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labwork [276]
3 years ago
9

A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s

. Determine the wheel's average rotational acceleration. Express your answer in radians per second squared. Enter positive value if the rotational speed increases and negative value if the rotational speed decreases.

Physics
2 answers:
Likurg_2 [28]3 years ago
7 0

Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)

Explanation: Please see the attachments below

sesenic [268]3 years ago
5 0

Answer:

The average rotational acceleration is -0.4 rad/s²

Explanation:

Given:

w₁ = initial rotational speed = 4 rad/s

w₂ = final rotational speed = 2 rad/s

t = time = 5 s

The average rotational acceleration is:

w_{average} =\frac{w_{2}-w_{1}  }{t} =\frac{2-4}{5} =-0.4rad/s^{2}

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Hi there!

To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.

Recall:

\large\boxed{F_D = \mu N}}

The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.

\large\boxed{N = Mgcos\theta + Fsin\theta}}

We can plug in the known values to solve for one part of the normal force:

N = (1)(9.8)(cos30)  + F(.5) = 8.49  + .5F

Now, we can plug this into the equation for the dynamic friction force:

Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F

For a block to move with constant speed, the summation of forces must be equivalent to 0 N.

If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:

Fcosθ = 1.697 + .1F

Solve for F:

Fcos(30) - .1F = 1.697

F(cos(30) - .1) = 1.697

F = 2.216 N

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Answer:

36.22 mA

Explanation:

i1 = I , i2 = I, d = 8.2 cm = 0.082 m

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