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vagabundo [1.1K]

2 years ago

10

what is the largest, low-cost airline carrier in the world, which recently canceled thousands of flights for reasons partially d

isputed by the federal aviation administration?

1 answer:

IgorC [24]2 years ago

8 0

The largest, low-cost airline carrier in the world that recently canceled thousands of flights for reasons partially disputed by the federal aviation administration (FAA) is Southwest Airlines.

Southwest Airlines canceled 2,500 flights in just a few days.

According to Southwest Airlines, the cause of the cancellation is due to air traffic control problems and disruptive weather.

While FAA claimed that nothing like a reported case of air traffic and staffing shortages.

Southwest Airlines is the world's largest low-cost carrier and has its main office is located in Dallas, Texas.

Hence, in this case, it is concluded that the correct answer is "Southwest Airlines."

Learn more here: brainly.com/question/24782945

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Write a program that asks the user to input a vector of integers of arbitrary length. Then, using a for-end loop the program exa

ELEN [110]

Answer:

%Program prompts user to input vector

v = input('Enter the input vector: ');

%Program shows the value that user entered

fprintf('The input vector:\n ')

disp(v)

%Loop for checking all array elements

for i = 1 : length(v)

%check if the element is a positive number

if v(i) > 0

%double the element

v(i) = v(i) * 2;

%else the element is negative number.

else

%triple the element

v(i) = v(i) * 3;

end

end

%display the modified vector

fprintf('The modified vector:\n ')

disp(v)

4 0

3 years ago

For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu

bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6 mo1^-2

Now according isothermal work of one mole methyl gas is

W=- $\int\limits^a_b {P} \, dV$

a= $v_2\\$

b= $v_1$

from virial equation

$\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\$

And

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

Substitute given values $P_1\\$ = 1 x 10^5 , T = 373.15 and given values of coefficients we get

$10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2$

Solve for V1 by iterative or alternative cubic equation solver we get

$v_1=30780 cm^3/mol$

Similarly solve for state 2 at P2 = 50 bar we get

$v_1=241.33 cm^3/mol$

Now

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a=241.33

b=30780

After performing integration we get work done on the system is

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get

dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

$W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP$

a= $v_2\\$

b= $v_1$

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series

8 0

2 years ago

How do you make a robotic dog

fredd [130]

In order to create a robotic dog, you are needing the necessary parts to create Goddard from Jimmy nutreon boy genius

7 0

3 years ago

Read 2 more answers

I have a question.What does DIY mean?

alexdok [17]

Answer: It means "Do it yourself".

Explanation: You're welcome!

6 0

2 years ago

Read 2 more answers

If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas

jarptica [38.1K]

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine ( $\eta$ ), no unit, is defined by this formula:

$\eta = \frac{\dot W}{\dot Q}$ (1)

Where:

$\dot Q$ - Heat input, in watts.

$\dot W$ - Power output, in watts.

If we know that $\dot W = 600\,W$ and $\eta = 0.3$ , then the heat input from the combustion phase is:

$\eta = \frac{\dot W}{\dot Q}$

$\dot Q = \frac{\dot W}{\eta}$

$\dot Q = \frac{600\,W}{0.3}$

$\dot Q = 2000\,W$

The heat input from the combustion phase is 2000 watts.

8 0

2 years ago