The force applied on the spring to stretch it is 13.2 N.
Hooke's law is a law of elasticity discovered by the English scientist Robert Hooke in 1660 that states that the displacement or size of a deformation is directly proportional to the deforming force or load for relatively small deformations of an object. When the load is removed under these conditions, the object returns to its original shape and size.
According to Hooke's law, F = k*e
where F is the force on the spring
k is force constant
and e is extension
F = (110)*(0.12)
F = 13.2 N
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Becuse your weighting with chalk that has pigment
To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

Here



Mass inside the orbit in terms of Volume and Density is

Where,
V = Volume
Density
Now considering the volume of the star as a Sphere we have

Replacing at the previous equation we have,

Now replacing the mass at the gravitational acceleration formula we have that


For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

At the same time the general expression for the centripetal acceleration is

Where
is the orbital velocity
Using this expression in the left hand side of the equation we have that



Considering the constant values we have that


As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.
So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J