GNPC has three refineries that produce gasoline, which is then distributed to four large storage facilities. The total quantities (1000 barrels) produced by each refinery and the total requirements (1000 barrels) for each storage facilities, as well as the associated distribution costs are shown as follows. To (Cost, in GHS 100s) Refinery Accra Kumasi Bawku Aflao Refinery Available Tema 90 80 60 70 25 Takoradi 55 85 35 75 20 Saltpond 50 45 90 85 15 Storage Requirement 10 40 10 20 Due to recent challenges with storage facilities in Kumasi, the warehouse can only operate at 50% of its current storage capacity. a) Based on the information above, develop a network graph of this problem showing all costs and decision variables. Determine the initial feasible solution using Northwest Corner Rule and the total Sensitivity Analysis AP 7 14 cost under this method. Major Topic Transportation Blooms Designation EV Score 7 b) determine the initial feasible solution using the Minimum Cell Cost and the total cost under this Method. Compare with the results in (a) and comment on the results based on the two approaches Major Topic Transportation Model: Minimum Cell cost Blooms Designation AN Score 7 c) Due to the bad nature of the transportation channels, distribution is prohibited from Takoradi to Bawku. Formulate the mathematical model to incorporate this in the problem Major Topic Transportation Model: Blooms Designation AP Score 6 TOTAL Sc

Answer:

Base current increases and collector current decreases

Explanation:

Base current increases and collector current decreases

The collector current does reduces because electrons that moves from emitter to base and the electrons that moves from collector to base are both cancelled out. Some of these electrons enter from the emitter to the collector because, in a way, the emitter has been heavily döped

An increase in base current also means that there is more bias to the transistor

Focus primarily on highly precise specifics of what Cornell offers and how it matches with your interests and values. You can regard this as a "Why us?" essay with some optional "Why major" spice.

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Excitons play a crucial role in silicon solar cells and are involved in several processes that contribute to the generation of electricity. Here are some key roles of excitons in silicon solar cells:

1. Absorption of Photons: When photons from sunlight strike the silicon material of a solar cell, they can be absorbed by silicon atoms, promoting an electron from the valence band to the conduction band. This process creates an exciton—a bound electron-hole pair.

2. Exciton Diffusion: After absorption, excitons can diffuse through the silicon material, moving towards the region of the solar cell where charge separation occurs. This diffusion process allows excitons to reach the vicinity of the p-n junction, where the separation of charges takes place.

3. Exciton Dissociation: At the p-n junction of a silicon solar cell, excitons can undergo dissociation. The electric field created by the junction separates the electron and hole of the exciton, allowing them to move freely in opposite directions as charge carriers.

4. Electron and Hole Transport: Once the exciton is dissociated, the free electron and hole can move independently within the solar cell. They are transported through the silicon material to the respective electrodes, creating an electric current that can be harnessed for external use.

5. Recombination: Excitons can also undergo recombination, where the electron and hole recombine, releasing energy in the form of light or heat. Recombination is undesirable in solar cells as it reduces the overall efficiency of the device.

To enhance the efficiency of silicon solar cells, various strategies are employed to minimize exciton recombination and improve exciton dissociation and charge carrier transport. These include the use of anti-reflection coatings, surface passivation techniques, and optimization of the device structure.

Overall, excitons play a vital role in the absorption and conversion of sunlight into electrical energy in silicon solar cells. Understanding and controlling exciton dynamics are essential for improving the performance of solar cells and advancing the field of photovoltaics.

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The function of the microscope diaphragm is c. regulates the amount of light passing through the specimen.

The microscope diaphragm is a component located below the stage of a microscope. Its primary function is to control the amount of light that passes through the specimen being observed. It consists of an adjustable aperture with different settings that allow the user to regulate the size of the opening.

By adjusting the diaphragm, the amount of light reaching the specimen can be controlled. Opening the diaphragm wider allows more light to pass through, while closing it down reduces the amount of light. This feature is particularly important when observing specimens that require specific lighting conditions, such as those that are highly transparent or have low contrast.

The microscope diaphragm helps achieve optimal illumination by balancing the brightness and contrast of the specimen. By controlling the amount of light, it assists in enhancing the visibility of the specimen's details and structures.

To summarize, the function of the microscope diaphragm is to regulate the amount of light passing through the specimen, allowing for better control of illumination and optimizing the observation conditions.

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Answer:

Given, diameter of pin head d = 1 mm = 1/25.4 = 0.0394 in Surface area of a pinhead, A = 4pr^2 =

Explanation:

eesh

To avoid becoming drowsy on an expressway, there are several steps you can take. Firstly, it is important to keep your windows closed while driving. This helps to reduce noise and distractions from outside, creating a more conducive environment for alertness. Additionally, opening the windows can create a draft that may make you feel drowsier. Secondly, engaging in conversation with a friend or passenger can help keep your mind active and focused. Talking about various topics and sharing stories can help you stay mentally engaged and prevent drowsiness. Lastly, it is advisable to take regular breaks and stop driving every couple of hours. Pulling over at a rest area or gas station allows you to stretch your legs, get some fresh air, and give your mind a break from the monotony of driving. Taking short breaks can help you stay alert and maintain your energy levels throughout the journey. Remember, ensuring your safety and the safety of others on the road is of utmost importance, so it is crucial to take proactive measures to prevent drowsiness while driving on the expressway.

Answer:

-Computer going blank .

-Mouse pointer freezing.

-Keyboard getting stuck.

Explanation:

Any problems with the computer related to its hardware will consist of issues directly related to the physical components of the PC or computer. So any problem involving the physical component of the computer, then it can be termed as a hardware problem.

Based on the problems given in the option, the hardware issues will be problems with the computer, mouse, and keyboard. These form the physical parts of the computer so they will be a hardware issue.

Thus, the correct answers are the third, fourth, and fifth options.

Answer:

c d e

Explanation:

The thread is crucial for ensuring the quality and durability of the garment. It is formed by spinning and twisting fibers together, creating a continuous strand that is used to hold the various components of the garment together.

The quality of the thread directly affects the overall quality of the garment. A strong and well-constructed thread is essential for providing structural integrity and preventing the garment from unraveling or falling apart over time. It must be able to withstand the stresses and strains that occur during regular use, including stretching, pulling, and washing.

When choosing the appropriate thread for a garment, several factors are considered, including the type of fabric, the intended use of the garment, and the desired aesthetic. Different threads have varying levels of strength, elasticity, and resistance to abrasion. For example, a heavier fabric might require a thicker and more robust thread, while a delicate fabric may require a finer and more delicate thread to prevent damage.

The quality of the thread also relates to its consistency and uniformity. Irregularities in the thread's thickness or tension can result in uneven stitches, which may compromise the appearance and functionality of the garment. In addition, the thread's color and texture should be chosen carefully to complement the fabric and enhance the overall aesthetic appeal.

In the manufacturing process, quality control measures are implemented to ensure that the thread meets the required standards. This involves testing the thread for strength, colorfastness, and resistance to abrasion. By using high-quality thread and maintaining strict quality control, garment manufacturers can ensure that the final product meets or exceeds the expectations of customers in terms of both appearance and durability.

In summary, the thread plays a vital role in holding a garment together and contributes significantly to its quality and longevity. Choosing the right thread and ensuring its consistent quality throughout the manufacturing process are essential for creating garments that are durable, aesthetically pleasing, and able to withstand regular use.

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Answer:

The answer is below

Explanation:

The mining auxiliary operations in underground mining involve various activities that are important for a successful mining operation particularly in the areas of productive operating conditions.

The activities involved in the auxiliary operations in underground mining include the following: ventilation, haulage, drainage, power supply, lighting, delivery of compressed air, water, supplies to the working sections, and communications.

Answer:

I think it is the 2,3,5 and 1 ones

The values of h at the two internal boundaries are :

Given data :

Z₁ = Z₂ = Z₃ = 25 m

h top = 120 m

h bottom = 100 m

K₁ = 0.0001 m/s

K₂ = 0.0005 m/s

K₃ = 0.0010 m/s

we will apply the formula below since flow is perpendicular to the bedding plane

Keq = \(\frac{Z1 + Z2 + Z3 }{\frac{Z1}{K1}+\frac{Z2}{K2} + \frac{Z3}{K3} }\)  ----- ( 1 )

Insert values given above into equation 1

Therefore ; Keq = 2.307 * 10⁻⁴ m/s

Hydraulic gradient ( Ieq ) = head loss / length

                                          = ( 120 - 100 ) / 3 * 25

                                   Ieq  = 0.266

Given that the flow is perpendicular to bedding plane

q1 = q2 = q3

V₁ = V₂ = V₃ = V

K₁i₁ = K₂i₂ = K₃i₃ = Keq * ieq

Hence :

V = Keq* Ieq

   = 2.307 * 10⁻⁴ * 0.266

   = 6.15 * 10⁻⁵ m/s .

Also;

K₁i₁ =  Keq * ieq = K₂i₂ = K₃i₃

therefore :

  i₁ = 0.615

  i₂ =  0.123

  i₃ = 0.0615

Pressure at point 1 ( i.e. pressure between first two formations )

h₁ = h top - i₁L₁

   = 120 - 0.615 * 25

   = 104.625 m

Pressure at point 2 ( i.e. pressure between the 2nd and 3rd formation )

h₂ = h₁ - i₂L₂

    = 104.625 - 0.123 * 25

    = 101.55 m

Therefore we can conclude that The values of h at the two internal boundaries are :  h₁ = 104.625 m , h₂ = 101.55 m

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probably alot lol

Explanation:

it probably is

The three ways that research can facilitate the work of building technicians is that:

Building technicians are known to be people who help with some tasks around construction projects and building works.

Note that building technician overseeing a scope of tasks such as monitoring build progress and as such The three ways that research can facilitate the work of building technicians is that:

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Answer:

Scheduling is a process of create a structure that shows the way a schedule is organized before developing a complete and final schedule of a project.

Explanation:

This is the initial step in of developing a project schedule for a capital cost project. At this stage, the estimates for the project are done involving the various actions in the project. A lot of details are included in this  schedule showing the activities that are to be performed and the constraints involved.

To calculate the force required to lift a weight, we need to determine the mechanical advantage provided by the jackscrew. The mechanical advantage can be calculated as the ratio of the handle length to the pitch of the screw.

How to calculate the force required to lift a weight?

In this case, the mechanical advantage is:

255 mm / 5.00 mm = 51

This means that for every 51 turns of the handle, the screw will raise the load by a distance equal to the pitch of the screw, or 5.00 mm.

Now that we have the mechanical advantage, we can calculate the required force as follows:

Force = Weight / Mechanical Advantage

Force = 19,400 N / 51

Force = 380 N

So, a force of 380 N must be applied to the handle of the jackscrew to lift the 19,400 N weight.

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Answer:

Subsystem denotes to the installation of some subassemblies, which can be very small or gigantic in size, that serve this subsystem and the overall system. ... In addition, availability importance measures are employed to rank various subsystems with regards to their impact on the overall system availability.

Explanation:

Subsystem refers to the installation of a few subassemblies that support this subsystem and the entire system.

An assembly is a template for a corridor's cross section at a certain station.

A set of interconnected subassemblies that are each connected to a center point or to other subassemblies makes up an assembly. An assortment of shapes, linkages, and points make up a subassembly.

The installation of a few subassemblies that support this subsystem and the overall system is referred to as a subsystem.

These subassemblies can range in size from little to huge. In addition, subsystems are ranked according to their impact on the overall system's availability using availability importance criteria.

The primary memory, the input/output subsystem, and the central processor unit (CPU) are the three basic categories or subsystems that make up a computer. Data processes are carried out by the central processing unit (CPU).

A subassembly could be:

Thus, these are important subassemblies and subsystems.

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Personal watercraft, also known as water scooters or Jet Skis, are considered a type of recreational vessel.

What is recreational vessel?

A recreational vessel is any type of watercraft that is used for leisure and recreational activities, rather than for commercial or military purposes. Some examples of recreational vessels include sailboats, powerboats, personal watercraft, houseboats, and recreational fishing boats.

Recreational vessels come in a wide range of sizes, from small inflatable rafts to large luxury yachts, and can be powered by different means, such as sail, human power, internal combustion engines, or electric motors.

Personal watercraft, also known as water scooters or Jet Skis, are considered a type of recreational vessel. They are small, lightweight watercraft that are powered by an internal combustion engine or an electric motor and are designed for one or two people.

They are called personal watercraft because of their small size and the fact that they are intended for personal use, typically for recreation and leisure activities such as water skiing, wakeboarding, and leisure cruising. They are operated by standing or sitting on the watercraft, and steering and accelerating are typically done with handlebars, similar to a motorcycle.

Personal watercraft are subject to different regulations compared to other types of vessels, such as larger boats and ships, due to their size, speed, and maneuverability. For example, they may have specific rules regarding minimum age of the operator, required safety equipment, and speed limits in certain areas.

In summary, personal watercraft are considered a type of recreational vessel designed for personal use, with specific regulations and rules due to their unique characteristics.

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The frequency of the AC supply is 0.036 Hz.

We know that an alternating current circuit is the kind of circuit in which we do have the resistor in addition to the to the capacitor or the inductor or both as the case may be.

Here we are told that; A 32x10^-6 F capacitor is connected to a 60V AC supply in series with a 56 ohm resistor. The current flowing in the circuit is 0.16 A.

We can tell that the impedance of the circuit can be obtained by the use of the formula;

V = IZ

V = voltage

I = current

Z = impedance

Z = V/I

Z = 60 V/0.16 A

Z = 375 ohms

Then we have;

Z = √R^2 - XC^2

375 = √(56)^2 - (1/2 * 3.142 * f *  32x10^-6)^ 2

140625 = 3136 - (1/2 * 3.142 * f *  32x10^-6)^ 2

140625 -  3136  =  (1/2 * 3.142 * f *  32x10^-6)^ 2

137489 = 1/2 * 10^-4 f

137489 * 2 * 10^-4 f = 1

f = 1/137489 * 2 * 10^-4

f = 0.036 Hz

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The COP of the refrigeration cycle is 3.83, and the power required to service a 150 kW cooling load is 39.2 kW.

The given problem involves an ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid. The condenser is maintained at 800 kPa, and the evaporator is at -20°C.

We need to determine the COP (Coefficient of Performance) and the power required to service a 150 kW cooling load. Firstly, we can determine the enthalpy change of the refrigerant during the cycle by using the saturation tables.

The enthalpy at the evaporator inlet is -129.2 kJ/kg, and at the condenser outlet is 214.5 kJ/kg. The COP can be calculated as the ratio of cooling effect to the work input, which is equal to the enthalpy change divided by the work input.

Thus, COP = (-129.2 - 214.5) / -129.2 = 3.83.

The power required to service the cooling load can be determined as the product of cooling load and COP.

Therefore, power = 150 kW / 3.83 = 39.2 kW.

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Answer:

V= 90h cm³ where h is the height of the rectangular prism.

Explanation:

The formula for volume of a rectangular prism is ;

V=l*w*h  where;

V=volume in cm³

l= length of prism=10cm

w =width of the prism = 9 cm

Assume the height of the prism as h cm then the volume will be;

V= 10* 9*h

V= 90h cm³

when the value of height of the prism is given, substitute that value with h to get the actual volume of the prism.

Answer: I believe the answer is 963.75 if I am reading the question right.

Explanation:

ACH=60L/V

ACH=2.25

V=25,700

25,700*2.25=57,825

57,825/60=963.75

L=963.75

60(963.75)/25,700=2.25

The maximum amount of air leakage that would result in an ACH of no more than 2.25 would be approximately 956.25 cfm (cubic feet per minute).

Given that ACH of a house is a measure of its​ air-tightness and the formula for ACH is ACH = 60L/V. It means that ACH is inversely proportional to air-tightness or leakage. Therefore, the lower the air leakage the higher the ACH and vice versa.Mathematically, it is represented as:ACH ∝ 1/L or L ∝ 1/ACH

We can solve the question as follows:ACH = 2.25 (since the maximum ACH is 2.25)The volume of the house, V = 25,700 cubic feet.Substituting in the formula,2.25 = 60L/25,700Dividing both sides by 2.25 gives us,L = (2.25 * 25,700)/60L = 956.25 cubic feet per minute (cfm)

Therefore, the maximum amount of air leakage that would result in an ACH of no more than 2.25 is approximately 956.25 cfm (cubic feet per minute).

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The research project/topic proposal focuses on the applications of hydrographic surveying. It involves conducting research and reviewing related literature to explore the various uses and advancements in hydrographic surveying techniques and technologies.

Hydrographic surveying plays a crucial role in various fields such as marine navigation, coastal engineering, offshore resource exploration, and environmental monitoring. This research project aims to investigate the applications of hydrographic surveying and analyze its impact on these industries.

The proposal will begin with a comprehensive literature review to gather existing knowledge and understanding of hydrographic surveying techniques, equipment, and methodologies. This review will encompass both academic research papers and industry reports to gain insights into the current state of the field and identify any gaps or areas that require further investigation.

The research project will then delve into specific applications of hydrographic surveying, such as the use of multibeam sonar for seafloor mapping, the application of LiDAR technology for coastal zone management, the role of hydrographic surveys in the installation and maintenance of underwater infrastructure, and the use of remote sensing techniques for monitoring and assessing coastal erosion and sediment transport.

By conducting thorough research and reviewing relevant literature, this project aims to contribute to the understanding of the applications of hydrographic surveying and highlight its importance in various industries. It will also identify potential areas for further research and advancements in hydrographic surveying technologies and methodologies.

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Answer:

Explanation:

9514 1404 393

Answer:

  (x, y) = (5.76, 1 5/7)

Explanation:

The location of the centroid in the x-direction is the ratio of the first moment of area about the y-axis to the total area. Similarly, the y-coordinate of the centroid is the first moment of area about the x-axis, divided by the area.

For the moment about the y-axis, we can define a differential of area as ...

  dA = (y2 -y1)dx

where y2 = √(x/k2) and y1 = k1·x^3

The distance of that area from the y-axis is simply x.

So, the x-coordinate of the centroid is ...

  \(\displaystyle c_x=\frac{a_x}{a}=\frac{\int{x\cdot dA}}{\int{dA}}\\\\a_x=\int_0^{12}{x(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{5k_2^{1/2}}\cdot12^{5/2}-\frac{k_1}{5}12^5\\\\a=\int_0^{12}{(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{3k_2^{1/2}}\cdot12^{3/2}-\frac{k_1}{4}12^4\\\)

For k1 = 4/12^3 and k2=12/4^2, these evaluate to ...

  \(a_x=115.2\\a=20\\c_x=5.76\)

The y-coordinate of the centroid requires we find the distance of the differential of area from the x-axis. We can use (y2 +y1)/2 for that purpose. Then the y-coordinate is ...

  \(\displaystyle c_y=\frac{a_y}{a}\\\\a_y=\int_0^{12}{(\frac{y_2+y_1}{2}(y_2-y_1))}\,dx=\frac{1}{2}\int_0^{12}{(\frac{x}{k_2}-(k_1x^3)^2)}\,dx\\\\a_y=\frac{12^2}{4k_2}-\frac{k_1^212^7}{14}=\frac{240}{7}\\\\c_y=\frac{12}{7}\approx1.7143\)

The centroid of the shaded area is ...

  (x, y) = (5.76, 1 5/7)

Answer:

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

Explanation:

We must estimate the monthly storage change of the lake by considering inflows, outflows, seepage and evaporation losses and precipitation. That is:

\(\Delta V_{storage} = V_{inflow} -V_{outflow}-V_{seepage}-V_{evaporation}+V_{precipitation}\)

Where \(\Delta V_{storage}\) is the monthly storage change of the lake, measured in cubic feet.

Monthly inflow

\(V_{inflow} = \left(30\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{inflow} = 77.76\times 10^{6}\,ft^{3}\)

Monthly outflow

\(V_{outflow} = \left(27\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{outflow} = 66.98\times 10^{6}\,ft^{3}\)

Seepage losses

\(V_{seepage} = s_{seepage}\cdot A_{lake}\)

Where:

\(s_{seepage}\) - Seepage length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{seepage} = 1.5\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{seepage} = (1.5\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{seepage} = 2.86\times 10^{6}\,ft^{3}\)

Evaporation losses

\(V_{evaporation} = s_{evaporation}\cdot A_{lake}\)

Where:

\(s_{evaporation}\) - Evaporation length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{evaporation} = 6\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{evaporation} = (6\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{evaporation} = 11.44\times 10^{6}\,ft^{3}\)

Precipitation

\(V_{precipitation} = s_{precipitation}\cdot A_{lake}\)

Where:

\(s_{precipitation}\) - Precipitation length gain, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{precipitation} = 4.25\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{precipitation} = (4.25\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{precipitation} = 8.10\times 10^{6}\,ft^{3}\)

Finally, we estimate the storage change of the lake during the month:

\(\Delta V_{storage} = 77.76\times 10^{6}\,ft^{3}-66.98\times 10^{6}\,ft^{3}-2.86\times 10^{6}\,ft^{3}-11.44\times 10^{6}\,ft^{3}+8.10\times 10^{6}\,ft^{3}\)

\(\Delta V_{storage} = 4.58\times 10^{6}\,ft^{3}\)

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

The volume of water gained and the loss of water through flow,

seepage, precipitation and evaporation gives the storage change.

Response:

Given parameters:

Area of the lake = 525 acres

Inflow = 30 ft.³/s

Outflow = 27 ft.³/s

Seepage loss = 1.5 in. = 0.125 ft.

Total precipitation = 4.25 inches

Evaporator loss = 6 inches

Number of seconds in a month is found as follows;

\(30 \ days/month \times \dfrac{24 \ hours }{day} \times \dfrac{60 \, minutes}{Hour} \times \dfrac{60 \, seconds}{Minute} = 2592000 \, seconds\)

Number of seconds in a month = 2592000 s.

Volume change due to flow, \(V_{fl}\) = (30 ft.³/s - 27 ft.³/s) × 2592000 s = 7776000 ft.³

1 acre = 43560 ft.²

Therefore;

525 acres = 525 × 43560 ft.² =  2.2869 × 10⁷ ft.²

Volume of water in seepage loss, \(V_s\) = 0.125 ft. × 2.2869 × 10⁷ ft.² = 2,858,625 ft.³

Volume gained due to precipitation, \(V_p\) = 0.354167 ft. × 2.2869 × 10⁷ ft.² = 8,099,445.123 ft.³

Volume evaporation loss, \(V_e\) = 0.5 ft. × 2.2869 × 10⁷ ft.² = 11,434,500 ft.³

Which gives;

ΔV = 7776000 - 2858625 + 8099445.123 - 11434500 = 1582823.123

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Answer:

Explanation:

Complements are used in digital circuits, because it is faster to subtract by adding complements than by performing true subtraction. The binary complement of a number is created by reversing all bits and adding 1. The carry from the high-order position is eliminated.