There is an even better approximation algorithm, a so-called PTAS1.
What is the algorithm with the LPT rule?we are given m machines, and 2m+ 1 jobs. There are three jobs with processing time m, and
2 jobs with processing times m + 1, m + 2, . . . , 2m − 1 each. In case of LPT, all but one of the machines
get two jobs with a total processing time of 3m − 1, and a single machine gets three jobs with a total
of 4m − 1 processing time. Thus, the makespan is 4m − 1. OPT schedules the three m jobs on a single
machine, and the remaining jobs on the remaining m−1 machines, such that each of those machines get
with the exception of one machine receiving three jobs with a total processing time of 4m 1.
the approximation ratio approaches 4/3.
There is an even better approximation algorithm, a so-called PTAS1
for this problem, but it is not
covered by this lecture.
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if the ignition switch ground wire becomes disconnected the magneto
The magneto will lose spark if the ignition switch ground wire becomes disconnected.
What happens to the magneto if the ignition switch ground wire becomes disconnected?If the ignition switch ground wire becomes disconnected, the magneto will not receive the necessary grounding connection, leading to a loss of spark and preventing the engine from starting or running.
The ignition switch ground wire serves as a vital electrical pathway for the proper functioning of the ignition system.
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Who is Robert Goldard
Answer:
An am American engineer
(Multiple Choice) Which of the following is most likely to be considered an overarching goal of the International Society of Technology in Education (ISTE) standards?
a. To enable teachers to create dynamic learning experiences that integrate our understanding and the relevance of technological tools that provide educational support.
b. To provide roles for government and professional organizations in the development of standards.
c. To introduce computers and technology into the classroom.
d. To create consequences for compliance and noncompliance for use of technology in the classroom.
The International Society for Technology in Education (ISTE) Standards for Learning, Teaching, and Leading in the Digital Age provide us with a comprehensive roadmap for effective use of technology in schools around the world.
What is ISTE Standards?The International Society for Technology in Education (ISTE) Standards, formerly known as the National Educational Technology Standards, are guidelines for using technology in teaching and learning. The International Society for Technology in Education, a nonprofit membership organization for educators focused on educational technology, publishes them.ISTE teacher standards place high expectations on technology educators. These standards prepare teachers to teach advanced concepts and technology in the classroom. It also encourages teachers to share ideas for engaging their students with one another.These standards include learning how to be responsible and safe online, understanding and applying computational thinking, effectively curating and evaluating digitally acquired knowledge, and successfully communicating with others using digital tools.To learn more about ISTE standard refer to :
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what are the properties of advanced materials?
Sandpaper.
Rubble.
Pyramid.
Wrinkled.
Marbled.
One - tenth kilogram of air as an ideal gas with k= 1.4 executes a carnot refrigeration cycle as shown i fig. 5,16, the isothermal expansion occurs at - 23C with a heat transfer to the air of 3.4 kj. The isothermal compression occurs at 27C to a final volume of 0.01m. Using the results of prob. 5.80 adapted to the case, Determine (a) the pressure, in Kpa, at each of the four principal states (b) the work, in KJ for each of the four processes (c) the coefficient of performance
Answer:
Hello your question is incomplete attached below is the missing part
a) p1 = 454.83 kPa, p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa
b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ
c) 5
Explanation:
Given data:
mass of air ( m ) = 1/10 kg
adiabatic index ( k ) = 1.4
temperature for isothermal expansion = 250K
rate of heat transfer ( Q12 ) = 3.4 KJ
temperature for Isothermal compression ( T4 ) = 300k
final volume ( V4 ) = 0.01m ^3
a) Calculate the pressure, in Kpa, at each of the four principal states
from an ideal gas equation
P4V4 = mRT4 ( input values above )
hence P4 = 860.959kPa
attached below is the detailed solution
b) Calculate work done for each processes
attached below is the detailed solution
C) Calculate the coefficient of performance
attached below is detailed solution
What time will current of 10A transferred from a charges of 50C
Answer:
8×10^-19 seconds
Explanation:
i=∆q/∆t
=> ∆t=∆q/i=50×1.6×10^-19/10=8×10^-19(s)
New, derived traits of human evolution over the past five million years include all of the following. except: O encephalization O lack of body hair O reduced sexual dimorphism O elon
New, derived traits of human evolution over the past five million years include all of the following except: elongation. The correct answer to the given question is option D .
Derived traits are new traits that have been passed down to a given organism by their ancestors. Some of the derived traits of human evolution over the past five million years include:
Encephalization: Encephalization is the evolutionary trend of brain size increasing relative to body size over geological time in a lineage. This evolutionary trend was observed in the human lineage.
Lack of body hair: One of the derived traits of human evolution is the lack of body hair. Humans have a far less dense body hair than other apes and primates. The loss of body hair facilitated an increase in sweat gland density, which helps us regulate our body temperature.
Reduced sexual dimorphism: The reduced differences between the sexes in size and appearance over time in a given population is referred to as reduced sexual dimorphism.
Elongation: Elongation is not a derived trait of human evolution over the past five million years, so it is the correct answer.
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Determine the enthalpy, volume and density of 1.0 kg of steam at a pressure of 0.5 MN/m2 and with a dryness fraction of 0.96
Answer:
Enthalpy, hsteam = 2663.7 kJ/kg
Volume, Vsteam = 0.3598613 m^3 / kg
Density = 2.67 kg/ m^3
Explanation:
Mass of steam, m = 1 kg
Pressure of the steam, P = 0.5 MN/m^2
Dryness fraction, x = 0.96
At P = 0.5 MPa:
Tsat = 151.831°C
Vf = 0.00109255 m^3 / kg
Vg = 0.37481 m^3 / kg
hf = 640.09 kJ/kg
hg = 2748.1 kJ/kg
hfg = 2108 kJ/kg
The enthalpy can be given by the formula:
hsteam = hf + x * hfg
hsteam = 640.09 + ( 0.96 * 2108)
hsteam = 2663.7 kJ/kg
The volume of the steam can be given as:
Vsteam = Vf + x(Vg - Vf)
Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)
Vsteam = 0.3598613 m^3 / kg
From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3
air enters a nozzle steadily at 2.05 /3 and 50 / and leaves at 0.657 /3 and 200 /. if the inlet area of the nozzle is 90 2 , determine determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.
The mass flow rate through the nozzle is 0.05535 kg/s, and the exit area of the nozzle is approximately 0.00422 m^2.
(a) To determine the mass flow rate through the nozzle, we can use the equation:
mass flow rate = density * velocity * area
Given:
Inlet conditions:
Density at inlet (ρ1) = 2.05 kg/m^3
Velocity at inlet (V1) = 3 m/s
Inlet area (A1) = 90 cm^2 = 0.009 m^2 (converting from cm^2 to m^2)
Using the given values, we can calculate the mass flow rate as follows:
mass flow rate = ρ1 * V1 * A1
mass flow rate = 2.05 kg/m^3 * 3 m/s * 0.009 m^2
mass flow rate = 0.05535 kg/s
Therefore, the mass flow rate through the nozzle is approximately 0.05535 kg/s.
(b) To determine the exit area of the nozzle, we can use the equation of continuity:
mass flow rate = density * velocity * area
Given:
Exit conditions:
Density at exit (ρ2) = 0.657 kg/m^3
Velocity at exit (V2) = 200 m/s
Exit area (A2) = ?
Using the given values and the known mass flow rate, we can rearrange the equation of continuity to solve for the exit area:
mass flow rate = ρ2 * V2 * A2
A2 = mass flow rate / (ρ2 * V2)
A2 = 0.05535 kg/s / (\(0.657 kg/m^3\) * 200 m/s)
A2 ≈ \(0.00422 m^2\)
Therefore, the exit area of the nozzle is approximately \(0.00422 m^2\).
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On-site oil storage containers must be marked used oil.a. Trueb. False
It is accurate to say that used oil must be clearly marked on on-site oil storage tanks.
What kind of container must be used to store leftover oil?Oils and other liquids have long been stored in steel containers, but normal steel is known to rust, corrode, and degrade faster than stronger materials. More protection is provided by a stainless steel container, and storing oil within one won't cause rust issues.
Where do you put the oil container?The best containers are made of glass or metal, and glass jars are excellent because they are airtight and simple to reuse. Always keep cooking oils in a tidy jar or container while storing them.
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Which of the following refers to the process in which engineers complete and repeat a sequence of steps to come closer to the project goal?
The process in question is referred to as the iterative design process. Engineers utilize this method by completing and repeating a sequence of steps in order to continually improve and refine their designs as they work towards achieving the project goal. This approach allows for flexibility and adaptability in the design process, as engineers can make adjustments and modifications based on feedback and testing, ultimately leading to a more successful outcome.
Given a 5 stage pipeline with stages taking 1, 2, 3, 1, 1 units of time, the clock period of the piple is
The clock period of the pipeline is 2 units of time.
Given a 5-stage pipeline with stages taking 1, 2, 3, 1, and 1 units of time
The clock period of the pipeline is equal to 3 units of time.
For a pipeline with 'n' stages, the clock period is equal to the sum of the time taken by each stage divided by 'n'.
The time taken by each stage of the pipeline is given as:
Stage 1: 1 unit of time
Stage 2: 2 units of time
Stage 3: 3 units of time
Stage 4: 1 unit of time
Stage 5: 1 unit of time
Therefore, the total time taken by all the stages is 1 + 2 + 3 + 1 + 1 = 8 units of time.
The number of stages in the pipeline is 5. Hence, the clock period of the pipeline is:
Clock period = (1 + 2 + 3 + 1 + 1)/5= 8/5= 1.6 units of time.
However, the pipeline must have integer clock cycles. Therefore, the clock period is rounded up to the nearest integer.
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Nitrogen (N2) at 1 bar, 300 K enters a compressor operating at steady state and is compressed adiabatically to an exit state of 10.5 bar, 645 K. The N2 is modeled as an ideal gas and kinetic and potential energy effects are negligible. Determine the work input, in kJ per kg of N2 flowing, the rate of entropy production, in kJ/K per kg of N2 flowing, and the isentropic compressor efficiency.
Answer:
a) 358.8 KJ/kg
b) 0.0977 KJ/K- kg
c) 83.28%
Explanation:
N2 at 300 k. ( use the properties of N2 at 300 k (T1) )
Cp = 1.04 KJ/kg-k , Cv = 0.743 KJ/Kgk , R = 0.1297 KJ/kgk , y = 1.4 ,
Given data:
T2 = 645 k
P1 = 1 bar , P2 = 10.5 bar
a)Determine the work input in KJ/Kg of N2 flowing
Winput = h2 - h1 = Cp( T2 - T1 ) = 1.04 ( 645 - 300 ) = 358.8 KJ/kg
b) Determine the rate of entropy in KJ/K- kg of N2 flowing
Rate of entropy ( Δs ) = Cp*InT2/T1 - R*In P2/P1
= 1.04 * In (645/300) - 0.1297 * In ( 10.5 / 1 )
= 0.0977 KJ/K- kg
c) Determine isentropic compressor efficiency
Isentropic compressor efficiency = 83.28%
calculated using the relation below
( h'2 - h1 ) / ( h2 - h1 ) = ( T'2 - T1 ) / ( T2 - T1 )
where T'2 = 587.314
In which of the following branches of engineering is the practice not restricted?
a) civil engineering.
b) mechanical engineering.
c) nuclear engineering.
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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Electricity is the flow of electrons from a negatively charged
material to a positively charged material.
Answer- TRUE
Cloud computing offers many significant advantages and opportunities for expansion, but it has its disadvantages as well. Which of the following options is the cheapest and simplest alternative to a cloud?a.Leased lineb.Remote access connectionc.internetd.VPN (virtual private network)
Note that cloud computing offers many significant advantages and opportunities for expansion, but it has its disadvantages as well. The option is the cheapest and simplest alternative to a cloud: Remote Access Connection. (Option B)
What is a Remote Access Connection?Remote access connections allow users to connect to a remote network or computer over the internet, allowing them to access resources and applications remotely.
This can be an effective and cost-effective way to access resources and applications without the need for cloud infrastructure.
Remote access works simply by connecting the remote user to the host computer through the internet. It does not necessitate any extra hardware. Distant access software must instead be downloaded and installed on both the local and remote PCs.
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The top down design process is sometimes called stepwise refinement
-overall task is broken down into a series of sub tasks
-each sub task is examined to see if it can be further broken down into more sub tasks
-the sub tasks are written in code
True. The top-down design process is indeed sometimes called stepwise refinement.
In this approach, the overall task or problem is broken down into a series of subtasks or smaller components. Each subtask is examined to determine if it can be further broken down into more detailed subtasks. This process continues until the subtasks are small enough to be written in code or implemented in a specific programming language. The stepwise refinement approach allows for a systematic and structured way of designing and implementing complex systems or programs.
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a universal chuck is able to hold square stock material securely on a lathe.
True or False
True. A universal chuck, also known as a three-jaw or four-jaw chuck, is a versatile clamping device used on a lathe. It is designed to hold various shapes of workpieces, including round, square, and hexagonal stock material.
The chuck jaws can be adjusted individually or simultaneously, depending on the chuck type, to securely grip the material during the machining process.
Three-jaw chucks, also called self-centering chucks, are commonly used for holding round or hexagonal stock. The jaws move in unison, automatically centering the workpiece. While they can hold square stock, their grip might not be as secure as with a four-jaw chuck.
Four-jaw chucks, also known as independent-jaw chucks, offer more versatility when holding irregularly shaped or square stock material. Each jaw can be adjusted individually, allowing precise positioning of the workpiece. This feature enables the operator to achieve a secure grip on the square stock, making it suitable for various machining operations on a lathe.
In summary, a universal chuck is capable of holding square stock material securely on a lathe. The four-jaw chuck, in particular, is best suited for this purpose due to its individually adjustable jaws. This versatility makes universal chucks essential tools in a machinist's toolbox.
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why is the electrical length of the so-called half-wave dipole actually taken to be slightly less than 0.5 λ at the design frequency?
The reason why the electrical length of a half-wave dipole is taken to be slightly less than 0.5 λ at the design frequency has to do with the way that the antenna is constructed and the properties of the materials that are used. While a half-wave dipole is theoretically supposed to be exactly 0.5 λ long, in practice it is difficult to achieve this length precisely due to the physical dimensions of the antenna elements and the way that they interact with the surrounding environment.
Additionally, the properties of the materials that are used to construct the antenna can also affect the electrical length of the dipole. For example, the velocity factor of the materials can cause the electrical length to be slightly shorter or longer than the physical length of the antenna. In order to compensate for these factors and ensure that the dipole operates at the desired frequency, the electrical length is typically adjusted to be slightly less than 0.5 λ.
Overall, while the half-wave dipole is a fundamental antenna design that is widely used in many applications, achieving precisely 0.5 λ electrical length can be challenging in practice. By adjusting the electrical length slightly, designers can ensure that the antenna operates as intended and achieves the desired performance characteristics.
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For a simply supported beam with a non-centered point loading, where would the approximate location of the maximum deflection be?
Answers Options:
at the end of the beam length
between the farther support and the load
at the center of the beam
between the closer support and the load
The maximum deflection for a simply supported beam with a non-centered point loading would be at the center of the beam. (option C)
What is the explanation for the above?When the point load is also at the mid-span, the absolute maximum deflection occurs. And it equals P l 3 48 E I. As a result, the distance between point 'P' and the right end for maximum deflection is equal to L/2, i.e. a = L/2.
Typically, support beams are built of wood, steel, or concrete. Among the numerous support, beams are: Cantilever - This beam only has an anchor at one end, leaving the other end unattached and swinging free. This beam is fixed because it is attached on both ends and will not rotate or move at the fulcrum.
The other types of beams are: simply supported beams, overhanging beams, fixed beams, and continuous beams.
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A helicopter landing pad is to be constructed over an area of 40 ft by 70 ft. It is to be 2 ft thick. How many cubic feet must be ordered?
The volume of cuboid in cubic feet which must be ordered is equal to 5,600 cubic feet.
Given the following data:
Length of helicopter landing pad = 40 ft.
Width of helicopter landing pad = 70 ft.
Height or thickness of helicopter landing pad = 2 ft.
How to calculate the volume of a geometric figure?Based on the information provided, we can reasonably infer and logically deduce that the shape of this helicopter landing pad is a cuboid.
Mathematically, the volume of a cuboid can be calculated by using this following formula:
Volume = l × w × h
Where:
l represents the length of a cuboid.w represents the width of a cuboid.h represents the height of a cuboid.Substituting the given parameters into the formula, we have;
Volume = 40 × 70 × 2
Volume = 5,600 cubic feet.
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There is an AC series circuit that is constructed of a 150.0-ohm resistor along with 300.0 ohm inductive reactance and 200.0 ohm capacitive reactance. What is the difference in phase between the current and resistor voltage of the circuit
Answer:
0°
Explanation:
The resistor voltage has the same phase as the circuit current. There is no phase difference.
Answer:
0° (zero degree)Explanation:
the difference in pjase between the current and resistor voltage of the given 150.0 ohm, 300.0 ohm and 200.0 ohmA wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.11 m2 and whose thickness is 4 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 11 m2 and 0.20 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window
Answer:
Explanation:
Given that,
The area of glass \(A_g\) = \(0.11m^2\)
The thickness of the glass \(t_g=4mm=4\times10^-^3m\)
The area of the styrofoam \(A_s=11m^2\)
The thickness of the styrofoam \(t_s=0.20m\)
The thermal conductivity of the glass \(k_g=0.80J(s.m.C^o)\)
The thermal conductivity of the styrofoam \(k_s=0.010J(s.m.C^o)\)
Inside and outside temperature difference is ΔT
The heat loss due to conduction in the window is
\(Q_g=\frac{k_gA_g\Delta T t}{t_g} \\\\=\frac{(0.8)(0.11)(\Delta T)t}{4.0\times 10^-^3}\\\\=(22\Delta Tt)j\)
The heat loss due to conduction in the wall is
\(Q_s=\frac{k_sA_s\Delta T t}{t_g} \\\\=\frac{(0.010)(11)(\Delta T)t}{0.20}\\\\=(0.55\Delta Tt)j\)
The net heat loss of the wall and the window is
\(Q=Q_g+Q_s\\\\=\frac{k_gA_g\Delta T t}{t_g}+\frac{k_sA_s\Delta T t}{t_g}\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j\)
The percentage of heat lost by the window is
\(=\frac{Q_g}{Q}\times 100\\\\=\frac{22\Delta T t}{22.55\Delta T t}\times 100\\\\=97.6 \%\)
How does adding substances to wastewater allow engineers to get rid of harmful substances?
Answer:
disaffectant pro to curements.
Explanation:
disaffectants controls affected water circles ,by the means of managing wastewater through treatments ,that avoids harmful substance.
5 Systems Modeling
es / SPE(2201 / General / Business System Modelling CAT
2. Business Process Modelling is important to a business due to the following advantages except:
(2 marks)
O a. None of the above
h
O b. Enhances Customization of Business Processes
O c Enhances Competitive advantage
O d. Enhances Process Communication
age
Next pag
Answer: None of the above
Explanation:
Business process modeling refers to the graphical representation of the business processes of a company, which is vital in the identification of potential improvements.
Business pticess modelling can be done through graphing methods, like data-flow diagram, flowchart etc. It is vital as business managers can effectively and quickly communicate their ideas.
It also enhances the customization of business processes, enhances the competitive advantage and enhances the process communication as well.
Therefore, the answer to the question will be "None of the above".
Hi im ***ar and im doing sculptural but what should it be about star wars or Marvel
Answer:
Cool I think u should do Marvel first
3. When a web page sends a request to its server, the session ID is always attached in the cookie section of the HTTP header. A web application requires all the requests from its own page to also attach the session ID in its data part (for GET requests, the session ID is attached in the URL, while for POST requests, the session ID is included in the payload). This sounds redundant, because the session ID is already included in the request. However, by checking whether a request has the session ID in its data part, the web server can tell whether a request is a cross-site request or not. Please explain why.
Including the session ID in both the cookie and data parts of the HTTP request is not necessary for identifying cross-site requests; CSRF protection is typically implemented separately.
When a web page sends a request to its server, the session ID is always attached in the cookie section of the HTTP header the web server can tell whether a request is a cross-site request or not. Please explain why?The statement you provided is incorrect. In general, web applications do not require the session ID to be included in both the cookie and the data part of the HTTP request. The session ID is typically sent in the cookie section of the HTTP header, and it is not necessary to include it in the data part of the request for the same purpose.
When a web page sends a request to its server, the session ID is usually attached as a cookie in the HTTP header. The server uses this session ID to identify the specific session associated with the client. The session ID is a unique identifier that is generated and assigned to the client when the session is initiated.
Including the session ID in the cookie allows the browser to automatically include it in subsequent requests to the same server. This eliminates the need to include the session ID in the data part of the request, whether it's a GET request (where the session ID is not typically included in the URL) or a POST request (where the session ID is not typically included in the payload).
The purpose of the session ID is to maintain the state of a user's session on the server-side. It helps the server associate subsequent requests from the same client with the correct session data. The server can retrieve the session ID from the cookie sent by the browser and use it to retrieve the corresponding session data.
Regarding cross-site requests, including the session ID in the data part of the request does not directly help determine whether a request is a cross-site request or not. Cross-site requests, also known as Cross-Site Request Forgery (CSRF) attacks, involve an attacker tricking a user's browser into making a request on their behalf to a different website where the user is authenticated.
These attacks are typically prevented by using anti-CSRF tokens or measures on the server-side.
In summary, the session ID is commonly included as a cookie in the HTTP header, and there is generally no need to include it in the data part of the request. The session ID helps maintain the session state on the server-side, but it does not directly relate to identifying cross-site requests.
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what were researchers alarmed to discover in the gulf of mexico in 1974?
In 1974, researchers were alarmed to discover that the Gulf of Mexico had an area of water that was almost devoid of oxygen and was thus unsuitable for marine life. The oxygen-deprived zone is known as the “Dead Zone,” and it is caused by excessive nutrient pollution from human activities.
Dead zones are hypoxic (low-oxygen) areas in the world's oceans and lakes, where aquatic life is unable to survive due to insufficient oxygen levels in the water. These hypoxic regions are caused by large quantities of nutrient pollution (particularly nitrogen and phosphorus) from human activities, such as fertilizer runoff from agricultural fields, sewage discharge from urban areas, and emissions from factories and vehicles.The oxygen-deprived area in the Gulf of Mexico was first discovered in the 1970s, and it has grown larger and more persistent over time.
The zone's size can fluctuate from year to year, but it typically covers a region of the Gulf of Mexico that is roughly the size of Connecticut.
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Water that has evaporated returns to earth as
Answer:
rain
Explanation:
evaoration causes clouds
clouds condense and rain
Answer:
rain
Explanation:
ihjpr2 ywjegnak'evsinawhe2'qwmasnh ngl,;snhy
Answer:
ummm ok?
Explanation:
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