Answer:
(a) 
(b) 
Solution:
As per the question:
Mass of the object, m = 1.30 kg
Length of the rod, L = 0.780 m
Angular speed, 
Now,
(a) To calculate the rotational inertia of the system about the axis of rotation:
Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:

(b) To calculate the applied torque required for the system to rotate at constant speed:
Drag Force, F = 

Answer:
Part a) When collision is perfectly inelastic

Part b) When collision is perfectly elastic

Explanation:
Part a)
As we know that collision is perfectly inelastic
so here we will have

so we have

now we know that in order to complete the circle we will have


now we have

Part b)
Now we know that collision is perfectly elastic
so we will have

now we have


Answer:
2274 J/kg ∙ K
Explanation:
The complete statement of the question is :
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
= mass of metal = 400 g
= specific heat of metal = ?
= initial temperature of metal = 100 °C
= mass of aluminum cup = 100 g
= specific heat of aluminum cup = 900.0 J/kg ∙ K
= initial temperature of aluminum cup = 15 °C
= mass of water = 500 g
= specific heat of water = 4186 J/kg ∙ K
= initial temperature of water = 15 °C
= Final equilibrium temperature = 40 °C
Using conservation of energy
heat lost by metal = heat gained by aluminum cup + heat gained by water
