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ivanzaharov [21]
2 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]2 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]2 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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For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same.
andrey2020 [161]

Answer:

he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values ​​where some natural frequency of the system coincides with the excitation frequency.

In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.

From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

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Arturiano [62]

Answer:

Unit Conversion

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An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis. 
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A 3.50-m-long, 380kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new b
LenKa [72]

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W_b*1.75 + W_w*2.4 = 3727.8 *1.75 + 725.94 * 3.5 = 9064.44 Nm

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