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An npn BJT has emitter, base, and collector doping levels of 1019 cm????3, 5 1018 cm????3, and 1017 cm????3, respectively. It is

Darina [25.2K]

Answer:

Explanation:

The answer to the given problem is been solved in the fine attached below.

8 0

3 years ago

Su prepares a presentation about the industrial revolution. She wants to change the font on the second slide of her presentation

choli [55]

Font of the slide can be edited in the presentations :-

Explanation

By following these steps Su can change the presentation font of her second slide.

1. Edit- As Su open her second slide , there will be a option of "Edit" on the top right corner of the screen.

2. Font- As the various options are available on the screen such as font size , add image etc ,click on the option "Font"

3. Drop Down- As you click on font option , it will be followed by a drop down option with various type of fonts. Su can select the font she wishes to add in her presentation and look it more precise by adjusting the Font Size.

4 Save- Save the changes made in the slide for future.

8 0

2 years ago

Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa a

Elena L [17]

Answer:

$S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

The entropy change of the carbon dioxide is -0.1104 kJ/kg.K

Explanation:

We are given that carbon dioxide undergoes a process in a closed system.

We are asked to find the entropy change of the carbon dioxide with the assumption that the specific heats are constant.

The entropy change of the carbon dioxide is given by

$S_2 - S_1 = C_p \ln (\frac{T_2}{T_1}) - R\ln (\frac{P_2}{P_1})$

Where Cp is the specific heat constant

Cp = 0.846 kJ/kg.K

R is the universal gas constant

R = 0.1889 kJ/kg.K

T₁ and T₂ is the initial and final temperature of carbon dioxide.

P₁ and P₂ is the initial and final pressure of carbon dioxide.

$S_2 - S_1 = 0.846 \ln (\frac{800}{400}) - 0.1889\ln (\frac{2000}{50})$

$S_2 - S_1 = 0.846(0.69315) - 0.1889(3.6888)$

$S_2 - S_1 = 0.5864 - 0.6968$

$S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

Therefore, the entropy change of the carbon dioxide is -0.1104 kJ/kg.K

6 0

2 years ago

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 5 mm diameter, 40 m long horiz

ra1l [238]

Answer:

Flow rate is $1.82\times 10^{-8} m^{3}/s$

Explanation:

Given information

Density of oil, $\rho_{oil}= 850 Kg/m^{3}$

kinematic viscosity, $v= 0.00062 m^{2} /s$

Diameter of pipe, D= 5 mm= 0.005 m

Length of pipe, L=40 m

Height of liquid, h= 3 m

Volume flow rate for horizontal pipe will be given by

$\bar v=\frac {\triangle P\pi D^{4}}{128\mu L}$ where $\mu$ is dynamic viscosity and $\triangle P$ is pressure drop

At the bottom of the tank, pressure is given by

$P_{bottom}=\rho_{oil} gh=850 Kg/m^{3}\times 9.81 m/s^{2}\times 3 m= 25015.5 N/m^{2}$

Since at the top pressure is zero, therefore change in pressure a difference between the pressure at the bottom and the top. It implies that change in pressure is still $25015.5 N/m^{2}$

Dynamic viscosity, $\mu=\rho_{oil}v= 850 Kg/m^{3}\times 0.00062 m^{2}/s=0.527 Kg/m.s$

Now the volume flow rate will be

$\bar v=\frac {25015.5 N/m^{2}\times \pi \times 0.005^{4}}{128\times 0.527 Kg/m.s \times 40}=1.82037\times 10^{-8} m^{3}/s\approx 1.82\times 10^{-8} m^{3}/s$

Proof of flow being laminar

The velocity of flow is given by

$V_{flow}=\frac {\bar v}{A}=\frac {1.82\times 10^{-8} m^{3}/s}{0.25\times \pi\times 0.005^{2}}=0.000927104 m/s$

Reynolds number, $Re=\frac {\rho_{oil} v_{flow} D}{\mu}=\frac {850 Kg/m^{3}\times 0.000927104 m/s\times 0.005}{0.527 kg/m.s}=0.007476648$

Since the Reynolds number is less than 2300, the flow is laminar and the assumption is correct.

5 0

3 years ago

The worst case signal-to-noise ratio at the output of an FM detector occurs when: ________

AlexFokin [52]

Answer:

a. the desired signal is 90 degrees out of phase with the intelligence signal.

Explanation:

The signal to noise ratio of FM detector is defined as function of modulation index for SSB FM signal plus narrow band Gaussian noise at input. The ratio is usually higher than 1:1 which indicates more signals than noise.

6 0

2 years ago