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Phantasy [73]
2 years ago
13

Help meh in this question​

Physics
1 answer:
V125BC [204]2 years ago
7 0

  • Radius=r=R_o/2
  • angular velocity=w=v_o
  • mass=m

We know

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

For Rotational motion

\boxed{\sf v=r\omega}.

Putting value

\\ \qquad\quad\sf{:}\dashrightarrow K.E=\dfrac{1}{2}m(r\omega)^2

\\ \qquad\quad\sf{:}\dashrightarrow K.E=\dfrac{1}{2}mr^2\omega^2

\\ \qquad\quad\sf{:}\dashrightarrow K.E=\dfrac{1}{2}m\left(\dfrac{R_o}{2}\right)^2v_o^2

\\ \qquad\quad\sf{:}\dashrightarrow K.E=\dfrac{1}{4}mv_o^2

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<span>6) On Earth, the gravity acceleration is </span>g=9.81 m/s^2<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is 
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Harlamova29_29 [7]

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