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iragen [17]
4 years ago
8

A 30-kg child starts at the center of a playground merry-go-round that has a radius of 2.1 m and rotational inertia of 500 kg⋅m2

and walks out to the edge. The merry-go-round has a rotational speed of 0.20 s−1 when she is at the center?
Physics
1 answer:
Sever21 [200]4 years ago
8 0

Answer:

\omega = 0.16 s^{-1}

Explanation:

As we know that there is no torque on the system of merry go round

so the angular momentum will remain conserved

so here we have

I_1\omega_0 = (I_1 + I_2)\omega

so we will have

I_1 = 500 kg m^2

I_2 = 30(2.1^2)

I_2 = 132.3 kg m^2

\omega_0 = 0.20 s^{-1}

now from above formula

500(0.20) = (500 + 132.3)\omega

\omega = 0.16 s^{-1}

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ahrayia [7]

Answer:

 Height of ramp = 17.49 m

Explanation:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Considering horizontal motion of skier

      Initial velocity = 33 m/s, Displacement = 63 m, acceleration = 0 and we need to find time taken to reach ground by the skier.

    63= 33t+\frac{1}{2} *0*t^2\\ \\ t=1.909seconds

The vertical distance traveled in 1.909 seconds is the height of  ramp

      Initial velocity = 0 m/s, acceleration = acceleration due to gravity =  9.8 m/s^2, time = 1.909 s and we need to find displacement.

      s= 0*1.909+\frac{1}{2} *9.8*1.909^2\\ \\ s=17.49 m

    So height of ramp = 17.49 m

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Explanation:

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Answer:

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