1. (40') An amplifier has a de gain of 10' and poles at 200kHz, 2MHz and 20MHz. Assume a phase margin of 30° is obtained, find the value of the maximum feedback ratio B. And also find the closed loop gain A, for an input signal of 3750Hz.

The roof modifications like rooftop gardens or it can be photovoltaic panels should be identified during the pre-incident survey's initial size-up. It should be done so that firefighters will be aware and have information about them in case the structure needs to be ventilated. Hence it will prevent the spread of fire between buildings your occupancy classifications.

The major functional requirements of a roof must be Strength and stability, Resistance to weather, Durability, freedom from maintenance, Resistance to the passage of heat, Resistance to the passage of sound, Resistance to air leakage, and most important fire safety.

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Answer:

defination is apromixely not available at this moment

Flash gas is a phenomenon that occurs when a liquid is suddenly released from a high-pressure vessel, causing some of it to rapidly vaporize into a gas. This sudden vaporization is caused by the decrease in pressure as the liquid is released into a lower-pressure environment.

Flash gas could be applicable in a variety of industries, such as oil and gas, where it can occur during the transportation of crude oil or natural gas. It can also occur in refrigeration systems, where it can cause a drop in efficiency and increased energy costs. To mitigate the effects of flash gas, various methods can be employed, such as using separators to remove the gas from the liquid stream or designing systems with lower pressure drops. Overall, understanding the phenomenon of flash gas is important in ensuring the safe and efficient operation of various industrial processes.

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Answer:

6.564 is the answers i learned that in the american school i live there

Answer:

Explanation:

Given that:

From process 1 → 2

\(P_1 = 10 bar \\ \\ V_1 = 1 m^3 \\ \\ V_2 = 4 m^3\)

\(PV^{1.5} = \ constant\)

\(\gamma = 1.5\)

Process 2 → 3

The volume is constant i.e \(V_2 =V_3 = 4m^3\)

\(P_3 = 10 \ bar\)

Process 3 → 1

P = constant  i.e the compression from state 1

Now, to start with 1 → 2

\(P_1V_1^{1.5} = P_2V_2^{1.5}\)

\(P_2 = P_1 (\dfrac{V_1}{V_2})^{1.5}\)

\(P_2 = 10 \times (\dfrac{1}{4})^{1.5}\)

\(P_2 =1.25\)

The work-done for the process  1 → 2 through adiabatic expansion is:

\(W = \dfrac{1}{1-\gamma}[P_2V_2-P_1V_1]\)

We know that 1 bar = \(10^5 \ N/m^2\)

∴

\(W = \dfrac{1}{1-1.5}[1.25 \times 10^5 \times 4- 10 \times 10^5 \times 1]\)

\(W =1000000 \ J\)

\(W_{1 \to 2} = 1000 kJ\)

For process 2 → 3

Since V is constant

Thus:

W = PΔV = 0

\(W_{2 \to 3} = 0\)

For process 3 → 1

W = PΔV

\(W _{3 \to 1} = P_3(V_1-V_3)\)

\(W _{3 \to 1} = 10 \times 10^5 (1-4)\)

\(W _{3 \to 1} = 10 \times 10^5 (-3)\)

\(W _{3 \to 1} = -3 \times 10^6 \ J\)

\(W _{3 \to 1} = -3000 \ kJ\)

The net work-done now  for the entire system is :

\(W_{net} = W_{1 \to 2} + W_{2 \to 3 } + W_{ 3 \to 1 }\)

\(W_{net} = (1000 + 0 + (-3000)) \ kJ\)

\(W_{net} =-2000 \ kJ\)

The sketch of the processes on p -V coordinates can be found in the image attached below.

A) The work done for each process are :

B) The net work for the cycle = -2000 kJ

Given Data :

For process (1 -2)      For process ( 2 - 3 )       process ( 3 - 1 )

P₁ = 10 bar                   P₃ = 10 bar                  constant pressure compression

V₁ = 1 m³                  constant volume heating

V₂ = 4 m³

PV\(^{1.5}\) = constant

A) Determine work done for each process

Calculate work done for process (1 - 2)

W₁ ₋ ₂ = \(\frac{P_{1}V_{1} - P_{2}V_{2} }{n -1 }\) * 100

         = [ ( 10*1 ) - ( 1.25 * 4 ) ] / 1.5 - 1

         = [ 10 - 5 ] / 0.5

         = 10 * 100 = 1000 kJ

Calculate work done for process ( 2-3 )

given that there is constant volume heating

W₂₋₃ = 0 kJ

Calculate work done for process ( 3-1)

W₃₋₁ = P ( Δ V )     given that p = constant

       = 10 * 100 ( -3 )

       = - 3000 kJ

B) The net work for the cycle

W₁ ₋ ₂  +  W₂₋₃  + W₃₋₁

= 1000 kJ  + 0 kJ  +  - 3000 kJ

= - 2000 kJ

Hence we can conclude that the ) The work done for each process are :

and The net work for the cycle = -2000 kJ

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Attached below is the P-V sketch of the process

Answer:

Trust me the answer is B I took the quiz last week

Answer: i believe it’s acoustics

Explanation:

b. A condition must involve a comparison.

A condition in the else statement does not need to be specified and does not require a comparison. The else statement is executed when the preceding if or elif statements are not satisfied.

The else statement is an integral part of conditional statements in programming languages like Python. It is used to define a block of code that executes when none of the preceding if or elif conditions evaluate to true. Unlike if and elif statements, the else statement does not require a specific condition or a comparison.

When the conditions in the if or elif statements are not met, the code within the else block is executed. This allows for handling alternative scenarios or default actions when none of the preceding conditions apply.

The absence of a condition in the else statement is intentional and necessary. It serves as a catch-all clause, ensuring that there is a defined course of action when all other conditions fail. The else statement is not contingent on a Boolean value or comparison but rather represents the default behavior when no other specific condition is met.

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Answer:

day if you workout without Zero billing that means you're not sweating. Sweating you're not losing anything that means you have zero outcomes

Explanation:

To simplify the given boolean expression f(a, b, c) = (ab'c)', we can apply De Morgan's theorems and other boolean algebra rules.

First, let's apply De Morgan's theorem to the inner expression (ab'c):

(ab'c)' = a' + b + c'

Next, let's apply De Morgan's theorem to the entire expression (a' + b + c'):

f(a, b, c) = (a' + b + c')'

Using De Morgan's theorem again, we can distribute the negation across the terms:

f(a, b, c) = (a')' · (b)' · (c)'

Now, we can simplify further:

(a')' = a

(b)' = b

(c)' = c

Therefore, the simplified expression is:

f(a, b, c) = a · b · c

So, the minimum (simplest) expression for f(a, b, c) is a · b · c.

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Without additional information on the raceway distance, ambient temperature, and voltage drop, it is impossible to determine the size of the copper service conductors required.

To determine the size of copper service conductors required for a multifamily building with a total demand load of 260 kW for a 120/208V wye, three-phase system, additional information is needed, such as the raceway distance, ambient temperature, and voltage drop. Without this information, it is impossible to accurately calculate the conductor size.

In general, conductor sizing is determined by the National Electrical Code (NEC), which provides guidelines based on various factors, including the load, the type of conductor, the length of the run, and the ambient temperature. To determine the conductor size for this particular situation, a licensed electrician or engineer would need to perform a detailed load calculation and take into account the specific installation requirements and local code regulations.

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Answer:

What specific fluid goes in the windshield wipers.

Distilled water

How much to put in fluid ounces?

There should be a tiny bit more than 3/4 of the way full.

Answer:

Explanation:

The maximum normal stress that can be applied to the specimen is given by the ratio of applied force to the area of the cross section of the specimen:

stress = force / area

The area of the cross section of a rod is given by the formula for the area of a circle:

area = πr^2

where r is the radius of the rod. We are given the diameter of the rod, which is 1.96 inches, so the radius is half of that, or 0.98 inches.

The maximum normal stress that the specimen can withstand is 68,787 psi, and the maximum force that can be applied is 797 lb. Substituting these values into the equation for stress gives:

68,787 psi = 797 lb / πr^2

Solving for the radius r gives:

r = √(797 lb / (68,787 psi × π))

Substituting in the value for π and calculating gives:

r ≈ 0.330 inches

The maximum diameter of the rod is twice the radius, so the maximum diameter is:

d = 2r ≈ 0.660 inches

Therefore, the maximum rod diameter that should be used for the specimen is approximately 0.660 inches.

In the radio communication system, multipath is the propagation phenomenon that causes diffusion or reflection in multiple different directions of a signal.

Multipath is a propagation mechanism that impacts the propagation of signals in radio communication. Multipath results in the transmission of data to the receiving antenna by two or more paths. Diffusion and reflection are the causes that create multiple paths for the signal to be delivered.

Diffraction occurs when a signal bends around sharp corners; while reflection occurs when a signal impinges on a smooth object. When a signal is received through more than one path because of the diffraction or reflection, it creates phase shifting and interference of the signal.

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Answer:

  A. Oil changes

Explanation:

It depends on the car and its usage and environment. Usually oil is supposed to be changed every few months, more often if the car is driven a lot. Coolant changes may be indicated as seasons change, so will generally occur less frequently than oil changes.

Tire and brake replacement depend on usage and driving habits. Some owners may never have to replace either one, if they trade their car every year or two. Folks who drive with their foot on the brake pedal may have to replace brakes relatively often.

The most frequent task is generally oil changes.

Answer:

Answer:

The belt in an open belt drive travels from one pulley's top to the top of another without intersecting. In cross belt drive, the belt crosses itself by moving from the top of one pulley to the bottom of another. Every revolution, the entire belt remains in the same plane.

Hope this helps!!!

The Robertson screwdriver has the largest tip size among screwdrivers because of its design.

The tip of a Robertson screwdriver has a square-shaped socket that fits perfectly with the corresponding square recess on the head of a Robertson screw. This design provides a secure and tight fit between the screw and the screwdriver, which prevents slippage and stripping of the screw head.

The square-shaped socket of the Robertson screwdriver also allows for greater torque transfer, meaning that the screwdriver can apply greater force to turn the screw, which is useful when working with larger or more stubborn screws.

In general, the size of a screwdriver tip is determined by the size of the screw head it is designed to fit. The Robertson screw was designed with a larger head to provide a stronger and more secure connection, and as a result, the corresponding Robertson screwdriver also has a larger tip size to match.

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Answer:

C. Count Noun

Explanation:

To find the new surface area, we need to consider the dimensions of the figure and how they are changed.  The original figure is an L-shaped complex solid composed of rectangular prisms.

so we need to find the surface area of each individual prism and then add them together. Let's break down the dimensions of each prism: The bottom face has a length of 9.8 ft. The shorter side of the L shape has a height of 6.5 ft. The height between the shorter side and taller side of the L shape is 8.1 ft. The taller section of the L shape has a length of 10.6 ft and a width of 4.3 ft. To calculate the surface area of each prism, we can use the formula: Surface Area = 2(length * width) + 2(length * height) + 2(width * height).

Now, let's add up the surface areas of all the prisms to get the original surface area: 84.04 + 55.9 + 172.08 = 312.02 ft². To find the new surface area, we need to multiply the original surface area by a factor that results in the new surface area being 14 times larger. So, we multiply the original surface area by 14: 312.02 * 14 = 4368.28 ft². Therefore, the new surface area is approximately 4368.28 ft², which can be rounded to 13.34 ft².

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The thing that  will happen to the Yield strength, the Tensile Strength, and the ductility. of a material if you increase the strain rate is option B: The Yield Strength and the Tensile Strength will increase, but the the ductility will decrease.

Boosting the strain rate or decreasing the temperature has the effect of raising the flow stress and decreasing ductility. The less uniform elongation before necking begins is what causes the poorer ductility; however, significant plastic flow may still be seen within the necked region.

Note that the primary distinction between yield strength and tensile strength is that the former refers to the lowest stress at which a material would permanently deform, whilst the latter refers to the highest tension at which a material will withstand before failing.

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See full question below

What will happen to the Yield strength, the Tensile Strength, and the ductility. of a material if you increase the strain rate? * The Yield Strength, the Tensile Strength, and the ductility (\%EL) will all increase. The Yield Strength and the Tensile Strength will increase, but the the ductility (\%EL) will decrease. The Yield Strength will increase, but the Tensile Strength and the the Ductility

(%EL)

will decrease. The Yield Strength will decrease, but the Tensile Strength and the the Ductility (\%EL) will increase. The Yield Strength and the Tensile Strength will decrease, but the the Ductility (\%EL) will increase. The Yield Strength, the Tensile Strength, and the ductility (\%EL) will all decrease.

In conclusion, the two-hand control switch is an important safety measure in factory environments, specifically designed to protect workers when they need to use both hands for assembling a product. Its function is to ensure that the machinery operates only when both buttons or switches are pressed simultaneously, keeping the worker safe from potential harm.

A two-hand control switch is used in factory environments as an emergency or contact switch when a worker is using both hands for assembling a product. This switch ensures the safety of the worker by requiring simultaneous and continuous pressure from both hands to activate the machinery.
When the worker presses and holds the buttons or switches of the two-hand control switch, it sends a signal to the machinery to start or continue the assembly process. If one hand releases the button or switch, the machinery will immediately stop, preventing any potential accidents or injuries.
The purpose of this switch is to minimize the risk of accidental operation and ensure that the worker's hands are safely away from hazardous areas of the machine during operation. By using both hands to operate the switch, the worker is forced to maintain a safe position and avoid any contact with moving parts.
In conclusion, the two-hand control switch is an important safety measure in factory environments, specifically designed to protect workers when they need to use both hands for assembling a product. Its function is to ensure that the machinery operates only when both buttons or switches are pressed simultaneously, keeping the worker safe from potential harm.

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Answer:

2)

a)  to the right of the dipole    E_total = kq [1 / (r + a)² - 1 / r²]

b)To the left of the dipole      E_total = - k q [1 / r² - 1 / (r + a)²]

c) at a point between the dipole, that is -a <x <a  

      E_total = kq [1 / x² + 1 / (2a-x)²]

d) on the vertical line at the midpoint of the dipole (x = 0)

E_toal = 2 kq 1 / (a ​​+ y)² cos θ

Explanation:

2) they ask us for the electric field in different positions between the dipole and a point of interest. Using the principle of superposition.

This principle states that we can analyze the field created by each charge separately and add its value and this will be the field at that point

Let's analyze each point separately.

The test charge is a positive charge and in the reference frame it is at the midpoint between the two charges.

a) to the right of the dipole

The electric charge creates an outgoing field, to the right, but as it is further away the field is of less intensity

           E₊ = k q / (r + a)²

where 2a is the distance between the charges of the dipole and the field is to the right

the negative charge creates an incoming field of magnitude

           E₋ = -k q / r²

The field is to the left

therefore the total field is the sum of these two fields

           E_total = E₊ + E₋

           E_total = kq [1 / (r + a)² - 1 / r²]

we can see that the field to the right of the dipole is incoming and of magnitude more similar to the field of the negative charge as the distance increases.

b) To the left of the dipole

The result is similar to the previous one by the opposite sign, since the closest charge is the positive one

E₊ is to the left and E₋ is to the right

          E_total = - k q [1 / r² - 1 / (r + a)²]

We see that this field is also directed to the left

c) at a point between the dipole, that is -a <x <a

In this case the E₊ field points to the right and the E₋ field points to the right

                      E₊ = k q 1 / x²

                      E₋ = k q 1 / (2a-x)²

                      E_total = kq [1 / x² + 1 / (2a-x)²]

in this case the field points to the right

d) on the vertical line at the midpoint of the dipole (x = 0)

    In this case the E₊ field points in the direction of the positive charge and the test charge

    in E₋ field the ni is between the test charge and the negative charge,

the resultant of a horizontal field in zirconium on the x axis (where the negative charge is)

                      E₊ = kq 1 / (a ​​+ y) 2

                      E₋ = kp 1 / (a ​​+ y) 2

                      E_total = E₊ₓ + E_{-x}

                      E_toal = 2 kq 1 / (a ​​+ y)² cos θ

e) same as the previous part, but on the negative side

                        E_toal = 2 kq 1 / (a ​​+ y)² cos θ

When analyzing the previous answer there is no point where the field is zero

The different configurations are outlined in the attached

3) We are asked to repeat part 2 changing the negative charge for a positive one, so in this case the two charges are positive

a) to the right

in this case the two field goes to the right

           E_total = kq [1 / (r + a)² + 1 / r²]

b) to the left

            E_total = - kq [1 / (r + a)² + 1 / r²]

c) between the two charges

E₊ goes to the right

E₋ goes to the left

            E_total = kq [1 / x² - 1 / (2a-x)²]

d) between vertical line at x = 0

E₊ salient between test charge and positive charge

           E_total = 2 kq 1 / (a ​​+ y)² sin θ

In this configuration at the point between the two charges the field is zero

Answer:

The greatest mass, that the weight C can have such that block A does not move is approximately 23.259 kg

Explanation:

The given parameters are;

The mass of the block A= 58 kg

The coefficient of static friction between the block and the surface, \(\mu _s\) = 0.300

The coefficient of static friction between the rope and the fixed peg, B \(\mu _B\) = 0.310

Let T represent the tension in the rope

Therefore, when the rope is static, T = The normal reaction at the peg, B, \(N_B\)

The angle of inclination of the rope holding the block A = arctan(3/4) ≈ 36.87°

The length of the rope = √(0.4² + 0.3²) = 0.5

∴ sin(θ) = 3/5 = 0.6

cos(θ) = 4/5 = 0.8

The vertical component of the tension in the rope = T × sin(θ) = 0.6·T

The horizontal component of the tension in the rope = T × cos(θ) = 0.8·T

The friction force = μ×(W - 0.6·T) = 0.300×(58×9.8 - 0.6·T) = 170.52 - 0.18·T

The block will start to move when we have;

The horizontal component of the tension in the rope = The friction force

∴ 0.8·T = 170.52 - 0.18·T

0.8·T + 0.18·T = 170.52

0.98·T = 170.52

T = 170.52/0.98 = 174

Therefore, the tension in the rope = T = 174 N = The normal reaction at the peg, B \(N_B\)

The frictional force at the peg, \(F_B\) = \(\mu _B\) × \(N_B\) = 0.310 × 174 N = 53.94 N

The weight of the mass, \(m_c\), \(W_c\) = The frictional force at the peg, \(F_B\)  + The tension in the rope

∴ The weight of the mass, \(m_c\), \(W_c\) = 53.94 N + 174 N = 227.94 N

Weight, W = Mass, m × The acceleration due to gravity, g, from which we have;

m = W/g

Where;

g = 9.8 m/s²

∴ \(m_c\) = \(W_c\)/g = 227.94 N/(9.8 m/s²) ≈ 23.259 kg.

The greatest mass, that the weight C can have such that block A does not move = \(m_c\) ≈ 23.259 kg.

Explanation:

150 divide by 150 and that how you do the is you what to divide together 15/ 150 you welcome have a good day is you need something else

Answer:

the range of length scales is ( 0.0602 -  0.1094 )

Explanation:  

  Given the data in the question;

\(P_{absolute\) = 1300 kPa

V\(_{prototype\) = 406 km/h

speed of model nit more than 29%

we know that Reynolds number Re = pVl/μ = constant

p\(_m\)V\(_m\)l\(_m\)/μ\(_m\) = pVl/μ  

such that;

l\(_m\)/l = ( p/p\(_m\) )( V/V\(_m\) )( μ\(_m\)/μ )  ----- let this be equation 1

Now, for an idea gas; P = pRT { with constant temperature }

p / p = constant; p/p\(_m\) = p/p\(_m\)  

assuming μ\(_m\) = μ\(_m\)  

Therefore, the relation becomes;

l\(_m\)/l = ( p/p\(_m\) )( V/V\(_m\) )

from the given data;

l\(_m\)/l = ( 101/1300 )( V/V\(_m\) )

where our V\(_m\) = ( 1 ± 29% )V

so

l\(_m\)/l = ( 101/1300 )( V / ( 1 ± 29% )V )

l\(_m\)/l = ( 101/1300 )( 1 / ( 1 ± 0.29  ) )

Now, The minimum limit will be;

l\(_m\)/l = ( 101/1300 )( 1 / ( 1 + 0.29  ) )

= ( 101/1300 ) × ( 1 / 1.29 )

= 0.0602

The maximum limit will be;

l\(_m\)/l = ( 101/1300 )( 1 / ( 1 - 0.29  ) )

= ( 101/1300 ) × ( 1 / 0.71 )

= 0.1094

Therefore, the range of length scales is ( 0.0602 -  0.1094 )

It means that the gas with the lowest molecular weight will have the highest effusion rate.

The given gases' effusion rates are listed in order from highest to lowest. The effusion rate of a hydrogen molecule is the highest, whereas that of a hydrocarbon is the lowest.

A gas will effuse faster when it is lighter and more slowly when it is heavier. Helium (He) will have the highest rate of effusion since it has the lowest molecular weight (atomic weight, in this example).

The following equation can be used to compare the rate of effusion for two gases: The effusion rates in this case are inversely related to the square root of the gas molecules' masses. A container contains an amalgam of neon and argon gas.

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Answer:

The idea is to protect the puddle while it cools

Explanation:

Answer:

Under cold air standard conditions, the thermal efficiency of this cycle is 56.9 percent.

Explanation:

From Thermodynamics we remember that thermal efficiency of the ideal Otto cycle (\(\eta_{th}\)), dimensionless, is defined by the following formula:

\(\eta_{th} = 1-\frac{1}{r^{\gamma-1}}\) (Eq. 1)

Where:

\(r\) - Compression ratio, dimensionless.

\(\gamma\) - Specific heat ratio, dimensionless.

Please notice that specific heat ratio under cold air standard conditions is \(\gamma = 1.4\).

If we know that \(r = 8.2\) and \(\gamma = 1.4\), then thermal efficiency of the ideal Otto cycle is:

\(\eta_{th} = 1-\frac{1}{8.2^{1.4-1}}\)

\(\eta_{th} = 0.569\)

Under cold air standard conditions, the thermal efficiency of this cycle is 56.9 percent.

Answer:

At temperature is and relative humidity is 86% therefore,  the humidity ratio is 0.0223 and the specific volume is 14.289

At temperature is and Relative humidity is 40% therefore, the humidity ratio is  0.0066 and the specific volume is 13.535.

To calculate the mass of air can be calculated as follows:

Now , we going to calculate the volume,

The time which is required to fill the cistern can be calculated as follows:

Now, putting the value in above formula we get,

Therefore, the hours required to fill the cistern is 4.65 hours.

Explanation:

The area occupied by one molecule of Red 40 dye on a 40 cm² wool piece can be found by dividing 0.004 m² by the number of dye molecules adsorbed on the surface.

To determine the area occupied by one molecule of Red 40 dye on the surface, we need to know the number of dye molecules that are present on the wool surface.

Let's assume that we have N molecules of Red 40 dye adsorbed on the surface of each 40 cm² wool piece.

The surface area of one wool piece is 40 cm² = 0.004 m².

The molecular weight of Red 40 dye is approximately 496 g/mol.

Assuming that the density of Red 40 dye is similar to that of water (1 g/cm³),

the volume occupied by one molecule of Red 40 dye is approximately 5.0 × 10⁻²³ m³ (496 g/mol divided by 6.02 × 10²³ molecules/mol).

To determine the area occupied by one molecule of Red 40 dye on the surface, we need to divide the surface area by the number of dye molecules, which gives:

Area occupied by one molecule = (0.004 m² / N)

Therefore, the area occupied by one molecule of Red 40 dye on the wool surface is (0.004 m² / N),

where N is the number of Red 40 dye molecules adsorbed on the surface of each 40 cm² wool piece.

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Answer: The maintenance factor of a lighting system reflects how much of the initial luminous flux is still available at the end of its useful life. The planned lighting engineer must compute the maintenance factor and multiply the new value of the light output by it.

Explanation:

Answer:

Explanation: A lighting system's maintenance factor indicates how much of the initial luminous flux remains available at the end of its service life. The maintenance factor must be determined by the planning lighting engineer and the new value of the luminous flux multiplied by it.