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bonufazy [111]
3 years ago
5

Place a small object on the number line below at the position marked zero. Draw a circle around the object. Mark the center of t

his circle with the symbol for “initial position”. Move the object 5.0cm to the right and stop. Label this circle with the correct symbol for “final position.”  (A) What was the initial position of the object?     (B) What is the final position of the object?     (C) What is the distance traveled by the object?     (D) What is the displacement of the object?     (E) Of the three underlined quantities, which are numerically equal?
​

Physics
1 answer:
Viefleur [7K]3 years ago
6 0
It’s is D bc it is and idk you should do it
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At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

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0.4823 m/s

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