Thank you,Thank you udahfsdf

The three most common software development methods are the Waterfall Approach, the Incremental Approach, and the SPIRAL Approach. These methods depend on the team size and specific goals.

Software development is the sequential procedure that involves the division of the work into smaller and parallel stages in order to improve software design and product management.

The software development methods depend on both the team size and specific objectives.

The most common methodologies for software development include:

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

Explanation:

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.

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:

general duty clause

Explanation:

Answer: It does make sense, because I've been involved in these careers and have a long family line of them. And other questions?

Explanation:

Lateral runout is felt in the steering wheel even at slower speeds. It leads to riding vibration and therefore poor ride quality.

Lateral runout refers to the side-to-side movement of the car's rotor as it rotates, which is caused by the off-center deviation of the rotor.

Lateral runout leads to riding vibration and therefore it is associated with poor ride quality.

Lateral runout can be properly quantified by using a dial indicator, which is a precision device used to measure linear displacements.

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A CNC machine stands for Computer Numerically Controlled Machine, they are used to Machine objects in hence create items, machine parts from prototypes. Hence the CNC machines are not used for testing/Analysis

D. CNC machines.

In Engineering the following items are basically needed for and during test/Analysis

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Answer: Well, that's a good question. When we look around us, we tend to see what technology has done for us, and how far it is taking us. It might be disruptive, but without it there would be no high tech military weapons, and then how would we defend ourselves? There would be no technology entertainment and without that, life would be very dull. In answering your question, we can be thankful for what it is and how far it has been taking us in life.

Explanation: Please mark a brainliest if correct it helps and have meery holidays!

Answer:

Answer:

a. Commutator:

the commutator reverses the current direction with each turn serving as a mechanical rectifier to convert the alternating current from the windings to unidirectional direct current in the external load circuit.

b. Armature core

Armature core is a physical structure made up of silicon steel stampings. The core is laminated to keep iron losses low. Armature winding is the main current carrying winding where we either provide(electric supply to motor) or extract e.m.f(collect electric power from generator).

c. Field system

The field winding basically form an electromagnet, that produces field flux within which the rotor armature of the DC motor rotates, and results in the effective flux cutting

d. Brushes

The function of brushes is that the brushes conduct current between stationary wires and moving parts of the electric Generator.

e. Armature winding

The winding wound on the core is given current through a DC supply. This current-carrying winding as under the influence of the magnetic field develops a force that develops a rotational torque. This torque enables the machine to rotate.

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Answer:

A-personal income in a year this is the answer

Answer: The elastic modulus and flexural strength are found to decrease with increased porosity. Fractography studies have been undertaken to identify the role of the pores on the fracture path. The deflection and cracking behavior of concrete structure depend on the flexural tensile strength of concrete. Many factors have been shown to influence the flexural tensile strength of concrete, particularly the level of stress, size, age and confinement to concrete flexure member, etc.

Explanation:

Answer:

the answer I know is A. different substances. hope this helps :).

Explanation:

just always study hard and God bless.

Answer:

  B. 10,000 ohms

Explanation:

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

The impedance is 10,000 ohms.

Answer:

umm okay for starters I have no clue lol.

what is aerodynamics????

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.

Answer:

manager and supervisor

Explanation:

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:

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:

I think it would be C but im not sure sorry if its wrong

Answer:

then back into the room

Explanation:

If you leave the door open, heat is merely recycled from the room into the refrigerator, then back into the room. A net room temperature increase would result from the heat of the motor that would be constantly running to move energy around in a circle.

Answer:

you know you could use like quora or yahoo answers

Answer:

Information technology is important in our lives because it helps to deal with every day's dynamic things. Technology offers various tools to boost development and to exchange information. Both these things are the objective of IT to make tasks easier and to solve many problems.

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:

  -0.1 inches

Explanation:

The net inflow is ...

  36 cfs -30 cfs = 6 cfs

The number of seconds in 30 days is ...

  (3600 s/h)(24 h/da)(30 da) = 2,592,000 . . . . seconds/(30 days)

Then the volume of inflow is ...

  (6 ft^3/s)(2,592,000 s) = 15,552,000 ft^3

The number of square feet in 1260 acres is ...

  (1260 ac)(43560 ft^/ac) = 54,885,600 ft^2

So, the increase in depth due to the inflow is ...

  (15,552,000 ft^3)/(54,885,600 ft^2) ≈ 0.283353 ft ≈ 3.4002 in

__

The net change in water level is then ...

  inflow - seepage + precipitation - evaporation

  3.4 in -1.5 in +4.0 in -6.0 in = -0.1 in

The water level change in the period is -0.1 inch.