A power station has to meet the following demand: Group-A: (200+10xZ) kW between 8 AM and 6 PM. Group-B: (100+2xZ) kW between 6 AM and 10 AM. Group-C: (50+Z) kW between 6 AM and 10 AM. Group-D: (100+3xZ) kW between 10 AM and 6 PM and then between 6 PM and 6 AM. Plot the daily load curve and load duration curve and determine: (i) Load Factor (ii) (iii) Diversity Factor Units generated per day.

The Pilot Car is the vehicle that alerts drivers to oversized loads. So, the correct answer to the question is a. Pilot car.

What exactly is a Pilot Car?

A pilot car is a vehicle that is used to guide and support convoys of large vehicles or trucks carrying heavy loads. Pilot cars can be sedans, SUVs, vans, or pickup trucks and must be equipped with lights, flags, and a CB radio in order to communicate with the oversized vehicle. ‍

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The components of the reactions are: vertical reaction at point A, horizontal reaction at point A, and reaction at point B. The tension in the cable is the force exerted along the length of the cable.

In this scenario, the bent rod ACDB is supported by a sleeve at point A and a ball-and-socket joint at point B. When analyzing the system, we need to determine the components of the reactions and the tension in the cable.

Firstly, at point A, there are two reaction components: the vertical reaction and the horizontal reaction. The vertical reaction counteracts the weight of the rod and any additional forces acting downward. It ensures equilibrium in the vertical direction. The horizontal reaction, on the other hand, prevents the rod from sliding or moving horizontally. It maintains equilibrium in the horizontal direction.

Secondly, at point B, there is a reaction that allows the rod to rotate or pivot around the ball-and-socket joint. This reaction balances the moment caused by the weight of the rod and any other external moments.

Lastly, the tension in the cable refers to the force exerted along the length of the cable. This tension arises from the need to balance the vertical and horizontal forces acting on the rod. It ensures that the rod remains in a stable position and prevents it from collapsing under its own weight.

To accurately determine the components of the reactions and the tension in the cable, specific calculations and analysis of the forces and moments involved in the system would be required.

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Thomas is checking an electrical wire in a washroom with a wet floor. Ground-fault circuit interrupters in the electrical wire will help prevent electrical shock.

The ground-fault circuit interrupter, or GFCI, is a quick-response circuit breaker that may cut off electricity in as little as 1/40 of a second in the event of a ground fault.

It compares the current flowing into and out of the machinery along the conductors of the circuit.

Therefore, electrical shock can be avoided with the aid of ground-fault circuit interrupters in the electrical line.

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Answer:

D

Explanation:

Your answer is D.There is nothing to do; if you do not own the car anymore, it is not your responsibility

The output time response of a control system is equal to the sum of the transient response and the steady-state response.

This can be represented by the equation: Output response = Transient response + Steady-state response. Therefore, the correct option is d) the transient response + the steady state response. The transient response of a control system represents the behavior of the system immediately after a disturbance or change in the input. It typically exhibits oscillations and decays over time until the system reaches a stable state. On the other hand, the steady-state response represents the long-term behavior of the system after it has settled down, where the output remains constant. The steady-state response is independent of the initial conditions and depends only on the input to the system. When these two components are combined, the resultant output time response of the control system captures both the initial transient behavior and the final steady-state behavior. It is important to consider both aspects to fully understand and analyze the system's performance.

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Answer:

Auto production stopped on Feb. 9, 1942, as factories switched over to making munitions for World War II. The industry retooled plants that could be adapted.

Answer:

9, 1942, as factories switched over to making munitions for World War II. The industry retooled plants that could be adapted. Car manufacturing equipment was replaced with what was needed to make planes and tanks, and floors were reinforced to carry the load of heavier machinery, according to Life magazine

Explanation:

Answer:

In the case of the production Line, we know that,

No of gasoline cans = 5

probability that 1st can is out of tolerance = 0.03

probability that 2nd can is out of tolerance = 0.03

.

.

probability that the 5th can is out of tolerance = 0.03

Therefore,

probability of 1st can out of tolerance + probability of 1st can not out of tolerance = 1

Probability of 1st can not out of tolerance = 1 -- 0.03 = 0.97

probability of 2nd can not out of tolerance = 0.97

.

.

probability of 5th can not out of tolerance = 0.97

Question A:

Probability that they are all out of tolerance

= P(1st can out of tolerance) * P(2nd can out of tolerance) * P(3rd can out of tolerance) * P(4th can out of tolerance) * P(5th can out of tolerance)  

= (0.03 ) * (0.03) * (0.03) * (0.03) * (0.03) =  2.43 E⁻⁸   (2.43 ˣ 10⁻⁸)

Question B:

Probability that exactly two are out of tolerance

= P(1st can is out of tolerance) * P(2nd can is out of tolerance) * P(3rd can is not out of tolerance) * P(4th can is not out of tolerance) * P(5th can is not out of tolerance)

= (0.03) * (0.03) * (0.97) * (0.97) * (0.97) = 0.0008214057

Explanation:

Answer:

(b) NAND

(d) NOR

(f) XNOR

Explanation:

This is best explained using a truth table.

A truth table containing 3 hypothetical inputs and their corresponding outputs using the AND, NAND, OR, NOR, XOR, and XNOR gates, has been attached to this response.

From the table, it can be seen that the NAND, NOR and XNOR gates produce a false value (0) when all inputs are true.

Note:

On the table,

For the AND column, a 1 is being output only when all three inputs are 1. Otherwise, a 0 results.

For the NAND column (which is the negation of the AND), a 0 is being output only when all three inputs are 0. Otherwise, a 1 results.

For the OR column, a 1 is being output when at least one of the three inputs is 1. Otherwise, a 0 results.

For the NOR column (which is the negation of the OR), a 0 is being output when at least one of the three inputs is 0. Otherwise, a 1 results.

For the XOR column, a 1 is being output when the number of inputs that are true (1) is odd. i.e when the number of 1s is 1 or 3. Otherwise, a 0 results.

For the XNOR column (which is the negation of the XOR), a 0 is being output when the number of inputs that are true (1) is odd. i.e when the number of 1s is 1 or 3. Otherwise, a 1 results.

The four mechanisms of adhesion are mechanical interlocking, diffusion, adsorption, and electrostatic attraction.


Mechanical interlocking: This mechanism involves physical interlocking of two surfaces, such as the interlocking of fibers in textiles or the interlocking of teeth in a zipper. The irregularities on the surface create a strong bond through intermolecular forces.

Diffusion: Diffusion occurs when molecules from one material penetrate the surface of another material. This mechanism is commonly observed in adhesives that consist of polymers. The adhesive molecules diffuse into the substrate, forming intermolecular bonds and creating adhesion.

Adsorption: Adsorption refers to the attraction of molecules from one material to the surface of another material. Van der Waals forces, dipole-dipole interactions, or chemical bonds can facilitate adsorption. Adhesives often rely on adsorption to adhere to substrates.

Electrostatic attraction: This mechanism involves the attraction between positively and negatively charged particles. It can occur between polar molecules or charged particles on the surfaces. Electrostatic attraction contributes to the adhesion of materials like plastics or rubber.

Adhesion can be achieved through various mechanisms. Mechanical interlocking relies on physical interlocking, diffusion involves the penetration of molecules into the substrate, adsorption occurs through the attraction of molecules to a surface, and electrostatic attraction relies on the attraction between charged particles. Understanding these mechanisms is crucial for developing effective adhesives and improving bonding techniques in various industries

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The four mechanisms of adhesion are mechanical interlocking, diffusion, adsorption, and electrostatic attraction. Each mechanism plays an important role in creating adhesion between two surfaces.

Mechanical interlocking is the process of creating a mechanical bond between two surfaces by physically interlocking them. This can be achieved by creating grooves or hooks in the surface of one of the materials and then pressing the two surfaces together. Diffusion is the process of atoms or molecules from one surface migrating into the other surface, creating a bond. This is typically seen in metal-to-metal bonding, where atoms from one surface diffuse into the other surface and create a solid-state weld. Adsorption is the process of molecules or atoms adhering to the surface of a material.

This can be achieved by creating a chemical bond between the surface of the material and the atoms or molecules. This type of bonding is often used in the creation of coatings and paints. Electrostatic attraction is the process of positively charged ions attracting negatively charged ions. This type of bonding is often seen in the bonding of two different materials, where one material has a positive charge and the other material has a negative charge.

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A gear drive system using two spur gears connected by an idler gear is an example of an intermediate gear drive.

An intermediate gear drive is a type of gear drive system that uses an idler gear to connect two spur gears. In this setup, the two spur gears are not directly connected but are instead connected through the idler gear, which is placed between them. The idler gear rotates in the opposite direction to the and in the same direction as the output gear. This arrangement allows the gear drive system to change the direction of rotation, increase or decrease the speed of rotation, or both. Intermediate gear drives are commonly used in automotive transmissions, industrial machinery, and other applications that require precise control of speed and torque.

In conclusion, a gear drive system using two spur gears connected by an idler gear is classified as an intermediate gear drive. This type of gear drive system is useful in a wide range of applications where precise control of speed and torque is necessary.

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In an entity supertype, you would store general data attributes shared by all subtypes, while in an entity subtype, you would store specific data attributes unique to each subtype.

An entity supertype is a higher-level, more abstract representation of entities in a database, encompassing common attributes among various entity subtypes. Entity subtypes, on the other hand, represent specific categories within the entity supertype and include unique attributes that are specific to that category.

For example, consider an entity supertype "Vehicle." Attributes like make, model, and year are common to all vehicles, so they would be stored in the supertype. Entity subtypes like "Car" and "Motorcycle" would then inherit these general attributes from the supertype.

Now, the "Car" subtype may have specific attributes like number of doors and seating capacity, while the "Motorcycle" subtype may have attributes like engine type and seat height. These unique attributes would be stored in their respective subtypes.

Using supertypes and subtypes in this way allows for a clear and efficient organization of data in a database. By separating general attributes from specific ones, you can reduce redundancy and maintain consistency in the data structure.

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The distance needed to stop an average train travelling at 55mph depends on various factors such as the weight of the train, the condition of the brakes, and the condition of the tracks.

However, on average, it can be estimated that a train travelling at 55mph will require a distance of almost 1000 yards to come to a complete stop. This is due to the massive weight of the train, which can range from several hundred to several thousand tons. The momentum generated by the train's speed is difficult to overcome, even with the most efficient braking systems. In fact, it is estimated that it takes almost one mile to stop a freight train travelling at 55mph. Therefore, it is crucial for train operators to maintain their equipment, adhere to speed limits, and keep a safe distance from other trains to ensure the safety of everyone involved. Additionally, it is important for motorists and pedestrians to be aware of the potential dangers of trains and to always exercise caution when crossing tracks or approaching railroad crossings.

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The answer is analyze

Answer:

a

Explanation:

that is what energy efficiency is

Based on the code given , the Step 1  is done by enforcing maximum() and min() in ARM7 assembly while Step 2  is done by enforcing the main while circle in ARM7 assembly.

To apply the maximum() and min() functions in ARM7 assembly, we can use some instructions to optimize the law and code speed. There is one possible way to do it based on the image attached.

Therefore, To apply the main while loop, we need to check if the left over of the division between result and d isn't zero, and if so, add c to affect. We can use the SWI 0x6 instruction to gain the balance of the division.

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See text below

We will translate the following C function, which calculates the Lowest Common Multiple (1cm) between two positive numbers, into ARM7 assembly:

unsigned int lcm(unsigned int a, unsigned int b)

{

(unsigned int c, d, result; // cannot be translated)

c = max(a, b); // i.e. c-a if a >= b, otherwise c=b d = min(a, b); // same, but vice-versa

result = c;

while ((result % d) != 0) {

}

result = result + c;

}

(return(result);

// cannot be translated)

In doing so, you are asked to optimize the assembly code for execution speed. i.e., a) fewest number of cycles, taking into account pipeline emptying/refilling on branch instructions for example, and b) fewest number of instructions.

Disregard the first/last lines of the lcm() function, which cannot be translated into assembly. This exercise is broken into these two steps:

1. First, implement the two successive lines min() and max() in ARM7 assembly:

c = max(a, b); // i.e. c-a if a>=b, otherwise c=b

d = min(a, b); // same, but vice-versa

[Hint 1]: Remember that the function's arguments, a and b, will be found in registers r0 and rl respectively. Every other register is free for you to use in your code. Remember to use conditional instructions whenever possible, to optimize code density and execution speed.

2. Implement the main while loop in ARM7 assembly.

[Hint 2]: The remainder of a division (i.e., the “modulo” operation %) can be obtained by relying on the BIOS, using the ARM7 instruction: SWI 0x6 (which cannot be conditional) after placing the input in the appropriate, predefined registers (see lecture notes), which cannot be selected/changed, and reading the output from the appropriate register.

Remember to comment each line of your program to explain what your code does.



Describe the changes to the memory and the registers, after the execution of each of the following five load/store instructions in the (five-lined) program below (i.e.. these five instructions are run as a sequence from the initial memory state shown below).

We assume big endian formatting.

Initial Memory State 0x420014 DE OC 63 20 0x420010 FF AE 10 00 0x42000c 13 46

FA 08

0x420008 0x420004 0x420000

24 AB

A0

22

00 00 CO

FF

00 1C OE 3B

Initial Registers

State

r0=0x00000000, r1=0x00000000, r2=0x00420008, 13=0x00000007,

r4=0x00000001

# Start of the program

LDR

r0, [r2, #-4]

LDRB

r1, [r2, r3]

STR

r1, [2], r4 LSL #2

SWP

STMDA

r4, r0, [r2]

# End of the program

r2!, (r4, r3, r0}

The maximum inventory of the part is limited to 3 units to match the demand rate of 12 units per hour.

To determine the maximum inventory of a part, we need to consider the processing time, maintenance time, and the demand rate. The processing time per part is given as 4 minutes, and after producing 190 units, a 34-minute maintenance process is required. The demand for the part is 12 units per hour.

First, let's calculate the total time required to produce 190 units:

Total production time = (Processing time per part) * (Number of units produced)

Total production time = 4 minutes * 190 units

Total production time = 760 minutes

Next, let's calculate the total time available for production:

Total time available = (Total production time) + (Maintenance time)

Total time available = 760 minutes + 34 minutes

Total time available = 794 minutes

Now, let's calculate the maximum inventory:

Maximum inventory = (Total time available) / (Time required to produce one unit)

Maximum inventory = 794 minutes / (4 minutes per unit)

Maximum inventory = 198.5 units

Since we cannot have fractional units of inventory, the maximum inventory of the part is rounded down to the nearest whole number, which is 198 units. However, since the demand for the part is 12 units per hour, we need to consider the time it takes to produce one unit. With a processing time of 4 minutes per unit, it takes 12 minutes to produce 3 units. Therefore, the maximum inventory of the part is limited to 3 units to match the demand rate of 12 units per hour.

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Answer:

263.69 W.

(None of the option).

Explanation:

So, from the question above we are given the following parameters or data or information which is going to allow us to solve this question and they are;

(1). diameter of 25 cm.

(2). Initial temperature of 300°C.

(3).temperature drops to 200°C = final temperature.

Step one: Calculate the Reynolds number.

Reynolds number = 3 × 0.25/1.562 × 10^-5 = 48015.365.

Step two: Calculate average heat transfer coefficient.

The average heat transfer coefficient = k/D { 2 + (0.4Re^1/2 + 0.06Re^2/3} px^0.4 × (u/uz)^1/4.

The average heat transfer coefficient = 0.10204 × [ 2 + (87.65 + 79.26) (0.8719) × 0.8909.

average heat transfer coefficient = 0.20204 ( 2 + 129.652).

average heat transfer coefficient = 13.43/m^2.k.

Step three: The rate of heat transfer loss due to convection = (average heat transfer coefficient ) × πD^2 × ( T1 - T2).

The rate of heat transfer loss due to convection= 13.43 × π(0.25)^2 × (300 - 200).

=>The rate of heat transfer loss due to convection = 263.69 W.

Answer:

According to the Principles of Project management, the three factors which dominate the lifecycle of any project are:

The relationship between the three is usually governed by trade-offs.

Explanation:

In simple term, in executing a project, one must deal with the factors mentioned above.

It is always desirous for a project to be finished within a stipulated time. If the time required is reduced inconsiderably, it will most likely incur more cost and even impact performance.

On the other hand, if the project is cost-sensitive and is executed to a very minimalistic budget, performance will be impacted and it may take a protracted amount of time.

In addition to the above, if the principal decides to change the original design of the project, the performance expected is altered. This will attract additional time as well as cost.

It is possible for any of the above factors to be renegotiated and readjusted at any time during the project. It usually is a trade-off.. that is one for the other.

Cheers!

In order to write the given expression as a decimal, we can simplify the expression and then evaluate it.

The given expression can be simplified as follows:$$
\begin{aligned}
\(\frac 15 \left(\frac 15\right)^2 \left(\frac 15\right)^3 \left(\frac 15\right)^4 &= \frac{1}{5^1} \cdot \frac{1}{5^2} \cdot \frac{1}{5^3} \cdot \frac{1}{5^4}\\\)
\(&= \frac{1}{5^{1+2+3+4}}\\\)
\(&= \frac{1}{5^{1+2+3+4}}\\\)
\(&= \frac{1}{9,765,625}\)
\end{aligned}
$$Now, we can evaluate the given expression by dividing 1 by 9,765,625 as follows:$$
\begin{aligned}
\(\frac{1}{9,765,625} &= \frac{1}{10,000,000 - 234,375}\\\)
\(&= 0.\overline{000}1\\&= 1 \cdot 10^{-7}\)
\(\end{aligned}$$Therefore, $\frac 15 \left(\frac 15\right)^2 \left(\frac 15\right)^3 \left(\frac 15\right)^4$ as a decimal is $1 \cdot 10^{-7}$\), which means that the decimal is a very small number.

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It is to be noted that all UHRS platforms are optimized for the popular kinds of internet browser applications.

The Universal Human Relevance System (UHRS) is a crowdsourcing platform that allows for data labeling for a variety of AI application situations.

Vendor partners link people referred to as "judges" to offer data labeling at scale for us. All UHRS judges are bound by an NDA, ensuring that data is kept protected.

A browser is a software tool that allows you to see and interact with all of the knowledgeon the World Wide Web. Web sites, movies, and photos are all examples of this.

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Heat energy is the known to be a product of the movement of tiny particles called atoms, molecules or ions. The true statement is by Technician A.

Internal combustion engine is dependent on the heat of combustion so as to make torque to move the vehicle and power the system.

A lot of heat made during combustion is not often used productively and therefore need to be removed to avoid overheating of the engine.

The heat energy that is not used for is wasted in three ways: They are:

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Answer:

Gasoline is injected directly into the cylinder.

Explanation:

In a direct injection system, the air and gasoline are not pre-mixed. Rather, air comes in using the intake manifold, while the gasoline is injected directly into the cylinder.

The classification of the working surfaces are as follows;

A working surface is a flat surface on which various activities can be carried out.

In a construction system, the duty carried on a particular working surface varies from another. They can be categorized into heavy-duty, medium-duty, light-duty, and very light-duty.

The classification of the working surfaces are as follows;

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Answer:

Explanation:

An engineered cover on landfills can reduce the decomposition of trash by creating a barrier between the waste and the environment. Here are a few ways in which an engineered cover can help:

1. Gas Management: Landfills produce gases such as methane as the waste decomposes. An engineered cover can include gas collection systems that capture and control these gases, preventing their release into the atmosphere. By managing the gas buildup, decomposition processes can be slowed down.

2. Moisture Control: Decomposition requires moisture. An engineered cover can include liners or barriers that limit the infiltration of water into the landfill. By reducing moisture levels, the activity of decomposing microorganisms can be inhibited.

3. Oxygen Limitation: Decomposition processes typically require oxygen. By designing the cover to limit oxygen infiltration, anaerobic conditions can be created. This can slow down decomposition as anaerobic bacteria, which decompose waste more slowly, dominate the environment.

4. Minimizing Leachate Generation: Leachate refers to the liquid that forms as water percolates through the waste. An engineered cover can help minimize leachate generation by incorporating liners and drainage systems that collect and redirect leachate for proper treatment and disposal. By reducing leachate, the conditions for decomposition can be less favorable.

5. Physical Barrier: The cover itself acts as a physical barrier, preventing direct contact between the waste and external elements. This reduces the availability of nutrients, moisture, and oxygen that are necessary for decomposition.

By implementing these strategies, an engineered cover on landfills can effectively reduce the decomposition of trash and mitigate its environmental impacts.

The EROEI (Energy Returned on Energy Invested) must be "a. positive" for any fuel to be worth harvesting.

A positive EROEI indicates that the energy gained from harvesting the fuel is greater than the energy expended to obtain it, making it a profitable and efficient energy source. EROEI is a measure of how much energy is obtained from a particular energy source compared to the amount of energy invested in obtaining that energy. If the EROEI is less than 1, it means that the energy invested in obtaining the fuel is greater than the energy obtained from it, making it an unsustainable source of energy.

Therefore, an EROEI of greater than 1 indicates that the energy source can sustain itself and is worth harvesting.

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To determine whether the vector field F(x, y, z) = e^(yz) i + xze^(yz) j + xye^(yz) k is conservative, we can check if its curl is zero.

The curl of F is given by the determinant of the curl operator applied to F:

curl(F) = (d/dy)(xye^(yz)) - (d/dz)(xze^(yz)) i

- (d/dx)(e^(yz)) + (d/dz)(xye^(yz)) j

+ (d/dx)(xze^(yz)) - (d/dy)(e^(yz)) k

Simplifying the above expression, we get:

curl(F) = (xe^(yz) + xyze^(yz)) i

+ (-e^(yz) - xyze^(yz)) j

+ (xze^(yz) - xe^(yz)) k

The curl of F is not zero, indicating that the vector field F is not conservative.

Therefore, there does not exist a function f such that the gradient of f is equal to F.

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Among cooperative processes, the action of process synchronization should be performed. Hence, option (a) i.e. ‘process synchronization’ holds the correct answer.

Process synchronization is the process to handle problems that arise when multiple processes execute simultaneously. Process synchronization refers to the way of coordinating the execution of processes so that no two processes can have access to the same shared resources and data.

On the basis of synchronization, the process is categorized into two types: independent process and cooperative process. If the execution of one process does not impact the execution of the other process, these two processes are known to be independent processes. In contrast, when one process execution is affected by the execution of another process, then these processes are said to be cooperative processes. These cooperative processes require synchronization in order to guarantee an appropriate order of execution.

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Answer:
The most common reason a cruise control stops working is due to a blown fuse or a defective brake pedal switch. It can also be caused by issues with the throttle control system or the ABS. In older cruise control systems it can be caused by a broken vacuum line.

The approximate average power output is mathematically given as

P=1097.6w

Question Parameters:

Iowa with a 10 m2 swept area and 50 m hub height

Assume 80% of the Betz limit, 80% conversion efficiency, and air density of 1.0 kg/m3. Wind speed is 7 m/s2

Generally, the equation for the average output power  is mathematically given as

\(P=0.5 \phi BAu^3*n\\\)

Where

B= Benz coefficient

n=0.8

Therefore

P=0.5*1*0.8*10*7^3*0.8

P=1097.6w

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Complete Question

What is the approximate average power output of a well-designed modern turbine in Des Moines, Iowa with a 10 m2 swept area and 50 m hub height? Assume 80% of the Betz limit, 80% conversion efficiency, and air density of 1.0 kg/m3. Wind speed is 7 m/s2

Answer:

While resistance welding you should wear clear grinding glasses, unbreakable plastic face shields or clear unbreakable plastic goggles. When resistance welding a #10, or more, shade lens should be worn. All hand and portable tools should be inspected for loose parts, cleanliness, or worn power cords.