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Komok [63]
3 years ago
11

1. What is an op-amp? List the characteristics of an ideal op-amp

Engineering
1 answer:
Alexus [3.1K]3 years ago
8 0

Answer:

An opamp is an operation amplifier. It takes an input signal and amplifies it on the output side.

An ideal opamp should have infinite impedance at its input, infinite gain on the output, and zero impedance on the output

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PLEASE HELP!! Its easy!!!
Rina8888 [55]

Answer:

C is tire

F is cassette

D is hub

4 0
3 years ago
Read 2 more answers
An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side
Aliun [14]

Answer:

total width bandwidth = 8kHz

Explanation:

given data

transmitter operating = 3.9 MHz

frequencies up to =  4 kHz

solution

we get here upper side frequencies that is

upper side frequencies = 3.9 × 10^{6} + 4  × 10³

upper side frequencies = 3.904 MHz

and

now we get lower side frequencies that is

lower side frequencies = 3.9 × 10^{6} - 4  × 10³

lower side frequencies =  3.896 MHz

and now we get total width bandwidth

total width bandwidth = upper side frequencies  - lower side frequencies

total width bandwidth = 8kHz

6 0
3 years ago
An office building is served by an air-cooled chiller currently operating at 115 tons (404.5 kW). The measured chilled water sup
Andrei [34K]

Answer:

B.197 gpm and 12.4 L/s

Explanation:

Given that

Load Q = 404.5 KW

Water inlet temperature= 6.1 °C

Water outlet temperature= 13.9°C

We know that specific heat for water

C_p=4.187\ \frac{KJ}{kg.K}

Now from energy balance

Q=\dot{m}C_p\Delta T

by putting the values

Q=\dot{m}C_p\Delta T

404.5=\dot{m}\times 4.187(13.9-6.1)

\dot{m}=12.38\ \frac{kg}{s}     (1 Kg/s = 15.85 gal/min)

We can say that

\dot{m}=196.31\ \frac{gal}{min}

We know that

\dot{m}=\rho\times volume\ flow\ rate

12.38=1000 x volume flow rate

volume\ flow\ rate\ = 12.38\times 10^{-3}\ \frac{m^3}{s}

So

volume flow rate = 12.38 L/s

So the option B is correct.

8 0
3 years ago
Write a script named dif.py. This script should prompt the user for the names of two text files and compare the contents of the
Ainat [17]

Answer:

1

Created on Nov 3, 2018 @author: ASLand

7import atexit

#Read, nanes of both files

Rrintll"Enter tvo files to be compared below

userliamel input ("Enter the nome of the first file: ")

userliame2 input("Enter the name of the second file: ")

ROpen each file

f1 - open(userNamel, r')

@17 f2 = opan(useriame 2, )

tread all the lines into a list

d1 f1.readlines ()

d2 f2.readlines()

re equivalent, print "Yes" else pri

oiterate, and conpare

#11

the

y

if dl == d2:

print("Yes")

atexit

elif for i in range(@, min(len (d1), len(d2))):

if di[i]!=d2[i]:

PCint("No")

print(d1[i])

pcint(d2[])

7 0
3 years ago
Air, at a pressure of 700 kPa and a temperature of 80°C, flows through a convergent– divergent nozzle. The inlet area is 0.005 m
vodomira [7]

Answer: The complete part of the question is to find the exit velocity

Explanation:

Given the following parameters

Inlet pressure = 700kpa

outlet pressure = 40kpa

Temperature = 80°C = 353k

mass flow rate =  1 kg/s

The application of the continuity and the bernoulli's equation is employed to solve the problem.

The detailed steps and the appropriate formula is as shown in the attached file.

7 0
3 years ago
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