency of a moving object to be in motion is called inertia of motion of the body.

Inertia of motion is the property of the body by virtue of which a moving body always tries to be in motion.

Hence, the tendency of an object in motion to remain in motion is not called as the orbital speed.

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When a deck of cards slips out of your hands and falls to the ground, gravity<br> does work on it
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