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3 years ago

Select the characteristics of an ideal operational amplifier.

Engineering

1 answer:

SpyIntel [72]3 years ago

3 0

Answer:

Numbers 4, 6, & 7 are correct

Explanation:

4- this allows the op amp to have zero voltage so that maximum voltage is transferred to output load.

6- this ensures that op amp doesn't cause loading in the original circuit, high input impedance would not deter the circuit from pulling current from it.

7- high difference between upper and lower frequencies.

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The mathematical relationship between

11Alexandr11 [23.1K]

4) Ohms law thats the answer

7 0

3 years ago

Viscous effects are negligible outside of the hydrodynamic boundary layer. (3 points) a. True b. False

Valentin [98]

Answer:

I would say false but I am not for sure

8 0

3 years ago

A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.

Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = $\frac{PL}{AE}$ ................1

here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = $\frac{PL}{AE}$

δ = $\frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}$

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0

3 years ago

The ice on the rear window of an automobile is defrosted by attaching a thin, transparent, film type heating element to its inne

pshichka [43]

Answer:

A)Q = 1208.33 W/m²

B)K = 0.138 W/m.K

Explanation:

We are given;

inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K

outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K

Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K

Thickness, L = 4mm = 0.004m

convection heat transfer coefficient ; hi = 25 W/(m².K)

A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;

Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]

Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;

Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]

Q = 583.33 + 625

Q = 1208.33 W/m²

B) The formula for thermal conductivity is;

K = (QL)/(AΔT)

Where;

K is the thermal conductivity in W/m.K

Q is the amount of heat transferred through the material

L is the distance between the two isothermal planes

A is the area of the surface in square meters

ΔT is the difference in temperature in Kelvin

ΔT = 298K - 263K = 35K

Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;

k = (Q/A) x (L/ΔT)

K = 1208.33 x (0.004/35)

K = 0.138 W/m.K

5 0

3 years ago

Let's model this housing price data! Before we can do this, however, we need to split the data into training and test sets. Reme

Lilit [14]

The program reads in a dataset into a pandas dataframe, and uses the train_test_split function in the sklearn library to split the data into training and test sets. The code goes thus :

import pandas as pd

#import the pandas dataframe and alias it as pd

from sklearn.model_selection import train_test_split

#import the train_test_split function

housing_df = pd.read_csv('housing price.csv')

#read in the housing data

features_df = df.iloc[:,1:]

#seperate the features from the label ;

target_df = df.iloc[:,0]

#put the label into a seperate dataframe as well.

X_train, X_test, Y_train, Y_test = train_test_split(features_df, target_df, test_size = 0.1, random_state = 1)

#uses tuple unpacking to randomly assign the data each of the 4 variables.

#Test size is test percent of the entire dataset

Learn more :brainly.com/question/4257657?referrer=searchResults

3 0

3 years ago