<h3><u>Answer</u>;</h3>
= F0 L ( 1 - 1/e )
<h3><u>Explanation;</u></h3>
Work done is given as the product of force and distance.
In this case;
Work done = ∫︎ F(x) dx
= F0 ∫︎ e^(-x/L) dx
= F0 [ -L e^(-x/L) ] between 0 and L
= F0 L ( 1 - 1/e )
Answer:

Explanation:
Given that,
Radius of a spherical shell, r = 0.7 m
Torque acting on the shell, 
Angular acceleration of the shell, 
We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

I is the rotational inertia of the shell

So, the rotational inertia of the shell is
.
Explanation:
Period of a mass on a spring is:
T = 2π√(m/k)
T is inversely proportional with the square root of k. So as the spring constant increases, the period decreases.