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marta [7]
2 years ago
7

A 5.5Kg block is hanging from a rope that is wrapped around the outside of a 13Kg flywheel disk witha radius of 33cm that is hag

ning form the ceilin. Friction in the flywheel provides a constant torque of 2.5Nm. When the block is released what is the magnitude of its acceleration as it falls
Physics
1 answer:
Sergeeva-Olga [200]2 years ago
8 0

Answer:

3.9m/s^{2}

Explanation:

Using second law of motion

a =\frac {m1 * g - \frac {T}{r}}{m1 + 0.5 * m2} where m1 is mass of block, m2 is mass of flywheel, g is acceleration due to gravity whose value is taken as 9.81 m/s^{2}, T is torque and r is radius

Substituting 5.5 Kg for m1, 13 Kg for m2, 0.33 m for r, 2.5 Nm for T we obtain

a = \frac {5.5 \times 9.81 - \frac {2.5}{0.33}}{(5.5 + 0.5 \times13)}=3.9m/s^{2}

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3 0
2 years ago
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xenn [34]
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

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6 0
2 years ago
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Answer:

\mu=0.74            

Explanation:

It is given that,

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