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Sloan [31]
3 years ago
13

A ball is dropped out of a window and falls for 8.75 s. What is the ball's final velocity?

Physics
1 answer:
Kipish [7]3 years ago
3 0

Answer:

Final velocity of ball is 85.75 m/s

Explanation:

According to the problem, ball is dropped from the window and it experience an acceleration due to Earth's gravity which is denoted by g.

Initially, the ball is at rest, so its initial velocity is zero. The ball takes 8.75 s to fall from the window and hits the ground.

The equation of motion for constant acceleration is :

v = u + at

Here v is final velocity, u is initial velocity, t is time and a is acceleration.

In this case, u is zero and a is equal to g. So, the equation becomes :

v = gt

Substitute 8.75 s for t and 9.8 m/s² for g in the above equation.

v = 9.8 x 8.75

v = 85.75 m/s

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A 20-liter container contains 2.0 moles of oxygen at a pressure of 92 kpa. the average kinetic energy of translation of oxygen m
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Solution A has a specific heat of 2.0 J/g◦C. Solution B has a specific heat of 3.8 J/g◦C. If equal masses of both solutions start
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Answer: 2. Solution A attains a higher temperature.

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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:

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  • 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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