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Oksanka [162]
3 years ago
10

A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio

n is required so that the bug stays on the turntable?
Physics
1 answer:
Agata [3.3K]3 years ago
3 0

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



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Answer:

2.68 hours

Explanation:

A.) Suppose the wind blows out from the west (with the air moving east). The pilot should then head her plane to northwest direction to move directly north.

B.) Given that plane flies at a speed of 102 km/h in still air. And the wind blows out from the west (with the air moving east) at a speed of 46 km/h.

The plan resultant speed can be calculated by using pythagorean theorem.

Resultant Speed = Sqrt( 102^2 + 46^2 )

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Resultant speed = 111.89 km/h

From the definition of speed,

Speed = distance/time

Where distance = 300 km

Substitute the resultant speed and the distance into the formula.

111.89 = 300/time

Time = 300/111.89

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Therefore, it take her 2.68 hours to reach a point 300 km directly north of her srarting point

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kiruha [24]

Answer:

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<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up without taking into consideration friction with the air.

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The tennis ball was thrown straight up with a speed of v0=22.5 m/s. The acceleration of gravity is g=9.81\ m/s^2, thus:

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A ball is thrown horizontally from the top of a tall cliff. Neglecting air drag, what vertical distance will the ball have falle
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The relevant equation to use here is:

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<span>y = 44.1 meters</span>

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