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tensa zangetsu [6.8K]
2 years ago
9

What type of simple machine is an electric fan?

Physics
1 answer:
GrogVix [38]2 years ago
5 0

Answer:

An electric fan is considered to be a mixture of several simple machines. It includes the Wheel and Axle type, wedges, and the Inclined plane types. The blades of an electric fan are the inclined planes and the wedges.

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To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit
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To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

R = 350ft

a_t = 1.1ft/s^2

PART A )

a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 5.25ft/s^2

a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

27.5625 = 1.21 + \frac{v^4}{122500}

v=42.3877ft/s

Now calculate the angular velocity of the motorcycle

v = r\omega

42.3877 = 350\omega

\omega = 0.1211rad/s

Calculate the angular acceleration of the motorcycle

a_t = r\alpha

1.1 = 350\alpha

\alpha = 3.1428*10^{-3}rad/s^2

Calculate the time needed by the motorcycle to reach an acceleration of

5.25ft/s^2

\omega = \alpha t

0.1211 = 3.1428*10^{-3}t

t = 38.53s

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a_r = \frac{v^2}{R}

a_r = \frac{21.5^2}{350}

a_r =1.3207ft/s^2

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a = \sqrt{a_t^2+a_r^2}

a = \sqrt{(1.1)^2+(1.3207)^2}

a = 1.7187ft/s^2

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

3 0
3 years ago
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