•THAT THE PROPAGATION OF SOUND WAVES NEED MEDIUM TO TRAVEL
•THE MEDIUM SHOULD POSSES ELASTICITY
•FOR THE FASTER PROPAGATION OF SOUND THE PARTICLES SHOULD BE VERY CLOSE TO EACH OTHER
Life darling
☆*:.。. o(≧▽≦)o .。.:*☆
A) 
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
(1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
(2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

Using (2) we can rewrite this as

And using (1), we find

Substituting
into the last equation, we find the value of x:

B) 
In this case, the kinetic energy is 1/10 of the total energy:

Since we have

we can write

And so we find:

It’s will be B because the circuit had a open or close so if that doesn’t work than it’s because it’s open
Answer:
1.70 J
Explanation:
The heat dissipated is the difference in the kinetic energies.
This is given by

and
are the initial and final velocities.
With <em>m</em> = 0.175 kg,

The negative sign appears because energy is lost.