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zavuch27 [327]
3 years ago
15

What is the voltage across a semiconductor bar if the current through it is 0.17 A? The electron concentration in the bar is 2.7

E18 cm^-3 and the electron mobility is 1000 cm^2/(V*s). Ignore the hole concentration. The bar length and the area are 0.1 mm and 500 um^2. (UNIT:V) 0.7870 V
Physics
1 answer:
Anastaziya [24]3 years ago
6 0

Answer:

The voltage across a semiconductor bar is 0.068 V.

Explanation:

Given that,

Current = 0.17 A

Electron concentration n= 2.7\times10^{18}\ cm^{-3}

Electron mobility \mu=1000 cm^2/Vs

Length = 0.1 mm

Area = 500 μm²

We need to calculate the resistivity

Using formula of resistivity

\sigma=n\times q\times \mu

\rho=\dfrac{1}{\sigma}

Put the value into the formula

\rho=\dfrac{1}{2.7\times10^{18}\times10^{6}\times1.6\times10^{-19}\times1000\times10^{-4}}

\rho=2\ \mu \Omega m

We need to calculate the resistance

Using formula of resistance

R=\dfrac{\rho l}{A}

R=\dfrac{2\times10^{-6}\times0.1\times10^{-3}}{500\times(10^{-6})^2}

R=0.4\ \Omega

We need to calculate the voltage

Using formula of voltage

V= IR

Put the value into the formula

V=0.17\times0.4

V=0.068\ V

Hence, The voltage across a semiconductor bar is 0.068 V.

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Answer:

The speed of the police car is 294 m/s

Explanation:

Given;

frequency of the siren in air, f = 280 Hz

speed of sound in air, v = 343 m/s

Determine the wavelength of the sound in air to the stationary car:

v = fλ

where;

λ is wavelength of the sound

λ = v/f

λ = 343 / 280

λ = 1.225 m

Now, determine the speed at which the police car is approaching the stationary car;

The actual frequency of the police car, F = 240 Hz

V = Fλ

Where;

V is speed of the police car

λ is the distance between the police car and the stationary car, (wavelength)

V = 240 x 1.225

V = 294 m/s

Therefore, the speed of the police car is 294 m/s

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Amy is in-line skating. Her mass is 50 kg. She is rolling forward (north) on a flat section of road at 10 m/s. The rolling frict
Licemer1 [7]

Answer:

The right solution is "0.50 m/s²". A further explanation is provided below.

Explanation:

The given values are:

Mass,

m = 50 kg

Speed,

v = 10 m/s

Rolling friction acting backward (south),

f = 10 N

Air resistance acting backward (south),

F_T = 15 N

The total force acting will be:

⇒  F = -f-F

On substituting the given values, we get

⇒      =-10-15

⇒      =-25 \ N

Now,

⇒  a = \frac{F}{m}

On substituting the given values, we get

⇒     =\frac{-25}{50}

⇒     =-0.50 \ m/s^2

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