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vodomira [7]
1 year ago
14

How could angela use the puzzle to model semiconductors? as an n-type semiconductor with the pegs representing electrons and the

holes representing positive charges as an n-type semiconductor with the holes representing electrons and the pegs representing positive charges as a p-type semiconductor with the pegs representing electrons and the holes representing positive charges as a p-type semiconductor with the holes representing electrons and the pegs representing positive charges
Engineering
1 answer:
bogdanovich [222]1 year ago
7 0

Angela can use the puzzle to model semiconductors as option C.  a p-type semiconductor with the pegs representing electrons and the holes representing positive charges.

<h3>What is semiconductor and its example?</h3>

A semiconductor is known to be any substance that is said to be one that has some specific electrical properties that helps it to be able to act as  the framework for computers and other kind of electronic devices.

Note that some examples of semiconductors are said to be silicon, germanium, and other elements such as metalloid staircase" that is shown on a periodic table.

Therefore, Angela can use the puzzle to model semiconductors as option C.  a p-type semiconductor with the pegs representing electrons and the holes representing positive charges.

Learn more about semiconductor from

brainly.com/question/9250618

#SPJ1

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Air enters the 1 m² inlet of an aircraft engine at 100 kPa and 20° C with a velocity of 180 m/s. Determine: a) The volumetric fl
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Answer:

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b) 213.4 kg/s

Explanation:

A_1 = 1 m²

P_1 = 100 kPa

V_1 = 180 m/s

Flow rate

Q=A_1V_1\\\Rightarrow Q=1\times 180\\\Rightarrow Q=180\ m^3/s

Volumetric flow rate = 180 m³/s

Mass flow rate

\dot{m}=\rho Q\\\Rightarrow \dot m=\frac{P_1}{RT} Q\\\Rightarrow \dot m=\frac{100000}{287\times 293.15}\times 180\\\Rightarrow \dotm=213.94\ kg/s

Mass flow rate = 213.4 kg/s

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A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces
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Answer:

a) volume flow rate of air at the inlet is 0.0471 m³/s

b) the velocity of the air at the exit is  8.517 m/s

Explanation:

Given that;

The electrical power Input W_elec = -1400 W = -1.4 kW

Inlet temperature of air T_in = 22°C

Inlet pressure of air p_in = 100 kPa

Exit temperature T_out = 47°C

Exit area of the dyer is A_out = 60 cm²= 0.006 m²

cp = 1.007 kJ/kg·K

R = 0.287 kPa·m3/kg·K

Using mass balance

m_in = m_out = m_air

W _elec = m_air ( h_in - h_out)

we know that h = CpT

so

W _elec = m_air.Cp ( T_in - T_out)

we substitute

-1.4 = m_air.1.007 ( 22 - 47 )

-1.4 =  - m_air.25.175

m_air = -1.4 / - 25.175

m_ air = 0.0556 kg/s

a) volume flow rate of air at the inlet

we know that

m_air = P_in × V_in

now from the ideal gas equation

P_in = p_in / RT_in

we substitute our values

= (100×10³) / ((0.287×10³)(22+273))

= 100000 / 84665

P_in = 1.18 kg/m³

therefore inlet volume flowrate will be;

V_in = m_air / P_in

= 0.0556 / 1.18

= 0.0471 m³/s

the volume flow rate of air at the inlet is 0.0471 m³/s

b) velocity of the air at the exit

the mass flow rate remains unchanged across the duct

m_ air = P_in.A_in.V_in = P_out.A_out.V_out

still from the ideal gas equation

P_out = p_out/ RT_out   ( assume p_in = p_out)

P_out = (100×10³) / ((0.287×10³)(47+273))

P_out  = 1.088 kg/m³

so the exit velocity will be;

V_out = m_air / P_out.A_out

we substitute our values

V_out = 0.0556 / ( 1.088 × 0.006)

= 0.0556 / 0.006528

= 8.517 m/s

 Therefore the velocity of the air at the exit is  8.517 m/s

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3 years ago
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