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inessss [21]
2 years ago
8

The p-type or N-type semiconductor is not of practical use. Why?​

Physics
1 answer:
Savatey [412]2 years ago
6 0

Explanation:

The p-type semiconductor doesn't have greater energy than the n-type semiconductor. This is because the semiconductor itself doesn't really have an energy. Charge carriers (electrons/holes) have energies. The relative positioning of the bands tell you how much energy an electron or a hole must have for it to exist.

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Rocket engineers use newton's third law during launch. identify the action force.
rosijanka [135]
The answer is A. Newton's third law of motion states that for every action, there is an equal and opposite reaction. A rocket exerts a large force on the gas that is in the rocket chamber (action). The gas thus exerts a large reaction force forward on the rocket (reaction). The large reaction force is called thrust.
6 0
3 years ago
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What solutes do not dissolve well in water?
hammer [34]

Answer:

Many substances do not dissolve in water and that is because they are non-polar and do not interact well with water molecules. A common example is oil and water. Oil contains molecules that are non-polar, thus they do not dissolve in water.

Explanation:

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3 years ago
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Saturated steam at 125 kpa is compressed adiabatically in a centrifugal compressor to 700 kpa at the rate of 2.5 kg⋅s−1. the com
Tpy6a [65]
M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h =  0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW
5 0
3 years ago
An object becomes positively charged when it
Montano1993 [528]

Answer:

D

Explanation:

8 0
3 years ago
The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would
natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a)v_{es} =v_{rms}= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

v_{rms} = \sqrt{ \frac{3RT}{M}}

So, v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}

multiply each side by M_{o2}, so we have

v_{rms,o2}^2 *M_{o2} =3RT_{o2}

solving for temperature T_{o2}

T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}

In the question given,v_{rms} =v_{es}

T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}

T_{o2} =1.58*10^5 K

7 0
3 years ago
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