A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The angular speed of the child has a constant value of 0.233 rad/s. At the instant the child spots the horse, one-quarter of a turn away, the merry-go-round begins to move (in the direction the child is running) with a constant angular acceleration of 0.0136 rad/s2. What is the shortest time it takes for the child to catch up with the horse?

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child:

For the horse:

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

muscle endurance

explanation:

i can do 50 in a row ( but i get so tired after that)

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.