Complete Question
The speed of a transverse wave on a string of length L and mass m under T is given by the formula

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
Answer:

Explanation:
From the question we are told that
Speed of a transverse wave given by

Maximum Tension is 
Generally making
subject from the equation mathematically we have




Therefore the Linear mass in terms of Velocity is given by

Answer:
a) 4.9*10^-6
b) 5.71*10^-15
Explanation:
Given
current, I = 3.8*10^-10A
Diameter, D = 2.5mm
n = 8.49*10^28
The equation for current density and speed drift is
J = I/A = (ne) Vd
A = πD²/4
A = π*0.0025²/4
A = π*6.25*10^-6/4
A = 4.9*10^-6
Now,
J = I/A
J = 3.8*10^-10/4.9*10^-6
J = 7.76*10^-5
Electron drift speed is
J = (ne) Vd
Vd = J/(ne)
Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)
Vd = 7.76*10^-5/1.3584*10^10
Vd = 5.71*10^-15
Therefore, the current density and speed drift are 4.9*10^-6
And 5.71*10^-15 respectively
The total flux through the cylinder is zero.
In fact, the electric flux through a surface (for a uniform electric field) is given by:

where
E is the intensity of the electric field
A is the surface
is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.
We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because
and
).
On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be

where r is the radius, and where we have taken
since the perpendicular to the surface is parallel to the direction of the electric field, so
. On the second surface, however, the perpendicular to the surface is opposite to the electric field, so
and
, therefore the flux is

And the net flux through the cylinder is

According to my examination I have confirmed that I do NOT repeat do NOT know this .