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nalin [4]
4 years ago
10

Determine how potential and kinetic energy changes at each position of the pendulum as the ball swings from A to E. Position A:

Position B: Position C: Position D: Position E:
Physics
2 answers:
slavikrds [6]4 years ago
5 0
<h3><u>Answer;</u></h3>

Position A: Maximum potential  and minimum kinetic energy

Position B: losing potential and gaining kinetic  

Position C: Maximum kinetic  and minimum potential energy.

Position D: is losing kinetic and gaining potential  

Position E: Maximum potential and minimum kinetic energy.

<h3><u>Explanation;</u></h3>
  • Position A is a point of maximum potential energy and minimum kinetic energy since it is the highest point of the pendulum.
  • Position B, which is half way between A and C, the ball is losing potential energy and gaining kinetic energy. This occurs as the ball descends from the highest position to the lower position thus looses potential energy and gains kinetic energy.
  • Position C is the point of maximum kinetic energy and minimum potential energy. Because this is the lowest point of the pendulum and a point where the ball has maximum velocity.
  • At position D which is half way between C and E, the potential energy is increasing as the kinetic energy is decreasing. At this point the ball has half of its maximum potential energy.
  • At position E, he potential energy of the pendulum reaches its maximum value and is neither increasing nor decreasing at that point.
worty [1.4K]4 years ago
5 0

Answer: position A is all potential) position B) losing potential and gaining kinetic) position C) is all kinetic) position D) is losing kinetic and gaining potential) position E) is all potential.

Explanation: I just did the questions trust me these are right try them! good luck.

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I have to say, i love this kind of problems.

So, we got the linear acceleration, as we all know, linear the acceleration its, in dimensional units:

[a]=[\frac{distance}{time^2}].

Now, we got the radius

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the angular frequency

[\omega] = \frac{1}{s}

and the mass

[m]=[mass].

Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.

The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write

a \propto r.

The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared

a \propto r \omega^2.

We are almost there. If this were any other problem, we would write:

a = A r \omega^2

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5 0
3 years ago
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Answer:

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4 0
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andrew-mc [135]

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<h3>What is the acceleration?</h3>

Let us recall that the acceleration is the change in the speed of a body with time. We have been told that the body accelerates for 3s and then decelerates to 2s. This implies that the total time that the object spent in motion is 5 s.

Thus;

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Again;

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Answer:

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