To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

Where,
B= Magnetic Field
l = length
= Vacuum permeability
= Vacuum permittivity
Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

Recall that the speed of light is equivalent to

Then replacing,


Our values are given as




Replacing we have,



Therefore the magnetic field around this circular area is 
I am sure it is frequency
The middle one on the list is the correct one.
The first one ... distance divided by time ... is Speed, not force.
The third one ... mass times velocity ... is Momentum, not force.
Answer:

Explanation:
According to the free body diagram, in this case, we have:

Recall that the force of friction is given by:

Replacing and solving for the coefficient of kinetic friction:

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

Finally, we calculate
:

Answer:
Diagram C
Explanation:
We are given that Sulfur with atomic number 16.
We have to find that which diagram shows the electronic configuration of sulfur.
S=16
Its Diagram C