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2 years ago

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A girl rolls a basketball across a basketball court. The ball slowly decelerates at a rate of -0.20m/s^2. If the initial velocit

y was 2.0m/s and the ball rolled to a stop at 5.0sec after. 12:00, at what time did she start the ball rolling?

1 answer:

Dafna1 [17]2 years ago

7 0

By equation of motion :

$2as=v^2-u^2\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{0^2-2^2}{2\times( -0.2)}\\\\s=10\ m$

Now, it is given that it stops after 5 seconds of motion and time at that point is 12:00.

So, time when it started is 11 : 59 : 55.

Therefore, distance travelled is 10 m.

Hence, this is the required solution.

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The torque generated by the weight of the meterstick around the pivot is:
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To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
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3 years ago

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