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julsineya [31]
3 years ago
12

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are

0.847 m and 1.27 kg. What constant-magnitude force acting at the other end of the rod perpendicularly both to the rod and to the axis will accelerate the rod from rest to an angular speed of 6.37 rad/s in 9.87 s?
Physics
1 answer:
igor_vitrenko [27]3 years ago
4 0

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

T = I\alpha = 0.303*0.6454 = 0.196 Nm

So the force acting on the other end to generate this torque mush be:

F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N

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Answer:

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If the final speed of an object as is strikes the ground is 77 m/s and it was in the air for 6.5 seconds. What was the initial d
azamat

Answer:

The answer to your question is: 13.2 m/s

Explanation:

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fs = is + gt     from this equation we clear "is" = fs - gt

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Process                               is = 77 - 63.8

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sergejj [24]

Answer:

The hottest temperature is  T_2 = 39^o C

Explanation:

From the question we are given

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