To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as

Both cover a magnitude of 8.32 ft, therefore

Now solving for x we have,





Therefore the shorter stick is 2.695ft long.
Answer:
a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²
b) I = 50.39 kg m²
c) I = 16.8 kg m²
Explanation:
a) Given data:
m = 0.98 kg
a = 4.14 * 4.14
The moment of inertia is:

For 4 particles:

b) Distance from top left mass = x = a/2
Distance from bottom left mass = x = a/2
Distance from top right mass = x = √5 (a/2)
The total moment of inertia is:

c)

Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.
Energy I believe. If there is no energy given or taken the object will not react.
Given mass= 1kg
Weight on earth = mg(gravity of earth) = 9.8N
weight on moon = mg(gravity of moon)= 1.62N
weight on outer space mg(gravity outer space = 0) = 0N