what does a flashing green light from the tower to an aircraft in flight indicate?

In the example, we have three bulbs with of r: A, B, and C. Let's examine the conditions of the circuit step by step:

1. When bulb C is installed, bulbs B and C are connected in parallel, creating an equivalent resistance of r/2.2. Bulb A is connected in series with the parallel arrangement of bulbs B and C, and the circuit's overall resistance is 3.3. The total resistance changes to 5r when bulb C is removed, putting bulbs A and B in series.4. Removing bulb C reduces the circuit's current flow since it raises the overall resistance.5. When bulb C is taken out, the battery's current drops to 7/12 of what it was before.

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Answer:

Scott was in a motorcycle accident and has injuries to his chest and his skull. The doctor has ordered Scott to have an ECG, which stands for ____, to check his heart rhythm

Answer:

2.35 + j8.34 Ω

Explanation:

Voltage = V\(_{L}\) = 240 V rms

supplying power = S\(_{s}\) = 8 kVA

power factor = pf\(_{s}\) = 0.6

Let P₁ represents one load draws 3kW at unity powder factor

The power angle is:

θ\(_{s}\) = cos⁻¹  pf\(_{s}\) = cos⁻¹  0.6 = 53.13°

Complex power supplied source is:

S\(_{s}\) =  S\(_{s}\) < θ\(_{s}\) = 8<53.13° kVA

Complex power for first load:

S₁ = P₁ = 3kVA

Since the power angle of first load is  θ₁ = 0°

According to principle of conservation of AC power, the power of second load is:

S₂ =  S\(_{s}\) - S₁

    = 8<53.13° - 3

    = 6.65<74.29° kVA

Since the second load is a Y connected load the phase voltage:

V\(_{p}\) =  V\(_{L}\) / \(\sqrt{3}\)

    = 240/1.732051

    = 138.564

    = 138.56 V

Complex power of second load:

S₂ = 3 V\(_{p}\)² / Z\(_{p}\)

impedance per phase of the second load:

Z\(_{p}\) =  3 V\(_{p}\)² / S₂

   = 3 (138.56)² /  6.65<74.29°

   = 3(19198.8736) / 6.65<74.29°

   = 57596.6208 / 6.65<74.29°

Z\(_{p}\) = 2.35 + j8.34Ω

Answer:

1.1⁰C

Explanation:

Width W = 5mm = 0.005

Thickness t = 1 mm = 0.001

K = thermal conductivity = 150W/m.K

P = q = heat transfer rate = 4W

We are to find the steady state temperature between the back and the front surface

We have to make these assumptions:

1. There is steady state conduction

2. The heat flow is of one dimension

3. The thermal conductivity is constant

4. The heat dissipation is uniform

We have:

∆t = t*P/k*W²

= (0.001m x 4W)/150x(0.005)²

= 0.004/0.00375

= 1.06667

This is approximately,

1.1⁰C

Thank you!

If Olivia hires a roofer to repair the leaks caused by the solar panel installation, the defect warranty for the solar panels may be voided.

The warranty typically covers defects specifically related to the solar panels themselves. By involving a roofer to address the leaking issue, Olivia is bypassing the opportunity for Sam Solar, the contractor responsible for the solar panel installation, to inspect and correct any defects associated with the panels. If the leaks are indeed caused by the solar panel installation, it is essential to allow the contractor the opportunity to investigate and rectify the issue, as stated in the warranty terms. Failure to do so may result in the defect warranty being voided, as the contractor should have the chance to address and resolve any potential defects.

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Data mapping is the process of matching fields from one data source to another. Hence option c is correct.

Data is defined as the information that has been altered to be more easily moved around or processed. Organizations can set baselines, benchmarks, and targets using good data in order to keep moving forward.

The process of matching fields from one database to another is known as data mapping. This is the initial step in making data transfer, integration, and other data management chores easier.

Thus, data mapping is the process of matching fields from one data source to another. Hence option c is correct.

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Answer:

Explanation:

i) In the attached truth table, TRUE means the respective key owner is present and their keys are inserted at the same time. The "Opens" column is TRUE when two owners are present, not including the case where the only two owners present are B and C.

__

ii) The second attachment is a Karnaugh map of the truth table. The circled terms are ...

  Opens = AB +AC

Primitive transportation and storage systems that make local distribution ineffective if not impossible, the lack of clean water, and the lack of effective sewer systems are all examples of this type of barrier: physical and environment.

Transportation can be defined as a process that involves the movement of humans, products, resources such as water, and other physical things from one geographical area to another, especially through various means (forms) such as:

A barrier can be defined as any form of obstacle, impediment or hindrance which makes it impossible to perform an action, function or task in a timely manner.

In this context, we can reasonably infer and logically deduce that all of the aforementioned barriers such as lack of clean water and effective sewer systems, primitive transportation and storage systems are all examples of a physical and environment barrier.

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Answer:

Phuong works on a research project and creates a report for her boss.

Answer:

B. Bart types information into a computer database.

C. Phuong works on a research project and creates a report for her boss.

D. Anton answers phone calls and greets guests who visit a company.

Molecular weight distribution is an important factor in the polymer’s physical properties. The molecular weight of a polymer, for example, influences its tensile strength, viscosity, and other properties. As a result, the average molecular weight and molecular weight distribution of a polymer must be determined before it can be used in any application.

A 50 g polymer sample was fractionated into six samples of different weights given in the table below. Estimate the number average and weight average molecular weights of the original sampleSo, let us first calculate the number average molecular weight and weight average molecular weight by using the formulae.Number average molecular weight is given by the formula, M_n = Σ(N_iM_i)/ΣN_iwhere N_i and M_i are the mole fraction and molecular weight of ith fraction of polymer respectively.From the table given above,ΣN_i = 1.0 (sum of all mole fractions of different fractions of polymers is always equal to 1)For Number average molecular weight, M_n = (4*10^20 * 2000 + 6*10^20 * 8000 + 10*10^20 * 20000 + 15*10^20 * 40000 + 18*10^20 * 80000 + 19*10^20 * 120000)/(4*10^20 + 6*10^20 + 10*10^20 + 15*10^20 + 18*10^20 + 19*10^20)M_n

= 1,207,527 g/molWeight average molecular weight is given by the formula, M_w =

Σ(N_iM_i^2)/Σ(N_iM_i)For Weight average molecular weight, M_w = (4*10^20 * 2000^2 + 6*10^20 * 8000^2 + 10*10^20 * 20000^2 + 15*10^20 * 40000^2 + 18*10^20 * 80000^2 + 19*10^20 * 120000^2)/(4*10^20 * 2000 + 6*10^20 * 8000 + 10*10^20 * 20000 + 15*10^20 * 40000 + 18*10^20 * 80000 + 19*10^20 * 120000)M_w = 2,715,479 g/molNow let us classify the molecular weight distribution of the original sample as narrow or broad.Now, Molecular weight distribution (Mw/Mn) = (2,715,479/1,207,527)Molecular weight distribution (Mw/Mn) = 2.25The distribution is broad since the molecular weight distribution of the original sample is greater than 1, which implies a broad range of molecular weights. This indicates that the polymer sample is made up of polymers with a wide range of molecular weights.

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The parts of machines that use abrasive wheels which must be protected by safety guards are the wheel periphery and adjustable tongue.

Abrasive wheels are commonly used in grinding machines, and they can cause severe injury if safety precautions are not taken. Safety guards are designed to protect operators from the hazards associated with these machines. The wheel periphery and adjustable tongue are the parts of the machine that come into contact with the abrasive wheel and can cause serious injury if not properly guarded.

The spindle end, nut, and flange projections must also be guarded. They are located on the machine's spindle, which holds the abrasive wheel in place. If the spindle is not guarded, an operator's hand or other body part could become caught in the moving parts, causing severe injury or even amputation.

Mounted wheels less than two inches in diameter may also require guards. These small wheels can break apart during use and cause flying debris that can injure the operator.

In summary, safety guards are an essential part of any machine that uses abrasive wheels. They must be designed to protect the operator from the wheel periphery and adjustable tongue, as well as the spindle end, nut, and flange projections, and mounted wheels less than two inches in diameter.

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In machines using abrasive wheels, the spindle end, nut, flange projections, mounted wheels less than two inches diameter, wheel periphery and adjustable tongue, and the horizontal plane of the spindle must all be guarded for safety and durability.

In machines that use abrasive wheels, several parts must be protected by safety guards to prevent mishaps or accidents. Firstly, the spindle end, nut, and flange projections must be guarded as they come into direct contact with the wheel and any fragmentation could lead to serious damage. Secondly, mounted wheels less than two inches in diameter must be protected as their small size makes them susceptible to quicker wear and tear. Thirdly, the wheel periphery and adjustable tongue need to be guarded as they are exposed areas during operation. Lastly, the horizontal plane of the spindle must also be protected as it is a crucial part of the wheel assembly process and any damages can compromise the functionality of the entire machine.

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A saturated soil with a mass of 43.2g. When dried in the oven its mass was 30g, the limit of contraction is 0.4.

For the limit of contraction, we know that:

Limit of contraction (Lc) = (Vw - Vd) / Vw

Lc = (20 - 12) / 20

Lc = 8 / 20

Lc = 0.4

Therefore, the limit of contraction is 0.4.

Now for the dry weight,

Dry unit weight (γd) = γw / (1 + e)

e = (Gs - 1) / Gs

n = e / (1 + e)

w = (wet weight - dry weight) / dry weight

In this case,

γw = 9.81 kN/\(m^3\)

Gs = 2.74.

So,

γd = 2010 / (1 + ((2.74 - 1) / 2.74))

γd = 2010 / (1 + 0.727)

γd = 2010 / 1.727

γd = 1163.54 kg/m^3

e = (2.74 - 1) / 2.74

e = 1.74 / 2.74

e ≈ 0.635

n = 0.635 / (1 + 0.635)

n = 0.635 / 1.635

n ≈ 0.388

w = (43.2 - 30) / 30

w = 13.2 / 30

w ≈ 0.44

Therefore, the dry unit weight (γd) is approximately 1163.54 kg/m^3, the void ratio (e) is approximately 0.635, the porosity (n) is approximately 0.388, and the moisture content (w) is approximately 0.44.

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The company should go with the out-house treatment option, as it has a lower annual worth value and will result in lower costs over the 11-year period.

GivenDataAn oil refinery finds that it is necessary to treat the waste liquids from a new process before discharging them into a stream. The treatment will cost $40,000 the first year, but process improvements will allow the costs to decline by $4,000 each year.An outside company will process the wastes for the fixed price of $20,000/year throughout the 11 year period, payable at the beginning of each year.MARR = 7%FormulaAnnual Worth (AW) = (P/A, i%, n)Annual Worth (AW) = Present Worth (PW) + Future Worth (FW)Where,P = Initial Cost (Present Worth)A = Capital Recovery Factori = InterestRaten = Life of the ProjectCalculationFirst of all, we calculate the AW of in-house treatment. The cash outflow would be $40,000 in year 0, then $36,000 ($40,000 – $4,000) in year 1, then $32,000 ($36,000 – $4,000) in year 2, and so on until year 10, and the cash inflow would be $0 as there is no salvage value.Annual Worth (AW) = (P/A, i%, n)Present Worth (PW) = $40,000Future Worth (FW) = $0Capital Recovery Factor (CRF) = (i(1 + i)n)/((1 + i)n – 1) = (0.07(1 + 0.07)11)/((1 + 0.07)11 – 1) = 0.122053Annual Worth (AW) = (P/A, i%, n)= ($40,000/0.122053)= $327,814.53Therefore, the AW of in-house treatment is $327,814.53.Now, we calculate the AW of out-house treatment. The cash outflow would be $20,000 in each year from year 0 to year 10, and the cash inflow would be $0 as there is no salvage value.Annual Worth (AW) = (P/A, i%, n)Present Worth (PW) = $20,000Capital Recovery Factor (CRF) = (i(1 + i)n)/((1 + i)n – 1) = (0.07(1 + 0.07)11)/((1 + 0.07)11 – 1) = 0.122053Annual Worth (AW) = (P/A, i%, n)= ($20,000/0.122053)= $163,907.27Therefore, the AW of out-house treatment is $163,907.27. AW in-house treatment = $327,814.53 AW out-house treatment = $163,907.27.

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Answer:

Explanation:

I do not have access to your diagram but based on what I know, a food web cannot sustain itself without decomposers. They are the main guys who break down dead animals, plants into organic or inorganic nutrients which are needed by the primary producers to grow. So if decomposers are not there then producers cannot produce and thus herbivores cannot live and carnovers die out as well. Hope this makes sense.

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A food web cannot sustain itself without decomposers. They are the main components who break down dead animals, plants into organic or inorganic nutrients which are needed by the primary producers to grow.

Decomposers are living organisms like fungi and bacteria. These decomposers are heterotrophic which means they must take the nutrients and it cannot make its own food. Decomposers play an eminent role in an ecosystem because of their process of nutrition.

They break down the dead plants and the animals. These break down the waste of the organisms. They are very eminent for all the ecosystem. If these were not in the ecosystem, the plants cannot get essential nutrients, dead matter and waste would deposit. So the nutrients in them gets recycled back so that they can be used again.

When fish dies, there is nothing which is ingested and the bladder air starts to release, which causes the fish to sink to the bottom. After few days, the internal organs of the dead fish gets decomposed, after which a gas is formed.

Therefore, A food web cannot sustain itself without decomposers.

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Answer:

The answer would most likely be c

Answer:

Voltage across resistor = 2 v

Explanation:

Given data

pulse height = 2 v

pulse width = 4s

calculate voltage across resistor ( the free hand sketch attached below explains more )

pulse height is also = amplitude of voltage ) = 2v

The voltage across the resistor = 2v  Since  the voltage from the source of the circuit is equal to the amplitude voltage in the circuit ( assuming no loss of voltage )

also the graphical representation of the problem is attached below

Housing in a Roman city was made up of apartment blocks called insulae.

In ancient Rome, housing was primarily composed of multi-story apartment buildings known as insulae. These insulae provided accommodation for the city's inhabitants and were constructed using a combination of brick, concrete, and wood. The insulae varied in size and quality, with some featuring multiple rooms and amenities while others were more basic.

The apartments were often cramped, lacking proper sanitation and ventilation, leading to challenges in terms of hygiene and living conditions. Despite these limitations, the insulae played a crucial role in shaping the urban landscape and accommodating the growing population of Roman cities.

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Answer:

Computer programming is where you learn and see how computers work. People do this for a living as a job, if you get really good at it you will soon be able to program/ create a computer.

Explanation:

Hope dis helps! :)

This problem involves redesigning the outer truss sections of a 3-D truss structure by determining the loads in various truss members, proposing a cross-sectional geometry for each member, calculating the safety factors, estimating the material cost, and discussing manufacturability.

To determine the structural loads in truss members AD, AE, BE, CD, CE, and DE, the loads acting on the truss should be analyzed, and the method of joints or method of sections can be used. Once the loads in each member are determined, a cross-sectional geometry can be proposed for each member that meets the safety factor requirements. The selection of standard stock sizes can be made from suppliers like OnlineMetals.

The actual FoS for outer truss sections under the maximum defined loads can be calculated using the proposed geometry. The material cost per truss sub-section can be estimated by considering the lengths and diameters of each member and the material cost per unit length of the selected stock size. The manufacturability of the redesigned outer truss sections can be discussed by considering factors like the availability of the selected stock sizes, the feasibility of joining the members, and the manufacturing processes involved.

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The man is developing approximately 0.57 horsepower while riding uphill at 5% grade and constant speed of 15 mi/hr.

To calculate the power P that the man is developing while riding uphill at Spercent grade and constant speed of 15 mi/hr, we can use the formula:

P = (F + mg) * v

Where F is the force exerted by the man on the pedals, m is the mass of the man and the bicycle (200 lb), g is the acceleration due to gravity (32.2 ft/s^2), and v is the velocity of the bicycle (15 mi/hr or 22 ft/s).

To determine F, we need to first calculate the total force required to overcome the uphill slope. This can be found using the following formula:

F_slope = m * g * sin(theta)

Where theta is the angle of the slope in radians. To convert Spercent grade to radians, we can use the formula:

theta = arctan(S/100)

Where S is the slope percentage. For example, if S is 5%, then theta = arctan(0.05) = 2.86 degrees or 0.05 radians.

So, for the given problem, let's assume S is 5%. Then:

theta = arctan(0.05) = 0.05 radians
F_slope = 200 * 32.2 * sin(0.05) = 33.23 lb

Now, we can calculate the power P as:

P = (F + mg) * v = (F_slope + 200 * g) * v

Substituting the values, we get:

P = (33.23 + 200 * 32.2) * 22 = 14984.4 ft-lb/s or 0.57 hp

Therefore, the man is developing approximately 0.57 horsepower while riding uphill at 5% grade and constant speed of 15 mi/hr.

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Cave-ins  known as the greatest danger associated with excavations. A cave-in happens when an excavation's walls give way. Cravings can be fatal.

What is Excavation?
Excavation is the process of uncovering and removing material from a given area. It is a process that typically requires the use of specialised machinery, such as shovels, bulldozers, backhoes, and other digging equipment. Excavation is most commonly associated with construction, mining, and archaeological activities, but it can also be used in the removal of earth, rock, and other materials to create space for roads, dams, canals, and other land development projects. Excavation is also used in exploration activities, such as oil and gas drilling, as well as in the restoration of damaged ecosystems. Excavation involves the careful removal of material, often in layers, to ensure the safety of workers and to preserve the integrity of the site.

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Answer:

Use cases are known to be a set of instruction  or processes between a User/Actor with the system to produce a desired input.

With the aid of a diagram, the set of use cases that are carried out in this ATM are given below:

Insert PIN

(1)Perform required transaction

(2)Withdrawal

(3)Deposit

(4)Transfer

(5)Change PIN

(6)Exit

Note: Kindly find an attached diagram of the Use case as part of the solution to process carried out at the ATM

Sources: The diagram of the Use case for ATM was researched and taken from Quizlet.

Explanation:

Solution

Use cases are normally a set of instruction  or processes between a User/Actor with the system to produce a desired input.

A use case diagram or image is a graphical representation of all the use case or processes that connects or interact with the system

The use case diagram is a part of Unified Modelling Language also called the UML.

The set of use cases that are carried out in this ATM use case diagram to know the requirements of the ATM is shown below:

Now both the customer/client and Bank are seen as Actors.

Actors are the ones or people that interface with the system.

The ATM is often used to withdraw money and it also have some requirement.

Some of the use-cases that could serve as a basis for understanding the requirements for an ATM system are;

Customer (actor)  often uses bank ATM to know or see the Balances of his/her bank accounts. They use it also to Deposit Funds and Withdraw Cash.

They can also use it to Transfer Funds (use cases). ATM Technician are known to give a form of Maintenance and Repairs. All these used cases is one that involve Bank as the actor  when it is liked to customer transactions or to the ATM servicing.

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Answer:

A. The force on the car was greater since it was moving

Explanation:

Hope this helps : )

Answer:  C. The forces exerted by the car and wall were equal

Explanation:  Isaac Newton’s third law states if an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A.

Explanation:

1. Ensure all circuits are de-energized before beginning work (29 CFR 1926.416(a)(3)).

2. Controls to be deactivated during the course of work on energized or de-energized

equipment or circuits must be tagged (29 CFR 1926.417(a)).

3. Employees must be instructed to recognize and avoid unsafe conditions associated with

their work (29 CFR 1926.21(b)(2)).

A. The number of binary digits required to encode a sample is 16 bits.

B. The resulting signal-to-quantization-noise ratio (SQNR) of the uniform quantizer output in part (a) is 26.8 dB

C. The number of binary digits per second (bit/s) required to encode the audio signal is 705,600 bits/s.

D. The number of bits per second required to encode the signal, and the minimum bandwidth required to transmit the encoded sign is 705,600 bits/s.

To determine the values above, the following solutions are provided:

(a) The number of binary digits required to encode a sample is the number of bits needed to represent L = 65,536 levels. This is equal to the base-2 logarithm of L, or log₂L. Therefore, the number of binary digits required to encode a sample is log₂L = log₂65536 = 16 bits.

(b) The signal-to-quantization-noise ratio (SQNR) is the ratio of the signal power to the quantization noise power. The quantization noise power is the difference between the signal power and the quantized signal power, normalized by the number of bits. Therefore, the SQNR is given by:

SQNR = 10 * log₁₀ (signal power / quantization noise power)

= 10 * log₁₀ ((signal power) / (signal power - quantized signal power))

= 10 * log₁₀ (1 + (quantized signal power / (signal power - quantized signal power)))

Since the signal power is 0.1 W and the peak voltage is 1 V, the signal power is (1 V)² / (2 * 1 Ω) = 0.5 W. The quantized signal power is (1 V)² / (2 * L) = (1 V)² / (2 * 65536) = 2.4 x 10^-5 W. Therefore, the SQNR is:

SQNR = 10 * log₁₀ (1 + (2.4 x 10^-5 W / (0.5 W - 2.4 x 10^-5 W)))

= 10 * log₁₀ (1 + 480)

= 10 * log₁₀ 481

= 10 * 2.68

= 26.8 dB

(c) The number of binary digits per second (bits/s) required to encode the audio signal is the number of bits per sample multiplied by the number of samples per second. Since each sample is encoded using 16 bits, and the audio signal is sampled at a rate of 44,100 samples per second, the number of bits per second required to encode the signal is 16 bits/sample * 44,100 samples/s = 705,600 bits/s.

(d) To transmit the encoded signal, the minimum bandwidth required is equal to the number of bits per second required to encode the signal. Therefore, the minimum bandwidth required to transmit the encoded signal is 705,600 bits/s.

Therefore, the correct answers are as given above

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You add up the individual resistances to determine the total overall resistance of several resistors connected in this manner.The formula used for this is as follows:Total R = R1 plus R2 plus R3 and so on

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