The force constant of the spring is approximately 24.038 newtons per meter.
As we are talking about Simple Harmonic Motion. In this exercise we need to determine the Spring Constant (
), in newtons per meter, from the equation of the Period (
), in seconds, which is described below:
(1)
Where 
is the mass of the moving element, in kilograms.
If we know that 
and 
, then the spring constant of the spring is:
The force constant of the spring is approximately 24.038 newtons per meter.
Please see this question related to Simple Harmonic Motion for further details: brainly.com/question/17315536
We are given with the following:
L = 25 cm
θx = <span>37.5</span>°<span>
</span>θy = 52.5°
vx = 6.8 m/s
B = 0.13i - 0.29j - 0.08k
And we are asked for the emf induced in the rod
E = B L v
Substitute the values for B and L into the equation. For v, use the given velocity along x and the angles to convert it to vector form. Then, solve for E.
Answer:
a)The approximate radius of the nucleus of this atom is 4.656 fermi.
b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Explanation:
= Constant for all nuclei
r = Radius of the nucleus
A = Number of nucleons
a) Given atomic number of an element = 25
Atomic mass or nucleon number = 52
The approximate radius of the nucleus of this atom is 4.656 fermi.
b) 
k=
= Coulombs constant
= charges kept at distance 'a' from each other
F = electrostatic force between charges
Force of repulsion between two protons on opposite sides of the diameter
The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Net force refers to the (vector) sum of all the forces acting on something. It's a mathematical construction, so it's not a single identifiable force. The forces themselves are real, but the net force is not an actual force.
Hope this helps you out.
