It is correct, the action is paddling, where you move the water backwards, and the reaction is the boat moving forwards.
Answer:
19.95 J
Explanation:
The center of mass of the ladder is initially at a height of:

The center of mass of the ladder ends at a height of:
=L/2
So, the work done is equal to the change in potential energy which is:
W = PE = 
now 
therefore
W = [mgL/2]×[1 - sin(theta)]
W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]
solving this we get
W = 19.95 J
Answer:

Explanation:
The equivalent of Newton's second law for rotational motions is:

where
is the net torque applied to the object
I is the moment of inertia
is the angular acceleration
In this problem we have:
(net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)
is the moment of inertia
Solving for
, we find the angular acceleration:

Answer:
I would say there is friction against the floor, air resistance, and gravity.
Explanation: