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Trava [24]
3 years ago
5

A small fan has blades that have a radius of 0.0600 m. When the fan is turned on, the tips of the blades have a tangential accel

eration of 20.0 m/s2 as the fan comes up to speed. What is the angular acceleration α of the blades?
Physics
1 answer:
yaroslaw [1]3 years ago
3 0

Answer:

α = 333 m/s^2

Explanation:

Radius = 0.0600m

Tangential acceleration (a) = 20.0m/s^2

Angular acceleration (α) = ???

Angular acceleration describes the motion of the entire object. The tangential acceleration describes the motion of a single point on the object. Tangential acceleration is proportional to the radius and the angular acceleration.

a = rα

α = a/r

α = 20/0.06

α = 333.33 m/s^2

α = 333m/s^2

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1 year ago
Bob walks 370 m south, then jogs 410 m southwest, then walks 370 m in a direction 28 degrees east of north.
wlad13 [49]
We will measure all angles from West, the negative x-axis and divide the journey into 3 parts:
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Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

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