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Anna [14]
3 years ago
5

An n- channel enhancement- mode MOSFET with 50 nm thick HfO2 high- k gate dielectric (Pr = 25) has a flat band voltage of 0.5 V,

and substrate doping of 1018 cm-3. The intrinsic carrier concentration is 1011 cm-3, effective electron channel mobility is 250 cm2/V­s, and Pr = 15. What is the drive current for a 50 om wide and 2 om long device at VG = 3 V and VD = 0.05 V? What is the saturation current at this gate bias?

Engineering
1 answer:
Alexeev081 [22]3 years ago
4 0

Answer:

Find the complete solution in the given attachments

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The first thing you should do is develop a ____________________ to determine what vehicle you can afford.
swat32

The first thing you should do is develop a <u>budget</u> to determine what vehicle you can afford.

<h3>What is an automobile?</h3>

An automobile is also referred to as a vehicle, car or motorcar and it can be defined as a four-wheeled vehicle that is designed and developed to be propelled by an internal-combustion (gasoline) engine, especially for the purpose of transportation from one location to another.

<h3>What is a budget?</h3>

A budget can be defined as a financial plan that is typically used for the estimation of revenue and expenditures of an individual, business organization or government for a specified period of time, often one year.

In this context, we can reasonably infer and logically deduce that the first thing anyone should do is to develop a <u>budget</u> in order to determine what vehicle they can afford.

Read more on budget here: brainly.com/question/13964173

#SPJ1

7 0
1 year ago
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0
3 years ago
A box-shaped aquarium has horizontal dimensions 0.5 m by 1 m, and depth 0.5 m, and is filled two-thirds of the way to the surfac
antoniya [11.8K]

Answer:

3270 N/m^2

Explanation:

we can calculate the pressure difference between the bottom and surface of the tank by applying the equation for the net vertical pressure

Py = - Ph ( g ± a )

for a downward movement

Py = - Ph ( g - a )  ------ ( 1 )

From the above data given will  be

p = 1000 kg/m^3, h = 2/3 * 0.5 = 0.33 m , a =2g , g = 9.81

input values into equation 1  becomes

Py =  -Ph ( g - 2g ) = Phg ------ ( 3 )

Py = 1000 * 0.33 * 9.81

    = 3270 N/m^2

6 0
3 years ago
11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t
lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2}  }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2}  }{4}) } =0.729m/s

It is necessary to get the Reynold's number:

Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517

The overall heat transfer coefficient:

Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} }  }

Here

h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C

Substituting values:

Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278}  } =1855.8923W/m^{2} C

5 0
3 years ago
A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5500 N weight. If the diameter decreases to 0.210 cm, determine th
ss7ja [257]

Answer:

1561.84 MPa

Explanation:

L=20 cm

d1=0.21 cm

d2=0.25 cm

F=5500 N

a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa

lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16

longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3

(assuming a poisson's ration of  0.3)

ε_l =0.16/0.3 = 0.5333

b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)

σ_true = 1561.84 MPa

ε_true = ln( 1+ε_l)= ln(1+0.5333)

ε_true= 0.4274222

The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.

7 0
3 years ago
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