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

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

Answer:

a) civil engineering.

Explanation:

Civil engineering is a professional engineering program that deals with the construction, design, and maintenance of all the natural and man-made environments including dams, buildings, railways, and roads.

Civil engineering is the branch of engineering that is the practice not restricted because civil engineer is not restricted to academic profession but practice in designing and construction can make someone a professional civil engineer.

Hence, the correct answer is "a)."

The branch of engineering in which the practice is not restricted is; Civil Engineering

             Civil Engineering is a branch of engineering that deals with the design, construction and maintenance of the physical and naturally built environment.

These physically and naturally built environment includes; houses, roads, bridges, airports, railways, canals, dams, sewage systems, pipelines e.t.c

             Now, mechanical engineering involves design, production, operation and maintenance of mechanical systems or machineries.

             In conclusion, we see that the mechanical branch of engineering is restricted to machinery unlike civil engineering that is not restricted to only buildings but also includes pipelines, bridges, roads, railways, dams, sewage systems e.t.c

             Finally, the nuclear engineering branch is also restricted to only nuclear fission and fusion applications.

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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

Answer:

"Test Phase " is the correct choice.

Explanation:

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

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

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Answer:

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

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:

oooiiiiijjjii niaje

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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Explanation:

a) Given B(n) is the number of the length 'n' bit sequences which have no three consecutive 0s(i.e., they does not contain substring “000”)

Any bit string which has no 000 should have a 1 in at least one of the 1st three positions. Then, the n we will break all the bit strings by avoiding the 000 by when the 1st 1 occurs. i.e., each of the bit of string of the length of n will avoid 000 falls into the exactly any one of these cases:

1 is followed by the bit string of the length of (n-1) avoiding the 000.

01 is followed by some bit string of the length of (n-2) avoiding a 000.

001 which is followed by any bit string of the length(n-3) avoiding the 000.

Therefore, recurrence is

B(n)=B(n-1)+ B(n-2)+ B(n-3), with B(0)=1, B(1)=2, B(2)=4

Or

B(n)=B(n-1)+ B(n-2)+ B(n-3), with B(1)=2, B(2)=4, B(3)=7

b) Let the S(n)={1,2,3,…,n}. We will say that the subset A of S(n) is very good if A does not have any of the two integers that are consecutive.

For any k, let a(k) be a number of any good subsets of a S(k).

There are two types of good subsets of S(n).

Type 1 of  good subsets of S(n) which  contain the element n,

Type 2 of good subsets of S(n)  which do not contain n.

We will first get an expression for a number of Type 1 of good subsets of the S(n) , where n≥2. This a subset does not contain n-1. So any Type 1 of the good subset of the S(n)  is obtainable when adding n to good subset of the S(n-2) . Also, on adding n to a good subset of the S(n-2) , we always get a Type 1 good subset of the S(n). Thus there are exactly as many good Type 1 of subsets of S(n) as there is good subsets of S(n-2) . By the definition there are a(n-2) good subsets of S(n-2) .

A good subset of a S(n)  is either Type 1 or Type 2. Thus the number of the good subsets of S(n)  is a(n-2)+a(n-1)

We have therefore shown that a(n) = a(n-2)+a(n-1)

c). Here firstly see that if the n is not the multiple of a 3, then there will be no chance to tile the rectangle. And also if n is a multiple of a 3, then there may be two ways to tile the 1st three columns:

And the rest of tilling is the tilling of  2x(n-3) rectangle for which there are T(n-3).

So, the recurrence is

T(n )= {2T(n-3), with T(0)=1     if n=0(mod 3)   or 0 else

We could use the base case T(3)=2

So, the following recurrence can be aslo equivalent to T(n)= 2T(n-3), with T(0)=1, T(1)=0,T(2)=0

Or

T(n)= 2T(n-3), with T(1)=0, T(2)=0,T(3)=2

d)A ternary string may be defined as a sequence of some digits, where each digit has either 0,1,or 2.

According to the problem given, we do not have a 2 anywhere after  0, and the dot which represents the binary string of length(n-1) with the property which we cannot use 2 anywhere after we use the 0.

Now for base case,  we should note that any of the ternary string which has a length of 1 satisfies the given required property. Hence the recurrence is

T(1)=3

T(n)=2T(n-1)+2^(n-1)

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]

Answer:

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.

Answer:

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

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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:

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.)

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.

Hence the option Increases is correct.

Learn more about the concentration in solid solution is increased.

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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

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

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

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.

Answer:

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

Answer:

External diameter = 158.15 mm mm

Internal diameter = 59.31 mm

Explanation:

We are given;

Diameter ratio; d_i = ⅜d_o

Where d_i is internal diameter and d_o is external diameter

Power;P = 375 KW = 375000 W

Rotational speed;N = 100 rpm

Max torque is 20% greater than mean torque; T_max = 1.2T_avg

Shear stress;τ = 60 N/mm²

Length; L = 4m = 4000 mm

Angle of twist; θ = 2° = 2π/180 radians

Modulus of rigidity;G = 0.85 X 10^(5) N/mm²

Formula for the power transmitted by the shaft is;

P = 2πNT_avg/60

Plugging in the relevant values, we have ;

375000 = 2π × 100T_avg/60

T_avg = (375000 × 60)/(2π × 100) = 35809.862 N.m = 35809862 N.mm

Since T_max = 1.20T_avg

Thus, T_max = 1.20(35809862) = 42971834.4 N.mm

Checking for strength, we'll use;

τ = Tr/J_p

Or since r = d/2

It can be written as;

τ = T(d_o)/2J_p - - - (1)

Where T is T_max

But Polar moment of inertia of hollow shaft is;

J_p = [π(d_o)⁴ - π(d_i)⁴]/32

Now, we are told that d_i = ⅜d_o

Thus;

J_p = [π(d_o)⁴ - π(⅜d_o)⁴]/32

J_p = (π/32) × d_o⁴(1 - 3⁴/8⁴)

J_p = 0.0926 d_o⁴

Plugging this for J_p in eq 1,we have;

τ = T(d_o)/2(0.0926d_o⁴)

Making d_o the subject gives;

d_o³ = T/(2 × 0.0926τ)

Plugging in the relevant values to give;

d_o³ = 42971834.4/(2 × 0.0926 × 60)

d_o³ = 3867155.7235421166

d_o = ∛3867155.7235421166

d_o = 156.96 mm

Thus, d_i = ⅜ × 156.96 = 58.86 mm

Checking for stiffness, we'll use;

T/J_p = Gθ/L

Again T is T_max

Plugging in the relevant values, we have;

42971834.4/0.0926 d_o⁴ = (0.85 × 10^(5) × 2π/180)/4000

464058686.825054/d_o⁴ = 0.7417649321

d_o⁴ = 464058686.825054/0.7417649321

d_o⁴ = 625614216.5028806

d_o = ∜625614216.5028806

d_o = 158.15 mm

d_i = ⅜ × 158.15 = 59.31 mm

So we will pick the highest valies.

Thus;

d_o = 158.15 mm

d_i = 59.31 mm

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.

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]

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

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.

Answer:

b

Explanation: because it says input file.read

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.

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

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.

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;

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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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.

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

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