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2 years ago

Help please!!! Physics circular motion

Physics

1 answer:

svp [43]2 years ago

6 0

Answer:

2. 3.1415 m/s

3. 0.63m

4. 0.006 m/s^2

Explanation:

2. v=(2*pi*r))/T, put in values and solve.

3. Circumference=2*pi*radius, radius is 0.1m, plug in and solve.

4. a=(v^2)/r, (don't worry about converting to meters, since units are the same in kilometers, they will cancel out) plug in values and solve.

That's it!

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How much work is done if a 15 newton suitcase is lifted 2 m ?

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Answer:

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2 years ago

If you travel 1.7 km north from your house at noon, and at 6:00 PM you travel 5.4 km south, what is your displacement? 3.7 km no

Yakvenalex [24]

<u>Answer</u>

3.7 Km south

<u>Explanation</u>

The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.

Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).

Now lets add the two values.

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2 years ago

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3 years ago

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viva [34]

Answer:

21.67 rad/s²

208.36538 N

Explanation:

$\omega_f$ = Final angular velocity = $\dfrac{1}{6}78=13\ rad/s$

$\omega_i$ = Initial angular velocity = 78 rad/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2$

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875$

Frictional force is given by

$F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N$

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

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2 years ago