Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum fill for that conduit is 40%?

Answer:

Those products are generally called Work in Process WIP

Explanation:

Work in process (WIP), or work in progress (WIP), goods in process, or in-process inventory in a manufacturing industry/company refer to the company's partially finished goods waiting for completion and eventual sale or the value of these items.

These items are either just being produced or require further processing (like purification, separation, packaging or handling) in a queue or a buffer storage.

Answer:b

Explanation:

A beam is subjected to an upward load in the middle and supported in both ends. The upper fibers i.e. a topmost layer of the beam is in tension while bottom fiber i.e. fiber of bottom layer is in compression because they are under compressive force.

Fibers above the neutral axis are in tension and fibers below the neutral axis are in compression.    

So option b is correct              

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer:

Explanation:

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Larry is exhibiting the defense mechanism known as Identification

Identification is a defense mechanism where an individual unconsciously takes on the characteristics, feelings, or behaviors of another person or group that they admire or identify with. In this case, Larry, who did not participate in sports himself, identifies closely with his school's sports teams.

By wearing the school's colors, cheering for the team, and feeling exhilarated when they win, Larry is psychologically aligning himself with the team and deriving a sense of belonging and accomplishment from their success, as if it were his own. This behavior allows Larry to cope with any feelings of inadequacy or unfulfilled desires he may have regarding his own athletic abilities.

Explanation:

Althought she referenced the article at the end, is imposible to know which part of the article is hers and which part is the expert's so that would be plagiarism.

If she used quotation marks in the words of the expert it would be clear and no plagiarism could be accused.

The scenario describes plagiarism, which occurs when one uses someone else's work or ideas without proper citation. Even though the student referenced the source at the end, not using quotation marks for direct quotes is considered plagiarism.

The scenario mentioned describes plagiarism. Plagiarism is a serious academic offense that occurs when one represents someone else's ideas, words, or work as their own without acknowledging the original source. In this case, the student has used the expert's words and ideas without proper citation (because she did not use quotation marks to indicate the direct quotes).

Although she mentioned the source at the end of her work, failing to use quotation marks for direct quotes is still considered plagiarism. To avoid this, it is essential to use quotation marks when directly quoting from a source and to paraphrase correctly when you are conveying the same idea in your own words.

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Answer:

The viscous shear stress is 968 Pa (or can be written as 968 [tex]N/m^2[/tex]) and the drag force is 968 N.

Explanation:

Viscous shear stress

For a parallel flow of a Newtonian fluid, the shear stress is proportional to the gradient of the velocity,

[tex]\tau_{yx}= \mu \cfrac{du}{dy}[/tex]

Considering a large square plate moving on a thin film of thickness h, the velocity profile is

[tex]u(y) = U\cfrac{y}{h}[/tex]

Thus its derivative will be just

[tex]\cfrac{du}{dy}= \cfrac U h[/tex]

So replacing on the viscous shear stress formula we get

[tex]\tau_{yx}= \mu \cfrac {U}h[/tex]

We can then replace the given information

[tex]\tau_{yx}= 0.968 \times \cfrac{kg}{m\times  s}\times \cfrac{0.1 \cfrac ms}{0.0001 \, m}[/tex]

Evaluating we get

[tex]\boxed{\tau_{yx}=968\, Pa}[/tex]

The viscous shear stress is 968 Pascals or [tex]N/m^2[/tex].

Drag force

Using the given equation for the drag force we have

[tex]D = \tau_{yx} A[/tex]

And since we have a large square plate of sides L we can write the area as

[tex]A = L^2[/tex]

So the drag force is

[tex]D = \tau_{yx} L^2[/tex]

Replacing values

[tex]D = 968\, Pa \times (1\, m)^2[/tex]

We get

[tex]\boxed{D= 968 \,N}[/tex]

The drag force is 968 N.

Answer:

Consider the following code.

Explanation:

Code:

Unix Terminal> cat loan_calc.py

#!/usr/local/bin/python3

import locale

from decimal import *

def main():

locale.setlocale(locale.LC_ALL, 'en_US')

print('Monthly Payment Calculator')

while True:

print('DATA ENTRY')

loan_amt = input('Loan amount: ')

loan_amt = float(loan_amt)

int_rate = input('Yearly interest rate: ')

int_rate = float(int_rate)

years = input('Years: ')

years = int(years)

mon_rate = int_rate / 12 / 100

months = years * 12

monthly_pay = loan_amt * mon_rate / ( 1 - 1/(1 + mon_rate) ** months)

monthly_pay = Decimal(monthly_pay).quantize(Decimal('.01'), rounding=ROUND_DOWN)

print()

print('FORMATTED RESULT')

print('Loan amount: %30s' %locale.currency(loan_amt))

print('Yearly interest rate: %20.2f' %int_rate + '%')

print('Number of years: %25d' %years)

print('Montly payment: %25s' %locale.currency(monthly_pay))

print()

print('Continue? (y/n): ')

choice = input().strip()

if choice.lower() == 'n':

break

if __name__=='__main__':

main()

Unix Terminal>

Code output screenshot:

Answer:

appropriate tunnel pressure is 384.64 ATM

Even at this (high) pressure, exact dynamic similarity is not achieved because the specific heat ratio of helium is approximately 1.66 and is not equal to that of air which is approximately 1.40

Explanation:

It is too long to type here.

Kindly check the attached file for the calculation.