Answer:
The kinetic energy of the merry-goround after 3.62 s is 544J
Explanation:
Given :
Weight w = 745 N
Radius r = 1.45 m
Force = 56.3 N
To Find:
The kinetic energy of the merry-go round after 3.62 = ?
Solution:
Step 1: Finding the Mass of merry-go-round


m = 76.02 kg
Step 2: Finding the Moment of Inertia of solid cylinder
Moment of Inertia of solid cylinder I =
Substituting the values
Moment of Inertia of solid cylinder I
=>
=> 
=> 
Step 3: Finding the Torque applied T
Torque applied T =
Substituting the values
T = 
T = 81.635 N.m
Step 4: Finding the Angular acceleration
Angular acceleration ,
Substituting the values,


Step 4: Finding the Final angular velocity
Final angular velocity ,
Substituting the values,


Now KE (100% rotational) after 3.62s is:
KE = 
KE =
KE = 544J
Volume will decrease if the heat remains constant
Answer:
0.198 s
Explanation:
Consider the motion of the block before collision
= initial velocity of block as it is dropped = 0 m/s
= acceleration = - g
= time of travel
= final velocity of block before collision
Using the kinematics equation

= mass of the bullet = 0.026 kg
= velocity of block just before collision = 750 m/s
= mass of the block = 5 kg
= final velocity of bullet block after collision = gt
Using conservation of momentum
