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ivanzaharov [21]
3 years ago
13

Explain the term majority and minority carriers of a material with respect to the acceptor and donor impurity.​

Physics
2 answers:
Mandarinka [93]3 years ago
4 0

Answer:

The majority charge carriers carry most of the electric charge or electric current in the semiconductor. Hence, majority charge carriers are mainly responsible for electric current flow in the semiconductor. The charge carriers that are present in small quantity are called minority

Ierofanga [76]3 years ago
3 0

Answer:

carriers and so are called majority carriers. A few thermally generated electrons will also exist in the p side; these are termed minority carriers. On the n side the electrons are the majority carriers, while the holes are the minority carriers. Near the junction is a region having no free-charge…

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If 7 ounces of a sports drink contains 110 milligrams of sodium, what is the total number of milligrams of sodium in 20 ounces o
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3 years ago
Two spherical objects have masses 447 kg and 285 kg. Their centers are separated by a distance of 27 m. Find the gravitational a
jonny [76]

Answer:

1.2 * 10' -8N (Check attachment

Explanation:

Check attachment

5 0
3 years ago
horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
3 years ago
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