In a certain semiconducting material the charge carriers each have a charge of 1.6 x 10-19C. How many are entering the semiconductor per second when the current is 2.0 nA?

Answers

Answer 1
Answer:

Answer:

1124923453 electrons

Explanation:

The formula for charge in coulomb ,C =Current in amperes, A * Time in seconds, s

Given in question ;

Charge = 1.6 * 10⁻¹⁹ C

Current = 2.0 nA = 2*e⁻⁹ A

Calculate time in seconds as

1.6 * 10⁻¹⁹ = 2*e⁻⁹ * t

1.6 * 10⁻¹⁹ /2*e⁻⁹  = t

6.48e⁻¹⁶  s = t

So using t=6.48e⁻¹⁶  s and current =2*e⁻⁹ A , the charge will be;

C = 2*e⁻⁹  * 6.48e⁻¹⁶  =1.799e⁻¹⁰ C

But 1 coulomb = 6.25 x 10¹⁸ electrons

so 1.799e⁻¹⁰ C = ?,,,,{ 6.25 x 10¹⁸} *{1.799e⁻¹⁰} =  1124923453.06 electrons


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A juggler throws a bowling pin straight up with an initial speed of 9.20m/s. How much time elapses before the pin reaches the juggler's hands?

Answers

Answer:

1.87 s

Explanation:

Initial speed of throw = 9.20 m/s

Net vertical displacement = 0

The bowling pin would be in free fall i.e. a = 9.8 m/s²

Use the second equation of motion:

s = ut + 0.5at²

0 = (9.20)t-0.5(9.8)(t²)

9.20 = 4.9 t

⇒t = 1.87 s

Thus, the total time of flight, the time elapses before the bowling pin falls in juggler's hand is 1.87 s.

A juggler throws a bowling pin straight up with an initial speed, the time that elapses before the pin reaches the juggler's hands is 1.88 s.

Given:
Initial speed, u = 9.2 m/s

The time can be calculated from the second equation of motion. The second equation of motion provides a relation between height, initial speed, acceleration, and time respectively.

From the second equation of motion:

h = ut + at²

When the ball reaches the hands, the distance becomes zero. Therefore, the time is:

0 = 9.2t -0.5 × 9.8t²

9.8t = 18.4

t = 18.4÷ 9.8

t = 1.88 s

Hence, the time that elapses before the pin reaches the juggler's hands is 1.88 s.

To learn more about time, here:

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Acid mine drainage is: pollution caused when water, wind or ice erodes the overburden forming erosion drainage channels.
dissolved toxic materials wash from mines into nearby lakes and streams.
a buildup of residual acidity in soils destroys large areas of vegetation.
the acidity of metals in mines exceeds level 9 on the pH scale.

Answers

Answer:

Acid mine drainage is dissolved toxic materials wash from mines into nearby lakes and streams.

Explanation:

Acid mine drainage is the flow of acidic water with pH typically between 2 and 4, and high concentrations of other dissolved toxic materials from mines into nearby lakes and streams. It mainly occurs during metal sulfide mining, when the metal sulfide ore such as pyrite (FeS2) is exposed to water and oxygen from air to produce soluble iron and sulfuric acid.

Microorganisms, especially acidophile bacteria like Acidithiobacillus ferrooxidans grow by pyrite oxidation, i.e., oxidizing the Fe²⁺ in pyrite to Fe³⁺, which again react with pyrite and water to produce sulfuric acid. Then the acidic water flows into nearby water sources and reduces the pH value of water in those sources. As a result, heavy metals such as copper, lead, mercury, etc in other mineral ores also get dissolved into the water. The action of acidophile bacteria also increases the rate and degree of acid-mine drainage process.

The acid mine drainage causes water pollution and adversely affect the aquatic plants and animals. It also results in the contamination of drinking water, corrosion of infrastructures such as bridges, etc.

If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?4.0 x 10^-1 A
1.0 x 10^2 A
1.0 x 10^-10 A
4.0 x 10^0 A

Answers

Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
I = 4 A

Q= I×t

Where Q is charge in Coulombs C,

I is current in Amperes A,

t is in seconds s.

The rate of current flow is defined as the charge passing through a point in the circuit per second.

Rearranging formula gives I=(Q)/(t)

⇒I = (2.0*10^-4)/(5.0*10^-5) = 4 Amperes

∴ current flow = 4.0×10^0 A

Calculate the resistance of a lamp that draw 0.75 A current when connected to 4.5 V.​

Answers

Calculate the resistance using the equation:

V=IR

In the question, the following values are given:

V = 4.5
I = 0.75

Substitute the values into the equation to find R, the resistance:

V = IR

4.5 = 0.75R

R = 4.5 / 0.75

R = 6 ohms

Ilya and Anya each can run at a speed of 7.30 mph and walk at a speed of 4.00 mph. They set off together on a route of length 5.00 miles. Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half. a) How long does it take Anya to cover the distance of 5.00miles? b) Find Anya's average speed. c) How long does it take Ilya to cover the distance? d) Now find Ilya's average speed.

Answers

Answer:

a)0.9675 h

b)5.168 mph

c)0.885 h

d)5.65 mph

Explanation:

The distance travelled by a moving object can be calculated as the velocity multiplied by the time, for this problem we can divide the problem in two stages, one where the are walking and another where they run,

Considering the total time for Anya is t_(a):

She first walks for half the distance (2.5 miles) so:

2.5 miles=V_(w)*t_(a1)

Solving for t_(a1):

t_(a1)=(2.5miles)/(V_(w))=(2.5miles)/(4 mph)=0.625 h

Then she runs the other half of the distance:

2.5 miles=V_(r)*t_(a2)

t_(a2)=(2.5miles)/(V_(r))=(2.5miles)/(7.3 mph)=0.3425 h

Adding this two durations we have the total time it takes for Anya to cover the distance:

t_(a)=0.3425 h+0.625 h=0.9675 h

The average velocity can be calculated as the total distance divided by the total time:

V_(a-av) =(5 miles)/(0.9675 h) =5.168 mph

Considering the total time for Ilya is t_(I):

She first walks for half the time so:

d_(I1)=V_(w)*t_(I)/2

Where V_(w) is the velocity when walking (same for both people). On the second half she runs:

d_(I2)=V_(r)*t_(I)/2

This two distances must sum the total distance 5 miles, so:

d_(I2)+d_(I2)=5 miles

replacing the expressions above:

V_(w)*t_(I)/2+V_(r)*t_(I)/2=(V_(w)+V_(r))*t_(I)/2=5 miles

t_(I)=(5miles*2)/(V_(w)+V_(r))=(10 miles)/(4mph+7.3mph)=0.885 h

so it takes Ilya 0.885 hours to to cover the distance

The average velocity can be calculated as the total distance divided by the total time:

V_(I-av) =(5 miles)/(0.885 h) =5.65 mph

As the speed of a fluid increases, ____.a. the pressure decreases
c. the force decreases
b. the pressure increases
d. the volume decreases

Answers

The correct answer for the question that is being presented above is this one: "a. the pressure decreases." As the speed of a fluid increases, the pressure decreases." The relationship of the speed and the pressure is inversely proportional. As the pressure increases, the speed decreases.

I think the answer is A.