As part of the overall systems engineering process, there are a variety of software development methods, but the three most common at DoD are the Waterfall Approach, the Incremental Approach, and the ______________ Approach.

Answer:

Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A​

Explanation:

Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A​

I can say it oxidizes gas pollutants such as hydrocarbons and carbon monoxide. It also reduces nitrogen oxides into water, hydrogen, and carbon dioxide.

Answer:

  4.5 m/s = 16.2 km/h

Explanation:

The speed is the ratio of distance to time:

  speed = (180 m)/(40 s) = 4.5 m/s

The first case where the power of synthesis was applied was in Chile when they had to put 100 families in houses around 40m².

It should be noted that the power of architecture is the fact that it can synthesize a complex entry to a problem.

Chilean architect Alejandro used this in building more than several houses for the poorest communities in Chile. His rigorous and innovative design approach used a social framework that laid a precedent within the profession.

During this period in Chile, a normal middle-class home was about 80m² but he built his in 40m² and it was a good home that managed the resources that were available.

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Answer:

False

Explanation:

Answer:

Explanation:

Galvanic corrosion occurs when two dissimilar metals are immersed in a conductive solution and are electrically connected. One metal (the cathode) is protected, whilst the other (the anode) is corroded. The rate of attack on the anode is accelerated, compared to the rate when the metal is uncoupled.

Answer: identify your decision. relize you need to make a decision.

Explanation:

Answer:

Alternator

Explanation:

A device which converts mechanical energy into electrical energy.

I hope it helps.

Answer:

you know you could use like quora or yahoo answers

Answer:

  B. 10,000 ohms

Explanation:

|8000 +j6000| = √(8000² +6000²) = 1000√(64 +36) = 1000√100 = 10,000

The impedance is 10,000 ohms.

Answer:

4.25

Explanation:

The mean deviation is how far, on average, all values are from the middle. To do that, we subtract each element from the mean of the elements, take their absolute values, and then divide by the number of elements in the set. Here's the formula: [tex]\sum\frac{|x-\mu|}{n}[/tex]

Here the mean is 9.5, the count n = 8, so the mean deviation is 4.25

Answer:

Battery Voltage

Voltage refers to the amount of electrical potential your battery holds. The standard automotive battery in today's vehicles is a 12-volt battery. Each battery has six cells, each with 2.1 volts at full charge. A car battery is considered fully charged at 12.6 volts or higher.

Answer:

umm okay for starters I have no clue lol.

what is aerodynamics????

The correct responses are;

(i) Weight percent of nitrogen: 58.6%

Weight percent of oxygen: 14.65%

Weight percent of carbon dioxide: 29.59%

(ii) Volume percent of nitrogen: 64.38%

Weight percent of oxygen: 14.1%

Weight percent of carbon dioxide: 21.53%

(iii) Partial pressure of nitrogen: 19.314 psia

Partial pressure of oxygen: 4.23 psia

Partial pressure of carbon dioxide: 6.459 psia

(iv) Partial pressure of nitrogen: 19.314 psia

Partial pressure of oxygen: 4.23 psia

Partial pressure of carbon dioxide: 6.459 psia

(v) The average molecular weight is approximately 32.02 g/mole

(vi) At 20 psia, 60 °F, the density is 0.11275 lb/ft.³

At 14.7 psia, 60 °F, the density is 0.0844 lb/ft.³

At 14.7 psia, 32 °F, the density is 0.0892 lb/ft.³

(vii) The specific gravity of the mixture 0.169

Reasons:

The volume of the tank = 40 ft.³ = 1.132675 m³

Content of the tank = Carbon dioxide and air

The pressure inside the tank = 30 psia = 206843 Pa

The temperature of the tank = 70 °F ≈ 294.2611 K

Mass of carbon dioxide placed in the tank = 2 lb.

Percent of nitrogen in the tank = 80%

Percent of oxygen in the tank = 20%

(i) 2 lb ≈ 907.1847 g

Molar mass of carbon dioxide = 44.01 g/mol

Number of moles of carbon dioxide = [tex]\displaystyle \mathbf{ \frac{907.1847 \, g}{44.01 \, g/mol}} \approx 20.613 \, moles[/tex]

Assuming the gas is an ideal gas, we have;

[tex]\displaystyle n = \mathbf{\frac{206843\times 1.132674}{8.314 \times 294.2611} }\approx 95.7588[/tex]

The number of moles of nitrogen and oxygen = 95.7588 - 20.613 = 75.1458

Let x represent the mass of air in the mixture, we have;

[tex]\displaystyle \mathbf{ \frac{0.8 \cdot x}{28.014} + \frac{0.2 \cdot x}{32}}= 75.1458[/tex]

Solving gives;

x ≈ 2158.92 grams

Mass of the mixture = 2158.92 g + 907.1847 g ≈ 3066.1047 g

[tex]\displaystyle Weight \ percent \ of \ nitrogen= \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{58.6 \%}[/tex]

[tex]\displaystyle Weight \ percent \ of \ oxygen = \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{14.65\%}[/tex]

[tex]\displaystyle Weight \ percent \ of \ carbon \ dioxide = \frac{907.184}{3066.105} \times 100 \approx \underline{29.59\%}[/tex]

ii.

[tex]\displaystyle Number \ of \ moles \ nitrogen= \frac{0.8 \times 2158.92 \, g}{28.014 \, g/mol} \approx 61.65\, moles[/tex]

[tex]\displaystyle Number \ of \ moles \ oxygen= \frac{0.2 \times 2158.92 \, g}{32 \, g/mol} \approx 13.5\, moles[/tex]

Number of moles of carbon dioxide = 20.613 moles

Sum = 61.65 moles + 13.5 moles + 20.613 moles ≈ 95.763 moles

In an ideal gas, the volume is equal to the mole fraction

[tex]\displaystyle Volume \ percent \ of \ nitrogen= \frac{61.65}{95.763} \times 100 \approx \underline{64.38 \%}[/tex]

[tex]\displaystyle Volume \ percent \ of \ oxygen = \frac{13.5}{95.763} \times 100 \approx \underline{14.1\%}[/tex]

[tex]\displaystyle Volume \ percent \ of \ carbon \ dioxide = \frac{20.613 }{95.763} \times 100 \approx \underline{21.53\%}[/tex]

(iv) The partial pressure of a gas in a mixture, [tex]P_A = \mathbf{X_A \cdot P_{total}}[/tex]

Partial pressure of nitrogen, [tex]P_{N_2}[/tex] = 0.6438 × 30 psia ≈ 19.314 psia

Partial pressure of oxygen, [tex]P_{O_2}[/tex] = 0.141 × 30 psia ≈ 4.23 psia

Partial pressure of carbon dioxide, [tex]P_{CO_2}[/tex] = 0.2153 × 30 psia ≈ 6.459 psia

(v) The average molecular weight is given as follows;

[tex]\displaystyle Average \ molecular \ weight = \frac{3066.105\, g}{95.7588 \, moles} = 32.02 \, g/mole[/tex]

(vi) At 20 psia 70 °F, we have;

Converting to SI units, we have;

[tex]\displaystyle n = \frac{137895.1456\times 1.132674}{8.314 \times 294.2611} \approx 63.84[/tex]

The number of moles, n ≈ 63.84 moles

The mass = 63.84 moles × 32.02 g/mol  ≈ 2044.16 grams ≈ 4.51 lb

[tex]\displaystyle Density = \frac{4.51\ lb}{40 \ ft.^3} \approx \underline{0.11275\, lb/ft.^3}[/tex]

When the pressure is 14.7 psia = 101352.93 Pa, and 60 °F

[tex]\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 288.7056} \approx 42.8247[/tex]

The mass = 47.8247 moles × 32.02 g/mol  ≈ 1531.35 grams = 3.376 lb.

[tex]\displaystyle Density = \frac{3.376 \ lb}{40 \ ft.^3} \approx \underline{ 0.0844\, lb/ft.^3}[/tex]

When the pressure is 14.7 psia = 101352.93 and 32 °F

[tex]\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 273.15} \approx 50.5482[/tex]

The mass = 50.5482 moles × 32.02 g/mol  ≈ 1618.55 grams = 3.568 lb.

[tex]\displaystyle Density = \frac{3.568 \ lb}{40 \ ft.^3} \approx \underline{ 0.0892 \, lb/ft.^3}[/tex]

(vii) [tex]\displaystyle The \ specific \ gravity \ of \ the \ mixture \ = \frac{\frac{3066.1047\, g}{40 \, ft.^3} }{1.00} = \frac{\frac{6.7596 \, lbs}{40 \, ft.^3} }{1.00} \approx \underline{0.169}[/tex]

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Answer:

oh this happened with me to ,this will be alright just wait a bit

Answer:

17A) 52mm

17B) 132mm

17C) 67mm

18.) 34

Explanation:

For 17)

Each distance is determined by the total distances surrounding it. For example, 17A is 73mm - 21mm, which is 52mm.

For 18)

Set up the problem as an algebraic expression:

a + b + c + d = 115

Since you know 3 of the values, you can plug them in and subtract them from 115, which turns into 34.