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Dennis_Churaev [7]
3 years ago
8

Using what you know about the NAND gate, draw how you could implement a three-input NAND gate using NMOS and PMOS transistors. T

his circuit should output logic low (0V) only when all three inputs (A, B, and C) are logic high (5V)
Engineering
1 answer:
Yuliya22 [10]3 years ago
4 0

Answer:

use this, it should help you understand

Explanation:

https://www.electronics-tutorials.ws/logic/logic_5.html

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Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w
ivanzaharov [21]

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

5 0
3 years ago
What should always be done before beginning any diagnosis?
vladimir2022 [97]

Answer:

c

Explanation:

if someone is wrong that they can help with

4 0
3 years ago
Ammonia enters the expansion valve of a refrigeration system at a pressure of 1.4 MPa and a temperature of 32degreeC and exits a
AveGali [126]

Answer:

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

Explanation:

given data

pressure p1 = 1.4 MPa = 14 bar

temperature t1 = 32°C

exit pressure = 0.08 MPa = 0.8 bar

to find out

the quality of the refrigerant exiting the expansion valve

solution

we know here refrigerant undergoes at throtting process so

h1 = h2

so by table A 14 at p1 = 14 bar

t1 ≤ Tsat

so we use equation here that is

h1 = hf(t1) = 332.17 kJ/kg

this value we get from table A13

so as h1 = h2

h1 = h(f2)  + x(2) * h(fg2)

so

exit quality  = \frac{h1 - h(f2)}{h(fg2)}

exit quality  = \frac{332.17- 9.04}{1382.73)}

so exit quality = 0.2337 = 23.37 %

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

5 0
3 years ago
4.116 The lid of a roof scuttle weighs 75 lb. It is hinged at corners A and B and maintained in the desired position by a rod CD
babunello [35]

Answer:

(a) The magnitude of force is 116.6 lb, as exerted by the rod CD

(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.

Explanation:

Step by step working is shown in the images attached herewith.

For this given system, the coordinates are the following:

A(0, 0, 0)

B(26, 0, 0)

And the value of angle alpha is 20.95°

Hope that answers the question, have a great day!

5 0
3 years ago
Regeneration can only increase the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than the worki
storchak [24]

Answer:

True, <em>Regeneration is the only process where increases the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than working fluid leaving the compressor</em>.

Option: A

<u>Explanation: </u>

To increase the efficiency of brayton cycle there are three ways which includes inter-cooling, reheating and regeneration. <em>Regeneration</em> technique <em>is used when a turbine exhaust fluids have higher temperature than the working fluid leaving the compressor of the turbine. </em>

<em>Thermal efficiency</em> of a turbine is increased as <em>the exhaust fluid having higher temperatures are used in heat exchanger where the fluids from the compressor enters and increases the temperature of the fluids leaving the compressor. </em>

6 0
3 years ago
Read 2 more answers
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