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11111nata11111 [884]
3 years ago
10

we want to make a schottky diode on one surface of an n-type semiconductor, and an ohmic contact on the other side. the electron

affinity is 5ev, band gap is 1.5ev, and the fermi potential is 0.25ev?

Engineering
1 answer:
RoseWind [281]3 years ago
7 0

Answer:

<h2>hope it helps you see the attachment for further information .....✌✌✌✌✌</h2>

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River models are used to study many different types of flow situations. A certain small river has an average width and depth of
slavikrds [6]

Answer: 7ft x21 I’d be right but yes I am

Explanation: because it is Welty

4 0
3 years ago
Key length is designed to provide desired factor of safety<br> a. True<br> b. False
zhuklara [117]

Answer: true

Explanation:

A key is a machine element that us used to connect the element of a rotating machine to a shaft. It should be noted that the key hinders the relative rotation that may take place between the two parts.

Key length is designed to provide desired factor of safety. It should also be noted that the factor of safety shouldn't be much and the key length is typically limited to the hub length.

8 0
3 years ago
The 2-lb block is released from rest at A and slides down along the smooth cylindrical surface. Of the attached spring has a sti
MA_775_DIABLO [31]

Answer:

L = 4.574 ft

Explanation:

Given:

- The weight of the block W = 2 lb

- The initial velocity of the block v_i = 0

- The stiffness of the spring k = 2 lb/ft

- The radius of the cylindrical surface r = 2 ft

Find:

Determine its unstretched length so that it does not allow the block to leave the surface until θ= 60°.

Solution:

- Compute the velocity of the block at θ= 60°. Use Newton's second equation of motion in direction normal to the surface.

                           F_n = m*a_n

Where, a_n is the centripetal acceleration or normal component of acceleration as follows:

                           a_n = v^2_2 / r

- Substitute:

                          F_n = m*v^2_2 / r

Where, F_n normal force acting on block by the surface is:

                          F_n = W*cos(θ)

- Substitute:

                          W*cos(θ) = m*v^2_2 / r

                          v_2 = sqrt ( r*g*cos(θ) )

- Plug in the values:

                          v^2_2 = 2*32.2*cos(60)

                          v^2_2 = 32.2 (ft/s)^2

- Apply the conservation of energy between points A and B where θ= 60° :

                      T_A + V_A = T_B + V_B

Where,

                      T_A : Kinetic energy of the block at inital position = 0

                      V_A: potential energy of the block inital position

                      V_A = 0.5*k*x_A^2

                      x_A = 2*pi - L            ..... ( L is the original length )

                      V_A = 0.5*2*(2*pi - L)^2 =(2*pi - L)^2

                      T_B = 0.5*W/g*v_2^2 = 0.5*2 / 32.2 *32.2 = 1

                      V_B = 0.5*k*x_B^2 + W*2*cos(60)

                      x_B = 2*0.75*pi - L            ..... ( L is the original length )

                      V_B = 0.5*2*(1.5*pi - L)^2 + 2*1 = 2 + ( 1.5*pi - L )^2

- Input the respective energies back in to the conservation expression:

                      0 + (2*pi - L)^2 = 1 + 2 + ( 1.5*pi - L )^2

                      4pi^2 - 4*pi*L + L^2 = 3 + 2.25*pi^2 - 3*pi*L + L^2

                      pi*L = 1.75*pi^2 - 3

                         L = 4.574 ft

                         

3 0
3 years ago
A reservoir manometer has vertical tubes of diameter D518 mm and d56 mm. The manometer liquid is Meriam red oil. Develop an alge
Nat2105 [25]

Answer:

Explanation:

Given that :

the diameter of the reservoir D = 18 mm

the diameter of the manometer d = 6 mm

For an equilibrium condition ; the pressure on both sides are said to be equal

∴

\Delta \ P = \rho _{water} g \Delta h_{water}  = \rho _{oil} g \Delta h_{oil}

\Delta \ P = \rho _{oil} g (x+L) ----- (1)

According to conservation of volume:

A*x = a*L

\dfrac{\pi}{4}D^2x = \dfrac{\pi}{4}d^2 L

x = ( \dfrac{d}{D})^2L

Replacing x into (1) ; we have;

\Delta \ P = \rho _{oil} g ( ( \dfrac{d}{D})^2L+L)

\Delta \ P = \rho _{oil} g \ L  ( ( \dfrac{d}{D})^2+1)

L = \dfrac{\Delta \ P}{\rho _{oil} g \   ( ( \dfrac{d}{D})^2+1)}

Thus; the liquid deflection is : L = \dfrac{\Delta \ P}{\rho _{oil} g \   ( ( \dfrac{d}{D})^2+1)}

when the applied pressure is equivalent to 25 mm of water (gage); the liquid deflection is:

L = \dfrac{\Delta \ P}{\rho _{oil} g \   ( ( \dfrac{d}{D})^2+1)}

L = \dfrac{\rho_{water} \g \Delta \ h}{\rho _{water} SG_{oil}g \   ( ( \dfrac{d}{D})^2+1)}

L = \dfrac{\g \Delta \ h}{SG_{oil}g \   ( ( \dfrac{d}{D})^2+1)}

L = \dfrac{25}{0.827    ( ( \dfrac{6}{18})^2+1)}

L = 27.21  mm

4 0
3 years ago
MATLAB can solve a variety of engineering problems including those requiring simulating differential equations and iterative num
frez [133]

Answer:

A. True

Explanation:

MATLAB may be defined as a programming platform that is designed specifically for the engineers as well as the scientists to carry out different analysis and researches.

MATLAB makes use of a desktop environment which is tuned for certain iterative analysis and the design processes with a programming language which expresses matrix as well as array mathematics directly.

Thus the answer is true.

3 0
3 years ago
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