hey guys a got a few problems i need help with im really desperate for help they're due soon online :(

1.Suppose that a constant current of 89 mA flows in a conductor.

Find the absolute value of the total charge that passes through the conductor in 3 minute(s).
Charge = ______ C

[b]b.) Number of electrons = ?

2.) A bar semiconduction material at 30 degrees Celsius has a density of 5E10 holes/cm3 and 5E10 electrons/cm3. The hole charge is 1.6E-19 C, and the charge of an electron is -1.6E-19 C.

The cross sectional area of the bar is [/B]9 cm2. When a voltage of 7V is applied across the ends of the bar, the holes travel with average velocity 8000 cm/s and the electrons travel in the opposite direction with velocity -100000 cm/s.

a) What is the flow rate of holes? ___ holes
b) What is the flow rate of electrons? __
c) What is the current? The current is referenced in the direction of the flow of holes ___ A

LAST ONE 3.) You wish to make 11ohm electric heating coil from 18 gauge nichrome wire. The resitivity of nichrome is 1E-4 ohm-cm and the cross-sectional area of 18 gauge wire is 0.001276 in2.

What length of wire in meters is needed? ___ meters

THANK YOU very much its very appreciated <3

1. Q (Coulombs) = [Current(Amps)]*180 s
(b) Divide Q by the electron charge e=1.6*10^-19 C to get the number of electrons.

2. a) hole flow rate = (hole density) * (hole drift velocity) * (bar area)
b) electron flow rate = (free electron density) * (electron drift velocity) * (bar area)
c) current = sum of (a)and (b), multiplied by the electron charge e.

3. R (ohms)= Length) * (resistivity) /Area

L = (R* Area)/(resistivity)
You must convert the area in in^2 to cm^2 before using the formula above. The answer will be in cm. Then convert that to meters by dividing by 100 cm/m

For problem 1:

To find the absolute value of the total charge that passes through the conductor in 3 minutes, you need to multiply the current (89 mA) by the time (3 minutes).

Since the current is given in milliamperes (mA), you first need to convert it to amperes (A). There are 1000 mA in 1 A, so the current is 0.089 A.

Next, you convert the time from minutes to seconds since the unit of current is in amperes (A). There are 60 seconds in 1 minute, so 3 minutes is 3 * 60 = 180 seconds.

Now you can calculate the charge by multiplying the current (in amperes) by the time (in seconds):

Charge = Current * Time
Charge = 0.089 A * 180 s
Charge = 16.02 C

So the absolute value of the total charge that passes through the conductor in 3 minutes is 16.02 Coulombs.

To find the number of electrons, you divide the charge by the charge of a single electron (1.6 * 10^-19 C):

Number of electrons = Charge / Charge of a single electron
Number of electrons = 16.02 C / (1.6 * 10^-19 C)
Number of electrons = 1.00125 * 10^20 electrons

So the number of electrons is approximately 1.00125 * 10^20 electrons.

For problem 2:

a) To find the flow rate of the holes, you multiply the hole density (5E10 holes/cm^3) by the hole drift velocity (8000 cm/s) and the cross-sectional area of the bar (9 cm^2):

Hole flow rate = Hole density * Hole drift velocity * Bar area
Hole flow rate = 5E10 holes/cm^3 * 8000 cm/s * 9 cm^2
Hole flow rate = 3.6E15 holes

So the flow rate of the holes is 3.6E15 holes.

b) To find the flow rate of the electrons, you multiply the electron density (5E10 electrons/cm^3) by the electron drift velocity (-100000 cm/s) and the cross-sectional area of the bar (9 cm^2):

Electron flow rate = Electron density * Electron drift velocity * Bar area
Electron flow rate = 5E10 electrons/cm^3 * -100000 cm/s * 9 cm^2
Electron flow rate = -4.5E16 electrons

So the flow rate of the electrons is -4.5E16 electrons.

c) To find the current, you add the flow rates of the holes and electrons, and then multiply by the charge of an electron (1.6E-19 C):

Current = (Hole flow rate + Electron flow rate) * Charge of an electron
Current = (3.6E15 holes + (-4.5E16 electrons)) * 1.6E-19 C
Current = -5.76E-4 A

So the current is approximately -5.76E-4 Amps (reference in the direction of the flow of holes).

For problem 3:

To find the length of wire needed to make an 11-ohm electric heating coil, you use the formula for the resistance of a wire:

Resistance (ohms) = (Length * Resistivity) / Area

You are given the resistance (11 ohms), the resistivity of the nichrome wire (1E-4 ohm-cm), and the cross-sectional area of the 18-gauge wire (0.001276 in^2).

First, you need to convert the cross-sectional area from square inches to square centimeters. There are 2.54 cm in 1 inch, so to convert square inches to square centimeters, you square the conversion factor:

Area (cm^2) = (0.001276 in^2) * (2.54 cm/in)^2
Area = 0.001276 in^2 * 6.4516 cm^2/in^2
Area = 0.0082493 cm^2

Next, you can use the formula to solve for the length of the wire:

Length (cm) = (Resistance * Area) / Resistivity
Length = (11 ohms * 0.0082493 cm^2) / 1E-4 ohm-cm
Length = 899.42 cm

So the length of wire needed to make an 11-ohm electric heating coil is approximately 899.42 centimeters.

To convert this to meters, divide by 100 cm/m:

Length (meters) = 899.42 cm / 100 cm/m
Length = 8.9942 meters

So the length of wire needed is approximately 8.9942 meters.