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Rashid [163]
3 years ago
6

A constant unbalanced force is applied to an object for a period of time. Which graph best represents the acceleration of the ob

ject as the function of elapsed time.

Physics
1 answer:
balu736 [363]3 years ago
5 0

Answer:

Here is an image attached with similar questions.

The correct answer is D where acceleration is function of time.

Explanation:

The force acting on the object is constant.

There is no change in the application of forces.

And we know that Force is the product of acceleration and masses.

Newtons second law: F=m\times a

Regarding mass we know that it can neither be created nor destroyed.

So in F=m \times a we have two constant terms, constant divided by constant will give the same result as a=\frac{Force}{mass}

There is no change in acceleration (a) with respect to time (t).

So the most appropriate graph where time (t) is changing on x-axis but acceleration doesn't changes is D.

The graph will be similar to y=any\ constant and will be horizontal to the x-axis.

Option D depicts the same.

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Answer: 2. Solution A attains a higher temperature.

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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:

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<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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