A. question: jessica stretches her arms out .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. what is the combined angular momentum of the masses?
work: for this question, i got l =2.64 kg*m^2/s. is this right?
b. question: if she pulls her arms in to 0.15 m, what is her linear speed if the angular momentum remains constant?
work: this is the equation i used, l/(m)(r^2)
which would make l= 2.64 from previous question. r= .15 m. m= 2 kg. v2= unknown
substitution:
v2= 2.64/(.15)(2)(2) is this step right?
what is the linear speed? (i tried to show all my work, so, you know where i got stuck.)

1. 2.64 kg m/s

The angular momentum of each mass is given by:

where

m = 2 kg is the mass

v = 1.1 m/s is the speed

r = 0.6 m is the distance from the center of rotation

Substituting the numbers, we get

Since we have two masses in two different hands, the combined angular momentum will be twice the angular momentum of each mass:

2. 4.4 m/s

The angular momentum must remains constant: this means that the only quantity that will change in the formula will be v, the speed.

The formula for the total angular momentum is:

In this case, we have:

L = 2.64 kg m/s

m = 2 kg

r = 0.15 m (because she pulled her arms to this distance)

Re-arranging the equation, we can find the new speed v:

when heated, the valance electrons increased their speed and transferred the heat from electron to electron.   i just took the test.

i think the answer is a and b

i wish you luck