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:
R = 545.38 m ; θ = 28.43°
Explanation:
given,
for town A : d₁ = 335 km at an direction of 20° north of east
for town B : d₂ = 245 km at 30.0° west of north from town A
x = 335 cos 20° + 245 sin 30°
x = 437.29 m
y = 335 sin 20° + 245 cos 30°
y = 326.75 m


R = 545.38 m

θ = 28.43°
Answer: Throwing some cargo out of the boa
Explanation:Pushing its mast, Pushing the front of the boat, Pushing another passenger are the forces in internal frame. Forces in the internal frame will not push the boat.
Throwing some cargo out of the boat will create force in external frame of the boat. This external force will push the boat and boat starts moving.