WILL MARK BRAINLIEST!!!!!!!!!!!!!!!!
3. Jessica stretches her arms out 0.6 m from the center of her body while holding a 2 kg mass in each hand. She then spins around on an ice rink at 1.1 m/s.
a. What is the combined angular momentum of the masses?
b. If she pulls her arms in to 0.15 m, what is her linear speed if the angular momentum remains constant?

Part A - We know that angular momentum is the measure of the amount of rotation of the body and is the product of of the moment inertia and the angular velocity. The equation for the moment of inertia can be expressed as l=mr^2 in which m is the mass and r is the radius of the circle. According to the problem, r is 0.6m. Then, l=2kg*(0.6)^2=0.72kg-m2/s2. We also know that angular velocity is the product of linear velocity and the radius which can be expressed as w=rv. Therefore, w=0.6*11 which equals to 0.66 rad/s. The angular momentum is 0.72 kg-ms/s2*0.66 rad/s. Which equals to 0.48 kg-m^2/s.

Part B - First we use l=mr^2, where m=2 kg and r=0.15m. 0.15 becomes 0.225 and when multiplied with 2 is 0.045, giving us the moment of inertia to plug in the following equation. L=l*w where l is 0.045 and w is 1.1 m/s. 1.1 times 0.045 is 0.0495 which is in other words, her linear speed.

I am rusty with this, but I hope it helps!