Suppose that the moment of inertia of a skater with arms out and one leg extended is 2.9 kg⋅m2 and for arms and legs in is 0.90 kg⋅m2 . If she starts out spinning at 4.3 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?

1.33 rev/s

Explanation:

I(arms and leg in) = I₁ = 0.9kgm²

I(arms out and one leg extended) = I₂ = 2.9kgm²

W₁ = 4.3 rev/s

W₂ = ?

Neglecting any external force acting on the skater (air resistance and friction), external torque isn't acting on the system and the angular momentum remains constant.

L(in) = L(out)

L₁ = L₂

L = angular momentum

L = Iw

I₁ w₁ = I₂ w₂

w₂ = (I₁ * w₁) / I₂

w₂ = (0.9 * 4.3) / 2.9

w₂ = 1.33 rev/s

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.

no

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