I think you should hold a stretch for 10-30 seconds
"Reflection" is the change direction of waves when blocked by obstacle
Hi there!
(a)
Recall that:

W = Work (J)
F = Force (N)
d = Displacement (m)
Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

To the nearest multiple of ten:

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.
Thus:

(c)
Similarly, the normal force is perpendicular to the displacement, so:

(d)
Recall that the force of kinetic friction is given by:

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:

In multiples of ten:

(e)
Simply add up the above values of work to find the net work.

Nearest multiple of ten:

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)



Nearest multiple of ten:

The answer is A. Hope this helps. :)
Answer:
2872.8 N
Explanation:
We have the following information
m =n72kg
Δy = 18m
t = 0.95s.
From here we use the equation
Δy=1/2at2 in order to solve for the acceleration.
So a
=( 2x 18m)/(0.95s²)
= 36/0.9025
= 39.9m/s2.
From there we use the equation
F = ma
F=(72kg) x (39.9)
= 2872.8N.
2872.8N is the average net force exerted on him in the barrel of the cannon.
Thank you!