Please help and do it step by step all parts of questions and use:

T = time(s)
I=current (amps)
Q=net charge on object (Coulombs)
N=no electrons (and write if added or removed from object and how)
E=elementary charge

A piece of fuse wire has a diameter of 0.22mm and carries a current of 5.0A. There are 1.0 x 10^29 electrons per metre cubed. Calculate the mean drift velocity of the electrons in the wire.

To calculate the mean drift velocity of electrons in the wire, we will use the formula:

v = I / (n * q * A)

Where:
- v is the mean drift velocity of the electrons
- I is the current through the wire (given as 5.0A)
- n is the number of electrons per unit volume (given as 1.0 x 10^29 electrons per cubic meter)
- q is the elementary charge (given as 1.6 x 10^-19C)
- A is the cross-sectional area of the wire

Let's calculate the cross-sectional area of the wire first:

1. The diameter of the wire is given as 0.22mm.
To calculate the radius, divide the diameter by 2: 0.22mm / 2 = 0.11mm.

2. Convert the radius to meters by dividing by 1000: 0.11mm / 1000 = 0.00011m.

3. Now, use the formula for the area of a circle: A = π * r^2.
The cross-sectional area of the wire is: A = π * (0.00011m)^2.

4. Calculate the value of A using the value of π as 3.14159: A = 3.14159 * (0.00011m)^2.

Now that we have calculated the cross-sectional area of the wire, we can continue with calculating the mean drift velocity of the electrons:

v = I / (n * q * A)
v = 5.0A / (1.0 x 10^29 electrons/m^3 * 1.6 x 10^-19C * A)

Substitute the calculated value of A into the formula and solve:

v = 5.0A / (1.0 x 10^29 electrons/m^3 * 1.6 x 10^-19C * 3.14159 * (0.00011m)^2)

Simplify and evaluate the expression to get the mean drift velocity of the electrons in the wire.