In a 1-phase UPS, Vd = 350 V, vo(t) = 170 sin(2π * 60t) V, and io(t) = 10 sin(2π * 60t - 30ᴼ) A.Calculate and plot da(t), db(t), vaN(t), vbN(t), Id, id2(t) and id(t). Switching frequency fs = 20 kHz.

Answers

Answer 1

Answer: provided in the explanation section

Explanation:

The question says;

In a 1-phase UPS, Vd = 350 V, vo(t) = 170 sin(2π * 60t) V, and io(t) = 10 sin(2π * 60t - 30ᴼ) A.Calculate and plot da(t), db(t), vaN(t), vbN(t), Id, id2(t) and id(t). Switching frequency fs = 20 kHz.

Answer

From this we have come to this;

da = Ṽan/Vd = 0.5 + (0.5 * 0.4857) sin w,t

db = Ṽbn/Vd = 0.5 - (0.5 * 0.4857) sin w,t

Ṽan(t) = da * 350 V

Ṽbn(t) = db * 350 V

Id = 0.5 * Ṽo/ Vd * Îo cosΦ1 = 0.5 * (170/350) * 10 * cos 30ᴼ = 2.429 A

Id2 = -0.5 * 170/350 * 10 * sin(2 w,t - 30ᴼ) = -2.429 sin(2 w,t - 30ᴼ)

īd = Id + id2 = 2.429 A + -2.429 sin(2 w,t - 30ᴼ).

Note: Attached is a copy of an image showing and explaining thr plots.

cheers i hope this has been helpful !!!!

In A 1-phase UPS, Vd = 350 V, Vo(t) = 170 Sin(2 * 60t) V, And Io(t) = 10 Sin(2 * 60t - 30) A.Calculate
In A 1-phase UPS, Vd = 350 V, Vo(t) = 170 Sin(2 * 60t) V, And Io(t) = 10 Sin(2 * 60t - 30) A.Calculate
In A 1-phase UPS, Vd = 350 V, Vo(t) = 170 Sin(2 * 60t) V, And Io(t) = 10 Sin(2 * 60t - 30) A.Calculate

Related Questions

One of the testing equipments used for inspection is? Test tubes Measuring jar Strain gauges None of the mentioned

Answers

Answer:

The answer is the strain gauges.

Explanation:

Inspection systems work or are performed to measure the characteristics of a product, to verify if it meets specified requirements, all using benchmarks and test equipment.

The strain gauges are part of the test equipment used for inspection. These are sensors that measure deformation, pressure and load in resistance tests of materials.

Which phase of DevSecOps emphasizes reliability, performance, and scaling

Answers

Answer:

Test Phase

Explanation:

DevSecOps is an organizational software engineering culture and practice that unifies software development (Dev), security (Sec) and operations (Ops). The main characteristic of DevSecOps is to improve customer outcomes and mission value by automating, monitoring, and applying security at all phases of the software lifecycle.

There are nine phases of the software lifecycle which are: plan, develop, build, test, release, deliver, deploy, operate, and monitor.

The Performance test in the test phase will ensure that applications will perform well under the expected workload. The test focus is on application response time, reliability, resource usage and scalability.

In development phases, database design, development, and testing activities generate database artifacts, which are data models, database schema files, trigger definitions, view definition, test data, test data generation scripts, test scripts, etc. These database artifacts must be under configuration management control. During test phase, database functional test is like application code unit test and functional test to validate the schema, triggers, and data compliance. The non-functional test includes load testing, stress test, and performance test. The security test focuses on vulnerability scan, user authentication and authorization, unauthorized access to data, data encryption, privilege elevation, SQL injection, and denial of service.

In a bid eliminate the vulnerability experienced during the traditional development process, DevSecOps emphasizes reliability, performance and Scaling with the integration of Security phase.

The integration of Security infrastructure into the Development operation(DevOps) process ensures that security challenges experienced by softwares are tackled immediately hence ensuring reliability and reduced vulnerability.

DevSecOps ensures that performance isn't sacrificed for security, hence, softwares are continously checked for security at every phase of the development process during testing.

Therefore, the security phase of the DevSecOps pipeline ensures that satisfactory security and Performance levels are met.

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Switches are placed only in the _ of a circuit?

Answers

the switch should always be placed immediately adjacent to the non-grounded terminal of the power supply.

Compute the repeat unit molecular weight of PTFE. Also compute the number-average molecular weight for a PTFE for which the degree of polymerization is 10,000

Answers

Answer:

a) the repeat unit molecular weight of PTFE MW = 100.015 g/mole

b) the number-average molecular weight for a PTFE for which the degree of polymerization is 10,000 щₙ = 1000150 g/mole

Explanation:

Given that;

PTFE which is also called Polytetrafluoroethylene

structure  of repeat unit of Polytetrafluoroethylene

(C2F4)n

a)

To compute the repeat unit molecular weight

we say

MW = 2( atomic weight of C ) + 4( atomic weight of F)

MW = 2 (12.0107) + 4 ( 18.9984)

MW = 100.015 g/mole

therefore the repeat unit molecular weight of PTFE MW = 100.015 g/mole

b)

To compute the number-average molecular weight for  PTFE of which the degree of polymerization is 10,000

we say

DP = щₙ / MW

where щₙ is the number of average molecular weight,

MW is the repeat unit molecular weight give as 100.015 g/mole

DP is degree of polymerization which is 10,000

Now we substitute

10,000 = щₙ / 100.015

щₙ = 10,000 × 100.015

щₙ = 1000150 g/mole

Therefore the number-average molecular weight for a PTFE for which the degree of polymerization is 10,000 щₙ = 1000150 g/mole

System grounding on a power system means electrically connecting the __?__ of every wye-connected transformer or generator to earth.

Answers

Answer: Neutrals

Explanation: System grounding on a power system is a term used to describe the entire processes involved when a neutral is used as the conductor to connect to the solid earth. This ensures that power is generated. This is usually done using either an inductor, an impendance or a resistor. It is very important and necessary to carry out a proper grounding of a power system in order to ensure the safety of the equipment and the personnel etc

The city park has a stream running through it. The city wants to make better use of the park area. What steps should the city planners take to design a park that has all areas available to the public?

Answers

Answer:

they could add a play structure, with the stream they can put ducks and fish in it and picnic places

brainliest plz

Explanation:

Answer:

just took the quiz (k12) answer is...

Ask questions to identify a problem, develop a model, and carry out the plan/desgn.

Explanation:

In the National Electrical Code, the current carrying abilities of conductors are called the ___________.

Answers

Answer:

Ampacity

Explanation:

Ampacity is a word used to expalin ampere capacity defined by National Electrical Codes.

Ampacity is defined as the maximum current, in amperes, that can flow in a conductor continuously under the conditions of use without exceeding the temperature rating of the conductor.

Therefore, in the National Electrical Code, the current carrying abilities of conductors are called the Ampacity.

The safety risks are the same for technicians who work on hybrid electric vehicles (HEVs) or EVs as those who work on conventional gasoline vehicles.

Answers

Answer:

Batteries are safe when handled properly.

Explanation:

Just like the battery in your phone, the battery in some variant of an electric car is just as safe. If you puncture/smash just about any common kind of charged battery, it will combust. As long as you don't plan on doing anything extreme with the battery (or messing with high voltage) you should be fine.

The safety risks are the same for technicians who work on hybrid electric vehicles (HEVs) or EVs as those who work on conventional gasoline vehicles: False.

Safety risks can be defined as an assessment of the risks and occupational hazards associated with the use, operation or maintenance of an equipment or automobile vehicle that is capable of leading to the;

Harm of a worker (technician).Injury of a worker (technician).Illness of a worker (technician).Death of a worker (technician).

Hybrid electric vehicles (HEVs) or EVs are typically designed and developed with parts or components that operates through the use of high voltage electrical systems ranging from 100 Volts to 600 Volts. Also, these type of vehicles have an in-built HEV batteries which are typically encased in sealed shells so as to mitigate potential hazards to a technician.

On the other hand, conventional gasoline vehicles are typically designed and developed with parts or components that operates on hydrocarbon such as fuel and motor engine oil. Also, conventional gasoline vehicles do not require the use of high voltage electrical systems and as such poses less threat to technicians, which is in contrast with hybrid electric vehicles (HEVs) or EVs.

This ultimately implies that, the safety risks for technicians who work on hybrid electric vehicles (HEVs) or EVs are different from those who work on conventional gasoline vehicles due to high voltage electrical systems that are being used in the former.

In conclusion, technicians who work on hybrid electric vehicles (HEVs) or EVs are susceptible (vulnerable) to being electrocuted to death when safety risks are not properly adhered to unlike technicians working on conventional gasoline vehicles.

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If you measure the flow conditions (velocity, pressure, temperature) at a single point and conditions do not change over time, the flow is characterized as

Answers

Answer:

Steady flow

Explanation:

Flows in fluids can be categorized into different classes depending on the type of flow and the variations in their characteristics such as velocity, pressure, density, temperature, e.t.c

When these characteristics do not change when measured over time at a single point, then the flow is said to be steady. For a steady flow, the mathematical expression, amidst other conditions, is given as follows;

[tex]\frac{dV}{dt} = 0, \frac{dP}{dt} = 0, \frac{dT}{dt} = 0[/tex]

Where;

V, P and T are the velocity, pressure and temperature of the fluid.

PS:

Other types of flows include:

i. Unsteady flow

ii. Laminar flow

iii. Turbulent flow

iv. Uniform flow

v Non-uniform flow

vi. Rotational flow

vii. Irrotational flow

A brittle material typically exhibits substantial plastic deformation with high energy absorption before fracture.a) trueb) false

Answers

Answer:

The answer is FALSE

Explanation:

Brittle material are materials that don not undergo plastic deformation, they have very low plasticity that is while cracks can form without plastic deformation

The major/ common examples are glass, ceramics, graphite

In other words brittle materials break instead of bending, they have very low energy absorption as they don not undergo plastic deformation

The motor of an electric vehicle runs at an average of 50 hp for one hour and 25 minutes. Determine the total energy. Write the result in engineering notation and with SI units.

Answers

Answer:

The total energy of the motor of the electric vehicle is 1.902 × 10⁸ joules.

Explanation:

Power is the rate of change of work in time, since given input is average power, the total energy ([tex]\Delta E[/tex]) of the motor of the electric vehicle, measured in joules, is determined by this formula:

[tex]\Delta E = \dot W \cdot \Delta t[/tex]

Where:

[tex]\dot W[/tex] - Average power, measured in watts.

[tex]\Delta t[/tex] - Time, measured in seconds.

Now, let convert average power and time into watts and seconds, respectively:

Average Power

[tex]\dot W = (50\,hp)\times \frac{746\,W}{1\,hp}[/tex]

[tex]\dot W = 3.730\times 10^{4}\,W[/tex]

Time

[tex]\Delta t = (1\,h)\times \frac{3600\,s}{1\,h} + (25\,min)\times \frac{60\,s}{1\,min}[/tex]

[tex]\Delta t = 5.100\times 10^{3}\,s[/tex]

Then, the total energy is:

[tex]\Delta E = (3.730\times 10^{4}\,W)\cdot (5.100\times 10^{3}\,s)[/tex]

[tex]\Delta E = 1.902\times 10^{8}\,J[/tex]

The total energy of the motor of the electric vehicle is 1.902 × 10⁸ joules.

A 400-MVA, 240-kV/24-kV, three-phase Y-A transformer has an equivalent series impedance of 1.2 + j6 N per phase referred to the high-voltage side. The transformer is supplying a three-phase load of 400-MVA, 0.8 power factor lagging at a terminal voltage of 24 kV (line to line) on its low-voltage side. The primary is supplied from a feeder with an impedance of 0.6 + ji.2 A per phase. Determine the line-to-line voltage at the high-voltage ter- minals of the transformer and the sending-end of the feeder.

Answers

Answer: the line-to-line voltage at the high-voltage terminals of the transformer and the sending-end of the feeder is 249.71∠1.8° kV

Explanation:

First we find the phase voltage per phase at the primary side connected in Y, so we say

V₂ = 240K/√3 = 138.56 kV

Now we find the primary current

I₁ = ((400 × 10⁶) / 3(138.56 × 10³)) ∠ -cos⁻¹ (0.8)

I₁  = 962.28∠ -36.87° A

To find the voltage V₁, we say

V₁ = ( 1.2 + j6) I₁ + V₂

we substitute

V₁ =  ( 1.2 + j6) 962.28∠ -36.87° + 138.56 × 10³

V₁  = 143∠1.57° kV

Now we find the phase voltage at the sending end

Vₓ = ( 0.6 + J1.2 )I₁ + V₁  

Vₓ = ( 0.6 + J1.2 ) 962.28∠ -36.87° + 143∠1.57° K

Vₓ = 144.17∠1.8° kV

So to Determine the line to line voltage at the sending end, we say:

Vₓ (line to line) = √3 × 144.17∠1.8° kV

Vₓ (line to line) = 249.71∠1.8° kV

A piston-cylinder device contains Xenon gas. During a reversible, adiabatic process, the entropy of the gas will (never, sometimes, always) increase.

Answers

Answer:

Never

Explanation:

In a reversible adiabatic process, there is not transfer of heat or matter between the system and its environment. An adiabatic reversible process is a  process with constant entropy, i.e ΔQ=0. The internal energy is solely dependent on the work done either due to compression or expansion. So the entropy of the gas will never increase.

Which statement below can be used to read data from a file one character at a time?
A) inputFile.get(1)
B) inputFile.read(1)
C) inputFile.split(1)
D) inputFile.open(1)

Answers

Answer:

b

Explanation: because it says input file.read

Consider a 2.4-kW hooded electric open burner in an area where the unit costs of electricity and natural gas are $0.10/kWh and $1.20/therm (1 therm = 105,500 kJ), respectively. The efficiency of open burners can be taken to be 73 percent for electric burners and 38 percent for gas burners. Determine the rate of energy consumption and the unit cost of utilized energy for both electric and gas burners.

Answers

Answer:

1.75 kW

$0.137 kWh

4.61 kW

$3.16 therm

Explanation:

Utilized power input of the burner is

P(ui) = total power input * efficiency

P(ui) = 2400 W * 0.73

P(ui) = 1752 W or 1.75 kW

Unit cost of utilized energy is

C(ui) = Unit cost of electricity/efficiency

C(ui) = $0.1 / 0.73 kWh

C(ui) = $0.137 kWh

Power input to the gas burner is

P(gi) = Utilized power input of the burner / efficiency of the burner

P(gi) = 1.75 / 0.38

P(gi) = 4.61 kW

Unit cost of utilized energy is

C(gi) = Unit cost of gas /efficiency

C(gi) = $1.2 / 0.38 kWh

C(gi) = $3.16 therm

Initially, a pump pressure of __________ pounds per square inch should be used to maintain a sprinkler or standpipe system.

Answers

Answer:

of 150 pounds per square inch

Explanation:

Note that the unit for measuring water pressure is called pounds per square inch (psi)

In the case of sprinklers and standpipe systems, a pressure of 150 pounds per square inch was used initially.

Consider a 2.4-kW hooded electric open burner in an area where the unit costs of electricity and natural gas are $0.10/kWh and $1.20/therm (1 therm = 105,500 kJ), respectively. The efficiency of open burners can be taken to be 73 percent for electric burners and 38 percent for gas burners. Determine the rate of energy consumption and the unit cost of utilized energy for both electric and gas burners.

Answers

Answer:

electric: 2.4 kW; $0.1370 per kWh utilizedgas: 4.61 kW; $0.1078 per kWh utilized

Explanation:

Electric burner

For a 2.4 kW electric burner, the rate of energy consumption is 2.4 kW.

If efficiency is 73%, the cost of utilized energy is ...

  ($0.10 /kWh) / 0.73 ≈ $0.1370 per kWh utilized

__

Gas burner

If the utilized energy provided by the gas burner is the same as the utilized energy of the electric burner, then the rate of energy consumption will be ...

  2.4 kW(0.73)/(0.38) = 4.61 kW

In terms of kWh, the cost of gas is ...

  $1.20/(105,500 kJ)·(1 kJ/(kW·s))·(3600 s/h) = $0.04095 /kWh

If efficiency is 38%, the cost of utilized gas energy is ...

  ($0.04095 /kWh) / 0.38 ≈ $0.1078 per kWh utilized

_____

The cost of gas is about 21% less per utilized joule.

_____

Comment on rates

The "unit rate" will depend on the unit chosen. In order to avoid unnecessary units conversions, we have elected to stick with kWh as the unit of energy for both cases. You may be asked for different units. We trust you or Google can make the necessary conversions.

A signal that cannot be faked and carries the most accurate information about a situation or individual is known as an ______ signal

Answers

Answer: Honest

Explanation: In other to establish communication between two or more species, the sender and receiver, It is required of the sender to showcase certain behavior, sound or other demonstrations which are capable of passing a message to the receiver. These demonstrations are often reffered to as signals. Signals could be honest or dishonest. Honest signals are characterized by it's usefulness to the receiver and the conveyance of the actual or true meaning of underlying signal being transmitted. This is the opposite of dishonest signals which are often used to trick the receiving party as the information being transmitted are inaccurate and unreliable.

Give the principle, construction and working of Bourdan tube pressure gauge.

Answers

Answer:

Working Principle Of Bourdon Gauge

If a tube having oval cross section is subjected to pressure its cross section tends to change from oval to circular.

Construction of a Bordon Gauge

Bourdon tube gauges consist of a circular tube.

One end of the tube is fixed while the other end is free to undergo elastic deformation under the effect of pressure.

Fixed end is open and pressure which is to be measured is applied at the fixed end.

Free end is closed and undergoes deformation under the effect of pressure.

Due to applied pressure the circular tube tends to uncoil and become straight along the dotted line.

Working of Bourdon Gauge

As the pressure is applied at the fixed end free end undergoes deformation.

The free end is attached with sector which further meshes with the pinion on which pointer is mounted.

Deformation of the pointer is transferred to pointer via this mechanism.

As a result point undergoes deflection and shows the pressure reading on calibrated dial.

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Answer:

oooo

Explanation:

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An electrical heater 150 mm long and 5mm in diameter is inserted into a hole drilled normal to the surface of a large block of material having a thermal conductivity of 10W/mK. The temperature of the electrical heater is 70 oC and the temperature of block surface is 40 oC. Estimate the power dissipation of the electrical heater. (in W)

Answers

Explanation:

From the table of conduction shape factor and dimensionless conduction rates for selected systems, for vertical cylinder in a semi-infinite medium

[tex]S=& \frac{2 \pi L}{\ln \left(\frac{4 L}{D}\right)}[/tex]

[tex]S=& \frac{2 \pi L}{\ln \left(\frac{4 L}{D}\right)}[/tex]

[tex]\frac{2 \pi \times 0.1}{\ln \left(\frac{4 \times 0.1}{0.005}\right)}[/tex]

[tex]=0.1433 \mathrm{m}[/tex]

Three 12-V, 100-A-hr batteries are connected in series. What are the output voltage and A-hr capacity of this connection

Answers

Answer:

36V, 100A-hr

Explanation:

Since the batteries are connected in series;

i. the output voltage will be the sum of the individual voltages

ii. the current rating (A-hr) will be the same as their individual current rating (A-hr). And this is because the same current flows through the batteries.

From i,

The output voltage, V, is given by the sum of the voltages of the three batteries;

V = 12V + 12V + 12V

V = 36V

From ii,

The A-hr capacity of the connection is the same as that of the individual batteries;

100 A-hr

Therefore, the output voltage and A-hr capacity of this connection is:

36V, 100A-hr

help out with that question

Answers

Answer:

  (523.74 lb)∠-48.59°

Explanation:

a) We can find the sum of the force vectors several ways, but first we need to know the direction of vector BT. The angle PBT is identical to the angle shown as α, which we can find from the lengths BD and AD.

  α = arctan(BD/AD) = arctan((3√3)/6)) ≈ 40.893°

Using the Law of Cosines, we can find the magnitude of PT from ...

  PT^2 = PB^2 +BT^2 -2·PB·BT·cos(α)

  PT^2 = 800^2 +600^2 -2·800·600·2/√7 = 1000000 -1920000/√7

  PT ≈ 523.74 . . . . lb

The direction of PT can be found from the Law of Sines.

  ∠BPT = arcsin(BT/PT·sin(α)) ≈ 1.145597sin(40.893°) ≈ 48.59°

Relative to the +x axis, the resultant force (R) is ...

  R = (523.74 lb)∠-48.59°

__

b) The graphical solution is shown in the attachment. The graphing tool measures the segment lengths and angles. Those measures confirm the above result. (The pound values shown are scaled up by a factor of 100 from the segment lengths on the diagram. They are "captions" for the respective vectors.)

Consider a 5 m long, air-filled section of a coaxial transmission line, given that the radius of the inner conductor is 10 cm and the inner radius of the outer conductor is 20 cm? If the current is 100 mA, how much magnetic energy is strored in the air medium between the conductors. This problem is similar to Example 2 in the Time Varying Fields folder in the Example Bank.
a) 200.89 nj
b) 3.47 nj
c) 10.45nj
d) 80.9nj

Answers

Answer: b) 3.47 nj

Explanation:

Given that;

length l = 5m

radius of inner conductor r = 10cm = 0.1m

radius of outer conductor D = 20cm = 0.2m

current I = 100A = 100×10⁻³ = 0.1

medium between conductor in air u₀ = 4π × 10⁻⁷

Energy in a coaxial cable transmission line is

w = u₀ /2π I² en(b/a)

we substitute

L = 4π × 10⁻⁷ /4π ×10⁻²× 5 en (20/10)

L =3.4657 × 10⁻⁹ J

L = 3.4657 nJ ≈ 3.47 nJ

Technician A says that kinked parts should be replaced. Technician B says that bent parts may be repaired. Who is right

Answers

Answer:

Both are right

Explanation:

It all depends on the customer. The technicians job is to inform the customer about what can be done, rest depends on the customer that what he wants to be done. If he prefers to get the parts replaced then he should do it, and he he thinks that after repair they will work well then he should go for the repair.

As the impurity concentration in solid solution is increased, the tensile and yield strengths ____________________.

increase

decrease

Answers

Answer:

As the impurity concentration in solid solution is increased, the tensile and yield strengths increases.

Explanation:

The addition of impurities in solid solutions shows an improved tensile and yield strength due to the grain refinement and obstacles to the motion of dislocation.

Example, the addition of carbon as impurity into iron, which forms steel shows a significant increase in the tensile and yield strengths of iron.

Blacksmiths also use work hardening to introduce dislocation into solid solutions in order to increase their  tensile and yield strengths.

Therefore, as the impurity concentration in solid solution is increased, the tensile and yield strengths increases.

A solid solution is a mixture of crystalline solids and is soluble over the partial or evenly complete range.

A solute may be replaced by a solvent particle. It may be used for heating the is related to the melting point. The tensile and strength of the solution form a solid increase as the concentration of the impurities increase.

Hence the option Increases is correct.

Learn more about the concentration in solid solution is increased.

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What is a semiconductor whose electrical properties are based on the electronic structure inherent in the pure materials? MATSE 81

Answers

Answer:

intrinsic semiconductors

Explanation:

An intrinsic semiconductor is also known as a pure conductor. In such a semiconductor there are no impurities, that is why it is said to be pure.

It has some of these properties:

1. Electrical conductivity is only based on temperature

2. The quantity of electrons is the same as the number of holes in the valence bond

3. Electrical conductivity is not on the high side

4. These materials exist in their pure forms.

If the equation for the velocity profile is given by: 3/2 = 4yv . Assuming v is in ft/s, what is the velocity gradient at the boundary and at y = 0.25 ft and 0.5 ft from boundary?

Answers

Answer:

When y = 0.25 ft, velocity gradient = -6 ft/s

When y = 0.5 f, velocity gradient = -1.5 ft/s

Explanation:

Given;

equation for the velocity profile, 3/2 = 4yv

Rearrange this equation, you will get;

[tex]4yv = \frac{3}{2}\\\\v = \frac{3}{2} *\frac{1}{4y} \\\\v = \frac{3}{2}(\frac{1}{4} )(\frac{1}{y} )\\\\v = \frac{3}{8}y^{-1}\\\\ gradient \ of \ velocity \ = \frac{dv}{dy} \\\\\frac{dv}{dy} = -1(\frac{3}{8})y^{-2}[/tex]

When y = 0.25 ft

[tex]\frac{dv}{dy} = -1(\frac{3}{8})(0.25)^{-2}\\\\\frac{dv}{dy} = -6 \ ft/s[/tex]

When y = 0.5 ft

[tex]\frac{dv}{dy} = -1(\frac{3}{8})(0.5)^{-2}\\\\\frac{dv}{dy} = -1.5 \ ft/s[/tex]

Which of the following Components Can't. be used as an amplifier Anless its Control electrode Conducts a Current? a) Ann-Channel Jfetb) An npn transistorc) A p. Channel ufet d) A mosfet

Answers

Answer:

d) A mosfet

Explanation:

MOSFET is the most common type of insulated gate Field Effect Transistor (FET),  used in electronic circuits and it stands for Metal Oxide Semiconductor Field Effect Transistor.

To configure MOSFET to act as an amplifier, a small AC signal is applied,   which is superimposed on to DC bias at the gate input, then the MOSFET will act as a linear amplifier.

Therefore, the correct option is (d) A mosfet

Which phase of DevSecOps emphasizes reliability, performance, and scaling

Answers

Answer:

"Test Phase " is the correct choice.

Explanation:

DevSecOps seems to be a community as well as experience of corporate data science which encompasses software design, regulation, including operational activities. This same main feature of DevSecOps has always been to strengthen customer achievement as well as expedition importance by computerizing, supervising as well as implementing data protection at all stages of the development including its development tools.The testing method throughout the test phase would then help make sure that the controller is designed mostly under the responsibilities forecasted. The test focuses on either the reaction times, dependability, use of resources but instead interoperability of applications.

The DevSecOps can be described as a software development life cycle which has seen security introduced into the continous development and operations pipeline. Hence, the phase of DevSecOps which emphasizes reliability, performance and scaling is the Security phase

DevSecOps can be broken down into a continous pipeline of processes which include Development, Security and Operation.

By integrating security into the continous software development process, it ensures that software security which is aimed at reducing the vulnerability of data and information is undertaken throughout the entire development lifecycle.

Therefore, ensuring that applications are reliable and performs well without having to sacrifice security in the process.

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A 10 mm diameter jet of water is deflected by a homogeneous rectangular block (15 mm by 200 mm by 100 mm) that weighs 6 N. Determine the minimum volume flowrate needed to tip the block.

Answers

Answer:

the minimum volume flow rate needed to tip the block is 2.66 × 10⁻⁴ m/s

Explanation:

Given that;

diameter of the jet d = 10 mm

weight W = 6 N

Now we say

Fₓ Lfₓ - Wlw= 0

horizontal force

Fₓ = W (lw/lfₓ)

Fₓ = 6 ( 0.015/2)

Fₓ = 0.9 N

X-component of momentum

v₁p(-v₁)A₁ = - Fₓ

pA₁v₁² = Fₓ

v₁² = Fₓ / pA₁

v₁ = √( Fₓ / pA₁ )

WE SUBSTITUTE

v₁ = √ ( 0.9 / ((999)(π/4)(0.01))²

v₁ = 3.39 m/s

Now Discharge Q = A₁v₁

Q = π/4 (0.01)² (3.39)

Q = 2.66 × 10⁻⁴ m/s

therefore the minimum volume flow rate needed to tip the block is 2.66 × 10⁻⁴ m/s

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