Since there is no friction between the ladder and the wall, there can be no vertical force component. That's the tricky part ;)
So to find the weight, divide the 100N <em>normal</em> force by earths gravitational acceleration, 9.8m/s^2

Then;
Draw an arrow at the base of the ladder pointing towards the wall with a value of 30N, to show the frictional force.
Newton's first law of motion says something like "An object remains
in constant, uniform motion until acted on by an external force".
Constant uniform motion means no change in speed or direction.
If an object changes from rest to motion, that's definitely a change
of speed. So it doesn't remain in the state of constant uniform
motion (none) that it had when it was at rest, and that tells us
that an external force must have acted on it.
Answer:
a) I = 464 kg m², b) K = 631 .6 J, c) v = 8.25 m / s
Explanation:
a) the moment of inertia of point particles is
I = ∑ m_i r_i²
in this case
I = 8 5² + 3 (-2) ² + 7 (-6) ²
I = 464 kg m²
b) The kinetic energy is
K = ½ I w²
K = ½ 464 1.65²
K = 631 .6 J
c) linear and angular velocity are related
v = w r
v = 1.65 5
v = 8.25 m / s
-Synodic period is the period of celestial bodies observed on the moving planet(mostly earth)
Sideral period is the period comparing to the fixed stars without motion of the earth involved.
(I will explain the second question with an example, so it's easier to understand)
-For Sideral month for example of the moon it cactually complete one revolution in around 27.3 days.
However, since the earth moves, for us it took some more time to see the moon the same as before (fullmoon to fullmoon) again. That make synodic month of the moon to be around 29.5 days.
Answer: 31.6ft
Explanation:
Check the attachment for the diagram.
According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|
that is 20ft - 10ft = 10ft
According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.
|AE|^2 + |EC|^2 = |AC|^2
10^2 + 30^2 = |AC|^2
100 + 900 = |AC|^2
|AC| = √1000
|AC| = 31.6ft
Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.