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ololo11 [35]
3 years ago
5

In a certain semiconducting material the charge carriers each have a charge of 1.6 x 10-19

Physics
1 answer:
Natali5045456 [20]3 years ago
4 0

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 car travelled 60 km west in 1.33 h along a straight line, then returned back east for 40 km in 0.67 h. Its average velocity is
Sholpan [36]

Answer:

Option A

Explanation:

Velocity is expressed as distance covered per unit time, with respect to direction. Therefore, v=d/t

Given distance west as 60 km and time as 1.33 then velocity will be

V=60/1.33=45.112781954887 km/h

Rounded off as 45.11 km/h West

Velocity in East will also be given by substituting 40 km for d and 0.67 h for h hence

V=40/0.67=59.701492537313 km/h rounded off as 59.70 km/h East

Taking East as positive then West as negative, the sum of two velocities will be (59.70+-45.11)/2=7.295 km/h East

Approximately 10 km/h East since it is positive

7 0
3 years ago
Two small plastic spheres are given positive electrical charges. When they are 16.0 cm apart, the repulsive force between them h
Drupady [299]

Answer:

q = 7.542 x 10⁻⁷ C = 754.2 nC

Explanation:

The Coulomb's Law gives the magnitude of the force of attraction or repulsion between two charges:

F = kq₁q₂/r²

where,

F = Force of attraction or repulsion = 0.2 N

k = Coulomb's Constant = 9 x 10⁹ N m²/C²

r = distance between charges = 16 cm = 0.16 m

q₁ = magnitude of 1st charge

q₂ = magnitude of 2nd charge

Since, both charges are said to be equal here.

q₁ = q₂ = q

Therefore,

0.2 N = (9 x 10⁹ N m²/C²)q²/(0.16 m)²

(0.2 N)(0.16 m)²/(9 x 10⁹ N m²/C²) = q²

q = √(5.88 x 10⁻¹³ C²)

<u>q = 7.542 x 10⁻⁷ C = 754.2 nC</u>

7 0
3 years ago
A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?​
Ilya [14]

The frequency of the wave is 1\cdot 10^9 Hz

Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

c=f \lambda

where

c is the speed of the wave

f is the frequency

\lambda is the wavelength

For the radio wave in this problem,

\lambda = 0.3 m

c=300,000,000 m/s = 3\cdot 10^8 m/s

Therefore, the frequency is:

f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz

Learn more about waves here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 0
3 years ago
Which of the following would be an example of basic research?
OverLord2011 [107]
<h2>Basic Research - Option B</h2>

Experimenting to determine the fundamental properties of x-rays would be an example of basic research. The experimental primary analysis tries to determine the laws of the events. The primary analysis is expected to know the stuff, its sort, its characteristics and its performance. It examines to get the speculation. To obtain wherewith the stuff performance yet not how to practice these forms for a firm determination.

The research of the characteristics of the particle to explain whatever it is, how it combines with other atoms, how it determines the characteristics of the material are some instances of primary analysis. For example examining to ascertain the basic features of x-rays. Spencer's research on WWII radar technology that drove to the discovery of the microwave furnace.

Edison's research and application of other scientists task is to create the light bulb, and Morrison and Franscioni's research was made to build the Frisbee models of applied research science.  


5 0
4 years ago
Read 2 more answers
Would you expect to find a high or low level of resistance in a copper wire and why?
marysya [2.9K]
Low. Copper has relatively high conductivity.
4 0
4 years ago
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