A disk-shaped merry-go-round of radius 2.93 m and mass 135 kg rotates freely with an angular speed of 0.621 rev/s . A 62.4 kg person running tangential to the rim of the merry-go-round at 3.21 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. What is the final angular speed of the merry-go-round? rad/s

2.55 rad/sec

Explanation:

Use conservation of angular momentum

The merry go round has

I = ½mr²

I = ½ * 135 * 2.93²

I = 579 kgm²

the person has

I = mr²

I = 62.4 * 2.93²

I = 536 kgm²

Converting 0.621 rev/sec to rad/sec we have 0.621 * 2π

0.621 * 2 * 3.14 rad/sec

3.9 rad/sec

for the person v/r = w

w = 3.21 / 2.93

w = 1.10 rad/sec

So

579 * 3.9 + 536 * 1.1 = (579+536)*w

2258 + 589.6 = 1115w

2847.6 = 1115w

solve for w

w = 2847.6 / 1115

w = 2.55 rad/sec

Thus, the final angular speed of the merry go round is 2.55 rad/sec

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..

the answer is the first electromagnet (a). this is because it has more coils around the ferromagnetic material. increasing the number of coils increases the surface area in which the metal interacts with the current in the coil. this enables more atoms in the metal to become aligned hence becoming more magnetized.