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Degger [83]
3 years ago
14

A flywheel in a motor is spinning at 530 rpm when a power failure suddenly occurs. The flywheel has mass 40.0kg and diameter 75.

0cm . The power is off for 39.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 250 complete revolutions. At what rate is the flywheel spinning when the power comes back on? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?
Physics
1 answer:
aleksley [76]3 years ago
5 0

Answer:

 w = 25.05 rad / s ,     α = 0.7807 rad / s² ,   θ = 1972.75

Explanation:

This is a kinematic rotation exercise, let's start by looking for the acceleration when the engine is off

            θ = w₀ t - ½ α t²

            α = (w₀t - θ) 2/t²

           

let's reduce the magnitudes to the SI system

 

w₀ = 530 rev / min (2pi rad / 1 rev) (1 min / 60 s) = 55.5 rad / s

θ = 250 rev (2pi rad / 1 rev) = 1570.8 rad

 

let's calculate the angular acceleration

           α = (55.5 39 - 1570.8) 2/39²

           α = 0.7807 rad / s²

having the acceleration we can calculate the final speed

           w = w₀ - ∝ t

           w = 55.5 - 0.7807 39

           w = 25.05 rad / s

the time to stop w = 0

           0 = wo - alpha t

           t = wo / alpha

           t = 55.5 / 0.7807

           t = 71.09 s

         

the angle traveled

       w² = w₀⁹ - 2 α θ

       w = 0

      θ = w₀² / 2α

let's calculate

      θ = 55.5 2 / (2 0.7807)

        θ = 1972.75

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3 years ago
4. A car, initially traveling east with a speed of
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150 m

Explanation:

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u=5m/s

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iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weig
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2 years ago
A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas
Radda [10]

Answer:

change in momentum, \Delta p=7.65 \,kg.m.s^{-1}

  • \Delta p_x= 6.6251 \,kg.m.s^{-1}
  • \Delta p_y= 3.825 \,kg.m.s^{-1}

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  • F_x=125.0018\,N
  • F_y=72.1698\,N

Explanation:

Given:

angle of kicking from the horizon, \theta= 30^{\circ}

velocity of the ball after being kicked, v=18 m.s^{-1}

mass of the ball, m=0.425\, kg

time of application of force, t=5.3\times 10^{-2}\,s

We know, since body is starting from the rest

\Delta p=m.v.....................(1)

\Delta p=0.425\times 18

\Delta p=7.65 \,kg.m.s^{-1}

Now the components:

\Delta p_x= 7.65\times cos 30^{\circ}

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similarly

\Delta p_y= 7.65\times sin 30^{\circ}

\Delta p_y= 3.825 \,kg.m.s^{-1}

also, impulse

I=F\times t.........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

F\times t=m.v

F\times 5.3\times 10^{-2}= 7.65

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Now, the components

F_x=144.3396\times cos 30^{\circ}

F_x=125.0018\,N

&

F_y=144.3396\times sin 30^{\circ}

F_y=72.1698\,N

6 0
3 years ago
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