Answer:
umm okay for starters I have no clue lol.
what is aerodynamics????
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.
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:
Waterfall Spiral Incremental Agile Continuous integrationLearn more about software development here:
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Tech A says that the voltage regulator controls the strength of the rotor s magnetic field. Tech B says that the voltage regulator is installed between the output terminal of the alternator and the positive terminal of the battery. Who is correct?
Answer:
Voltage Regulator
Technician A is correct.
Explanation:
Technician B is not correct. The voltage regulator is not installed between the output terminal of the alternator and the positive terminal of the battery as claimed by Technician B. Technician A's opinion that the voltage regulator controls the strength of the rotor's magnetic field is correct. The computer can also be used to control the output of the alternator by controlling the field current.
Which of the following is a measurement at the nominal level?
A-personal income in a year
B-student grades
C-the color of a person's hair (blond, brunette, redhead)
D-temperature
Answer:
A-personal income in a year this is the answer
what are the main subsystem of GSM
network
Explanation:
Network and Switching Subsystem (NSS)
Base-Station Subsystem (BSS)
Mobile station (MS)
Operation and Support Subsystem (OSS)
A Rankine power generation cycle operates with steam as the working fluid. The turbine produces 100 MW of power using 89 kg/s of steam entering at 700C and 5MPa. The steam exits the turbine at 0.10135 MPa. Saturated liquid water exits the condenser and is pumped back to 5Mpa before it is fed to an isobaric boiler. a. Draw a schematic of the cycle. Number the streams and label any constraints across the units. b. The turbine operates adiabatically but not reversibly. What is the temperature of the steam exiting the turbine
How to make this circuit for simulating on Proteus.
A tank with a volume of 40 cuft is filled with a carbon dioxide and air mixture. The pressure within the tank is 30 psia at 70oF. It is known that 2 lb of carbon dioxide was placed in the tank. Assume that air is 80% nitrogen and 20% oxygen and use the ideal gas laws. Calculate ,
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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The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is 41:1. The primary coil is plugged into a standard 120-V outlet. The current in the secondary coil is 1.2 x 10-3 A. Find the power consumed by the air filter.
Answer:
5.9 watts
Explanation:
The secondary voltage is the primary voltage multiplied by the turns ratio:
(120 V)(41) = 4920 V
The power is the product of voltage and current:
(4920 V)(1.2·10^-3 A) = (4.92)(1.2) W = 5.904 W
The power consumed is about 5.9 watts.
All of the following are employment requirements for a technician EXCEPT:
Select one:
a. Being drug and alcohol free
b. Having all of your ASE certifications
c. Having a clean driving record
O d. Having honesty, integrity, reliability, and quality
Answer:
I think it would be C but im not sure sorry if its wrong
Question 10 of 25: Select the best answer for the question. 10. In a series RC circuit, if the resistance is 8,000 ohms and X is 6,000 ohms, what is the impedance of the circuit O A. 8,600 ohms O B. 10,000 ohms O C. 14,000 ohms O D. 9,200 ohms Mark for review (Will be highlighted on the review page) << Previous Question Next Question >> M LL 5 - 1080 acer
Answer:
B. 10,000 ohms
Explanation:
|8000 +j6000| = √(8000² +6000²) = 1000√(64 +36) = 1000√100 = 10,000
The impedance is 10,000 ohms.
A bird is flying in air. It is stretching its wings all along the sky and is flying in a certain direction Flying a bird is an example of: a. Composition of vector b. Collinear vector c. Addition of vector d. Multiplication of vector
Answer:
a. Composition of Vector.
Explanation:
When a bird flies in the air, it stretches its wings into air and this movement helps it move in a certain direction. This is an example of composition of vector. Air strikes the wings in opposite direction and bird wing movement helps it move against the wind.
Explain the 11 sections that a typical bill of quantity is divided into
Answer:
The main sections included in the bill of quantities are Form of Tender, Information, Requirements, Pricing schedule, Provisional sums, and Day works.
Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domestic wastewater from an influent BOD concentration of 250 mg/L to an effluent concentration of 10 mg/L. X (MLVSS) = 3000 mg/L, and the kinetics are first order and not variable order. The first order equation you must use to calculate the specific substrate utilization rate is q = K S where S is the effluent BOD concentration and K is the first order BOD degradation rate constant. The value of K is 0.04 L/(day*mg). What is the required reactor volume in MG (millions of gallons)? All the choices below are in units of MG.
0.4
1.0
0.2
4.8
Answer:
1.0MG
Explanation:
to solve this problem we use this formula
S₀-S/t = ksx --- (1)
the values have been given as
concentration = S₀ = 250mg
effluent concentration = S= 10mg
value of K = 0.04L/day
x = 3000 mg
when we put these values into this equation,
250-10/t = 0.04x10x3000
240/t = 1200
we cross multiply from this stage
240 = 1200t
t = 240/1200
t = 0.2
remember the question says that 5MGD is required to be treated
so the volume would be
v = 0.2x5
= 1.0 MG
Can anyone solve this for me?sequential circuit ,flipflop
Explanation:
We assume the T flip-flop changes state on the rising edge of the clock input.
The first stage is connected to the clock. The second stage clock is connected to the inverse of the Q output of the first stage, so that when the first stage Q makes a 1 to 0 transition, the second stage changes state.
Problem 89:A given load is driven by a 480 V six-pole 150 hp three-phase synchronous motor with the following load and motor data. Determine the voltage E necessary for this operating condition. Note: assume that the rotational loss torque is negligible. Load: Tൌ0.05∗????ଶ????mMotor: Eൌ400 V; Xௗൌ1ΩAnswer: Eതൌ400∠-17.36° V
Answer:
[tex]E_f=400<-17.4volts[/tex]
Explanation:
From the question we are told that:
Load [tex]V=480[/tex]
Poles [tex]p=6[/tex]
Power [tex]P=150hp[/tex]
3-Phase
Load:
[tex]Tl=0.05*\omega_s^2Nm[/tex]
Motor:
[tex]Ef=400V\\\\X_d=1ohm[/tex]
Generally the equation for Synchronous speed is mathematically given by
[tex]N_s=\frac{120F}{p}=\frac{120*60}{6}[/tex]
[tex]N_s=1200rpm\\\\N_s=125.66 rads/sec[/tex]
Therefore
[tex]Tl=0.05*\omega_s2Nm[/tex]
With
[tex]\omega=N_s[/tex]
We have
[tex]Tl=0.05*(125.66)^2Nm[/tex]
[tex]T_l=789.52 Nm[/tex]
Therefore
Load Power
[tex]P_l=T_l*\omega_s\\\\P_l=789.52*125.66[/tex]
[tex]P_l=9922watts[/tex]
Generally the equation for Load Power is mathematically given by
[tex]P_l=\frac{\sqrt{3}*E_f.V_t}{x_d}*sin\theta\\\\9922=\frac{\sqrt{3}*480*400}{1}*sin\theta[/tex]
[tex]\theta=17.4 \textdegre3[/tex]
Therefore
Voltage
[tex]E_f=400<-17.4volts[/tex]
A person has driven a car 180 m in 40 seconds. What is the car’s speed?
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
Technician A says that to start a fuel-injected car, the accelerator should be depressed
once. Technician B says that to start a car with a carburetor, do not touch the
accelerator pedal until the engine is running. Who is correct?
Technician A
Technician B
Both A and B
Neither Anor B
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
An incompressible viscous fluid flows through a pipe with a flow rate of 1 mL/s. The pipe has a uniform diameter D0 and a length L0. A pressure difference of P0 between the ends of the pipe is required to maintain the flow rate. What would be the flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0
Answer:
[tex]Q_2 = 32[/tex] mL/s
Explanation:
Given :
The flow is incompressible viscous flow.
The initial flow rate, [tex]Q_1[/tex] = 1 mL/s
Initial diameter, [tex]D_1= D_0[/tex]
Initial length, [tex]L_1=L_0[/tex]
The initial pressure difference to maintain the flow, [tex]P_1=P_0[/tex]
We know for a viscous flow,
[tex]$\Delta P = \frac{32 \mu V L}{D^2}$[/tex]
[tex]$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$[/tex]
[tex]$Q \propto \Delta P \times D^4$[/tex]
[tex]$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$[/tex]
[tex]$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$[/tex]
[tex]$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$[/tex]
[tex]$\frac{1}{Q_2}= \frac{1}{32}$[/tex]
∴ [tex]Q_2 = 32[/tex] mL/s
The flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0 is; Q2 = 32 mL/s
We are given;
Initial flow rate; Q1 = 1 mL/s
Initial uniform diameter; D0
Initial Length; L0
Initial Pressure difference; P0
Relationship between pressure, flow rate and diameter for vicious flow is given by;
Q1/Q2 = (P1/P2) × (D1/D2)⁴
Where;
Q1 is initial flow rate
Q2 is final flow rate
P1 is initial pressure difference
P2 is final pressure difference
D1 is initial diameter
D2 is final diameter
We are told that the pressure difference was increased to 2P0 and the diameter was increased to 2D0. Thus;
P2 = 2P0
D2 = 2D0
Thus;
1/Q2 = (P0/2P0) × (D0/2D0)⁴
>> 1/Q2 = ½ × (½)⁴
1/Q2 = 1/32
Q2 = 32 mL/s
Read more about vicious flow at; https://brainly.com/question/2684299
The end of the industrial robotic arm extends along the path (r = 2 + 2 cos (5t)) m. At the instant ( 0 = 0.8t ) radians. When the arm is located at (t = 0.85) second Determine the velocity and acceleration of the object A at this instant.
Answer:
v = 8.95 rad / s, a = 22.3 rad / s²
Explanation:
This is an exercise in kinematics, where we must use the definitions of velocity and acceleration
v = dr / dt
we perform the derivative
v = 0+ 2 (-sin 5t) 5
v = -10 sin 5t
we calculate for t = 0.85 t,
remember angles are in radians
v = -10 sin (5 0.85)
v = 8.95 rad / s
acceleration is defined by
a = dv / dt
we perform the derivatives
a = -10 (cos 5t) 5
a = - 50 cos 5t
we calculate for t = 0.85 s
a = -50 are (5 0.85)
a = 22.3 rad / s²
An apple, potato, and onion all taste the same if you eat them with your nose plugged
Answer:
I didn't understand your question or is it a fun fact
At steady state, a thermodynamic cycle operating between hot and cold reservoirs at 1000 K and 500 K, respectively, receives energy by heat transfer from the hot reservoir at a rate of 1500 kW, discharges energy by heat transfer to the cold reservoir, and develops power at a rate of (a) 1000 kW, (b) 750 kW, (c) 0 kW. For each case, apply Eq. 5.13 on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
Answer:
a. impossible
b. possible and reversible
c. possible and irreversible
Explanation:
a. 1000kw
Qh - Wnet
we have
QH = 1500
wnet = 1000
1500 - 1000
= 500kw
σcycle = [tex]-[\frac{QH}{TH} -\frac{QC}{TC} ][/tex]
Qh = 1500
Th = 1000
Tc = 500
Qc = 500
[tex]-[\frac{1500}{1000} -\frac{500}{500} ][/tex]
solving this using LCM
= -0.5
the cycle is impossible since -0.5<0
b. 750Kw
Qc = 1500 - 750
=750Kw
Qh = 1500
Th = 1000
Tc = 500
Qc = 750
σ-cycle
[tex]-[\frac{1500}{1000} -\frac{750}{500} ]\\= 1.5 -1.5\\= 0[/tex]
This cycle is possible and it is also reversible
c. 0 kw
Qc = 1500-0
= 1500
Qh = 1500
Th = 1000
Tc = 500
Qc = 1500
σ- cycle
[tex]-[\frac{1500}{1000} -\frac{1500}{500} ]\\-(1.5-3)\\-(-1.5)\\= 1.5[/tex]
1.5>0
so this cycle is possible and irreversible
Flat plate collector can provide temperature upto_____
Answer:
80⁰C
Explanation:
80°C.
Normal flat plate collectors can deliver heat at temperatures up to 80°C. Deficiency rates for normal flat plate collectors can be classified as visual losses, which produce with cumulative angles of the incident sunshine, and thermal losses, which upsurge fast with the working temperature intensities
Side milling cutter is an example of ______ milling cutter.
Answer:
special type
Explanation:
As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.
The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of hydrogen that pass per hour (in kg/h) through a 4.0-mm thick sheet of palladium having an area of 0.25 m^2 at 500°C. Assume a diffusion coefficient of 6.0 x 10^-8 m^2/s, that the concentrations at the high- and low-pressure sides of the plate are 3.5 and 0.25 kg/m^3 (kilogram of hydrogen per cubic meter of palladium), and that steady-state conditions have been attained. (clearly show your solution step by step, pay attention to units otherwise you will lose points!)
Who is responsible for shielding a welding arc to protect
the eyes and skin of all persons in the working area?
Answer:
manager and supervisor
Explanation:
Chemical engineers determine how to transport chemicals.
O True
False
Answer:
Chemical Engineers determine how to transport chemicals:
TRUE
Answer:
Explanation:
Trues
‘Politics and planning are increasingly gaining prominence in contemporary urban and regional planning debates’. Using relevant examples, discuss this assertion reflecting on the critical success factors for the successful implementation of the land reform program in South Africa.
Answer:
The governments receiving aid were generally experienced in industrial development. ... During the 1950s, little attention was given to differences in the Third World's conditions and needs, until these appeared to create obstacles to achieving high levels of industrial output
I hope it is helpful
anxiety: a. is never normal. b. is common of many psychological disorders c. is identical to fear d. is a modern development, unlikely to have roots in human history
Answer:
B
Explanation:
Anxiety is very common especially nowadays but it's especially common in psychological disorders
The customer you were waiting on is nodding while listening to you and gesturing when he speaks what type of language does this represent
Urgent!!!
List the assumptions that can be taken into account in torsion analysis.
Answer:
Explanation:
In the development of a torsion formula for a circular shaft, the following assumptions are made: Material of the shaft is homogeneous throughout the length of the shaft. Shaft is straight and of uniform circular cross section over its length. Torsion is constant along the length of the shaft.
