Question 5 a) Explain how an induction motor can be simplified to an equivalent circuit. You must explain the importance of any quantities. (8 Marks) b) A 20kW, 4-pole induction motor is designed to operate from a 440V, 50Hz, three-phase supply, and when operating at full power on this supply it runs at 1470RPM. The motor efficiency is 90% under both conditions. (i) What supply frequency will be needed to make this motor run at 1270RPM while delivering a shaft power of 12.5kW? (7 Marks) (ii) If the motor were supplied from a sinusoidal variable frequency source, what voltage and current will need to be supplied to it when running at 1365RPM at 12.5kW if the power factor of the motor is 0.85? (10 Marks

Answer:

glue

Explanation:

The correct answer is the cost of the antenna stainless steel is $8.00 per cubic centimeter.

Antennas are much more than simple devices connected to every radio. They're the transducers that convert the voltage from a transmitter into a radio signal. And they pick radio signals out of the air and convert them into a voltage for recovery in a receiver.plural antennae : one of a pair of slender, movable, segmented sensory organs on the head of insects, myriapods, and crustaceans see insect illustration. : a usually metallic device (such as a rod or wire) for radiating or receiving radio waves.

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Answer:

The application of regulations in locations are very important.

Explanation:

The application of regulations in locations are very important in order to gain more benefit from it because people choose those places that are well regulated and having more facilities. If the location has baths, showers, electric floor heaters and other necessities so the people prefer the place over another and increase of clients occurs which give more benefits to the place owners.

USGS stream-gauging stations record water depth and velocity and channel width in order to calculate the discharge of the stream.

Water discharge is used as a support for people's lives in the form of liquid substances. Water has the most important role in human needs. Water discharge is very closely related to time & volume. Water discharge is the velocity of the flow of liquid through the cross-sectional distance per unit time. Water discharge uses volume units per time or ml/second, liters/second, m³/second, liters/hour, m³/hour, and various other units. Discharge units are often used in monitoring the capacity (capacity) of water in rivers or dams so that the existing water can be controlled.

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The application of robotics to industrial and commercial sewing tasks such as sewing leather, fabrics, and wool is known as automated sewing.

How Does Robot Sewing Work?

Indeed, robotic sewing is a specialised automation application with unique requirements. Many sewing robots, for example, are custom-built to a company's exact specifications—there is no one-size-fits-all solution as in other industries.

To sew materials, sewing robots require unique mechanics such as sewing heads, extra grippers, and multiple robotic arms and end effectors.

Each of these materials presents challenges, but the lure of increased production rates, efficiency, and reliability has led robot manufacturers to develop solutions for the industry's most difficult challenges.

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At 1.2mpa pressure and 50c

What is pressure?
By pressing a knife against some fruit, one can see a straightforward illustration of pressure. The surface won't be cut if you press the flat part of the knife against the fruit. The force is dispersed over a wide area (low pressure).

a)Mass flow rate of the refrigerant
Therefore h1= condenser inlet enthalpy =278.28KJ/Kg
saturation temperature at 1.2mpa is 46.29C
Therefore the temperature of the condenser

T2 = 46.29C - 5
T2 = 41.29C

Now,
d)power consumed by compressor W = 3.3KW
Q4 = QL + w = Q4
QL = mR(h1-h2)-W
= 0.0498 x (278.26 - 110.19)-3.3
=5.074KW
Hence refrigerator load is 5.74Kg

(COP)r = 238/53
(Cop) = 4.490

Therefore the above values are the (a) mass flow rate of the refrigerant, b) the refrigerant load, c) the cop, and d) the minimum power input to the compressor for the same refrigeration load.

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Answer:

Step On: Your foot forces the clutch pedal down and then causes it to take up the slack. This, in turn, causes the clutch friction disk to slip, creating heat and ultimately wearing your clutch out.

Step Off: When the clutch pedal is released, the springs of the pressure plate push the slave cylinder's pushrod back, which forces the hydraulic fluid back into the master cylinder.

Answer:

C. WordArt

Explanation:

WordArt feature of MS word is used for styling the text n order to make it look neater or more noticeable. These features are found in the top home menu. They feature text size, text style, text color, and options to make text bold, italic, underlined, etc. There are also many features to make text flashy, colorful, and add special effects. It can be shaded, neon, have shadows, etc.

Answer:

a) 158.4 HP.

b) 1235.6 °F.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:

\(m_{in}h_{in}=W_{out}+m_{out}h_{out}\)

Whereas the mass flow is just the same, which means we have:

\(W_{out}=m_{out}(h_{out}-h_{in})\)

And the enthalpy and entropy of the inlet stream is obtained from steam tables:

\(h_{in}=1860.7BTU/lbm\\\\s_{in}= 2.2096BTU/lbm-R\)

Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:

\(s_{out}= 2.2096BTU/lbm-R\)

Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.

Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:

\(h_{out}=1659.15BUT/lbm\)

Then, the isentropic work (negative due to convention) is:

\(W_{out}=2500lbm/h(1659.15BUT/lbm-1860.7BUT/lbm)\\\\W_{out}=-503,875BTU\)

And the real produced work is:

\(W_{real}=0.8*-503875BTU\\\\W_{real}=-403100BTU\)

Finally, in horsepower:

\(W_{real}=-403100BTU/hr*\frac{1HP}{2544.4336BTU/hr} \\\\W_{real}=158.4HP\)

Regards!

The mean temperature difference (ΔTm) between the water and steam:ΔTm = (steam temperature - water inlet temperature) / ln(steam temperature - water outlet temperature).

To determine the heat transfer rate, overall coefficient of heat transfer, and the length of the pipe required, we can follow these steps:Calculate the inner and outer surface areas of the pipe:Inner surface area = π * inner diameter * length of the pipe, Outer surface area = π * outer diameter * length of the pipe,Calculate the mean temperature difference (ΔTm) between the water and steam:ΔTm = (steam temperature - water inlet temperature) / ln(steam temperature - water outlet temperature).

Calculate the heat transfer rate (Q) using the formula: Q = overall coefficient of heat transfer * surface area * ΔTm  Calculate the hydraulic diameter (Dh) using the formula: Dh = 4 * (inner diameter * outer diameter) / (inner diameter + outer diameter),Substitute the known values into the equations to find the heat transfer rate, overall coefficient of heat transfer, and the length of the pipe required. Remember to double-check your calculations and units to ensure accuracy.

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Answer:

The original seven wastes, or muda, are transportation, inventory, motion, waiting, overproduction, overprocessing and defects. ... When manufacturers are able to identify the seven wastes, they can correct and prevent further loss of time, money and other resources.

Explanation:

Answer:

transportation, inventory,motion, waiting,overproduction,overprocessing,and defects

Explanation:

they are often referred to by the acronym TIMWOOD

Repairing advanced composites using materials and techniques traditionally used for fiberglass repairs is likely to result in suboptimal repairs and reduced performance.

Advanced composites and fiberglass are different materials with distinct properties and characteristics. While fiberglass repairs may work well for fiberglass components, using the same materials and techniques on advanced composites can lead to several issues:

Material compatibility: Advanced composites, such as carbon fiber-reinforced polymers (CFRP) or aramid fiber composites, have unique material properties, including a high strength-to-weight ratio and specific stiffness. These materials require specialized repair materials that match their properties for effective and durable repairs. Using traditional fiberglass repair materials may not provide the same compatibility, compromising the structural integrity and performance of the repaired composite.

Repair effectiveness: Advanced composites are often used in high-performance applications where precision and strength are critical. Repairing them using materials and techniques designed for fiberglass repairs may not adequately restore the original properties and strength of the composite. The repaired area may be weaker, more prone to delamination or cracking, and unable to withstand the same loads and stresses as the original composite.

Longevity and durability: Advanced composites are engineered to have excellent durability and fatigue resistance. They are designed to withstand specific environmental conditions and operational requirements. Using inappropriate repair materials and techniques can undermine these properties, leading to premature failures, reduced lifespan, and compromised safety.

To ensure effective repairs and maintain the performance of advanced composites, it is essential to use specialized repair materials and techniques specifically developed for these materials. This may involve utilizing compatible resin systems, reinforcement fabrics, adhesives, and processes tailored to the unique characteristics of advanced composites. Following manufacturer guidelines and seeking expert advice from professionals experienced in composite repairs can help ensure proper repairs and optimal performance of advanced composite structures and components.

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Answer:

Die of intoxication by water first

Explanation:

We assume that the weight of the man is 154.35 pounds which is 70 kg

LD50 water = 90g per kg

Maximum concentration = 90x70

= 6300grams

Convert grams to liters

6300/100

= 6.3 litres

From here we get amount of kool aid

6.3 x 1.97x10^-4

= 1.24x10^-3

= 1.24grams

1.24 grams is below 420 kool aid is lower than LD50 with about 6 grams for 1 kg (6x70kg = 420). So 420 is lethal dose. But 1.24 is less than this so the man has to die of water intoxication first.

The continuity equation allows finding the results for the speed and flow of the outlet pipe are:

Fluid mechanics studies the movement and fluid dynamics, it is described  by two fundamental relationships:

          \(P_1 + \frac{1}{2} \rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g y_2\)  

Where P is the pressure, v the velocity, y the height, the subscripts are two selected points of interest in the pipe.

            \(A_1v_1 = A_2 v_2\)

Where A₁ and A₂ is the area of ​​the inlet and outlet pipes and v₁ and v₂ are the velocities of the fluid at the inlet and outlet

They indicate the inlet and outlet diameters of the pipe are d₁ = 0.02 m and d₂ = 0.1 m, the inlet velocity v₁ = 10 m / s, with the continuity equation let's find the outlet velocity

         \(v_2 = ( \frac{A_1}{A_2} )^2 \ v_1\)  

the cross section of the tavern is

        A = \(\pi r^2 = \pi \ d^2/4\)  

Where r is the radius and d is the diameter

Let's substitutes

        \(v_2 = (\frac{d_1}{d_2} )^2 \ v_1\)  

        \(v_2 = (\frac{0.02}{0.1} )^2 \ 10\)  

        v₂ = 0.4 m / s

The flow  (Q) is defined as the amount of liquid that passes through the section of the pipe per unit of time, we can see that this is the same conservation of mass, therefore

        Q = A v

        Q = \(\pi \frac{d_2^2}{4} \ v_2\)π d2 ^ 2 v2

        Q = \(\pi \ \frac{0.1^2}{4} \ 0.4\)

        Q = 3.14 10⁻³ m ^ 3 / s

As the transported liquid is water

         1 liter water = 1 kg

Consequently

         1 m³ = 1000 liter = 1000 kg of water

Let's reduce the flow

         Q = 3.14 10⁻³ m³/s ( \(\frac{10^3 kg}{1 m^3}\) )

         Q = 3.14 kg / s

In conclusion using the continuity equation we can find the results for the speed and flow of the outlet pipe are:

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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! :)

Fits snugly and completely into the screw’s slot.

Phillips. A Phillips screwdriver features a cross-shaped head with pointy ends that fits snugly into the cross slots of the Phillips screw.

A screwdriver's tip is what distinguishes it from others since it is formed to meet the screw head's driving surfaces, such as slots, grooves, and recesses. In order to operate a screwdriver properly, the tip must make contact with the head of a screw that is the same size and type as the tip.

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Answer:

\(250\ \text{lbm/min}\)

\(625\ \text{ft/min}\)

Explanation:

\(A_1\) = Area of section 1 = \(10\ \text{ft}^2\)

\(V_1\) = Velocity of water at section 1 = 100 ft/min

\(v_1\) = Specific volume at section 1 = \(4\ \text{ft}^3/\text{lbm}\)

\(\rho\) = Density of fluid = \(0.2\ \text{lb/ft}^3\)

\(A_2\) = Area of section 2 = \(2\ \text{ft}^2\)

Mass flow rate is given by

\(m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}\)

The mass flow rate through the pipe is \(250\ \text{lbm/min}\)

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

\(m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}\)

The speed at section 2 is \(625\ \text{ft/min}\).

(a) The mass flow rate will be "250 lbm/min".

(b) At section 2, the speed will be "625 ft/min".

According to the question,

Section 1 area, A₁ = 10 ft²

Section 2 area, A₂ = 2 ft²

Water's velocity, V₁ = 100 ft/min

Volume at section 1, v₁ = 4 ft³/lbm

(a) We know the formula,

Mass flow rate, m = ρA₁V₁

                              = \(\frac{A_1 V_1}{v_1}\)

By substituting the values,

                              = \(\frac{10\times 100}{4}\)

                              = \(\frac{1000}{4}\)

                              = 250 lbm/min

(2) The speed will be:

→ m = ρA₂V₂

or,

  V₂ = \(\frac{m}{\rho A_2}\)

By substituting the values,

       = \(\frac{250}{0.2\times 2}\)

       = \(\frac{200}{0.4}\)

       = 625 ft/min

Thus the responses above are correct.  

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The most accurate answer would be option C (10%), as it falls within the typical range and is a reasonable estimate of organic matter content in an ideal or typical soil on a volume basis.

The amount of organic matter in soil is an important indicator of soil health and fertility. Organic matter is composed of plant and animal residues that have undergone decomposition, and it contributes to the soil's physical, chemical, and biological properties. In this question, we are asked to determine the percentage of organic matter in an ideal or typical soil on a volume basis. According to the United States Department of Agriculture (USDA), the typical range of organic matter content in agricultural soils is between 1% and 6%, with an average of 3%. However, this can vary depending on factors such as climate, soil type, land use, and management practices. Some soils may have higher organic matter content due to factors such as high vegetation cover, low disturbance, and frequent additions of organic materials such as manure or compost. On the other hand, some soils may have lower organic matter content due to factors such as erosion, tillage, and intensive cropping.

Based on the information provided, we can eliminate options A and D as they are outside the typical range of organic matter content in agricultural soils. Option B (5%) is within the typical range, but the average organic matter content is closer to 3%. Therefore, the most accurate answer would be option C (10%), as it falls within the typical range and is a reasonable estimate of organic matter content in an ideal or typical soil on a volume basis.

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Answer:

ICC

Explanation:

The International Building Code (IBC) is a model building code developed by the International Code Council (ICC). It has been adopted for use as a base code standard by most jurisdictions in the United States.

The probability that one telephone call passes through the switchboard in three minutes is 1.49%.

Using the Poisson probability formula, we can calculate the probability of exactly one call passing through the switchboard in 3 minutes as follows:

P(X = 1) = (e^(-6) * 6^1) / 1!

Where

6 = 2 per minute * 3 minute

X is the number of calls passing through the switchboard in 3 minutes.

So, we have

P(X = 1) = (e^(-6) * 6^1) / 1!

= (0.00248 * 6) / 1

= 0.0149

Therefore, the probability that one telephone call passes through is approximately 0.0149, or about 1.49%.

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Answer: Option B

Explanation:

In order to enhance net radiation exchange between a high temperature surface and a low temperature surface in an enclosure, one should: paint both surfaces so that they are close to a blackbody surface but don't polish both surfaces so that they are highly reflective

Option C, which states that the common to run terminals show the lowest resistance, is the correct answer to the question.

The correct answer to the question above is option C; The common to run terminals show the lowest resistance. Below is an explanation of the other options and why they are incorrect:Option A: The pair of terminals with the highest resistance will indicate that the other terminal is the start. This statement is incorrect. The highest resistance will be observed between the start and the run terminals of the single-phase motor. Option B: The common to start terminals show the lowest resistance. This statement is also incorrect. It's only the start and run terminals that show the lowest resistance, not the common to start terminals. Option D: The start to run terminals show the lowest resistance. Although this statement is true, it doesn't correctly answer the question because it doesn't mention the other terminals.

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Answer:

-Khái niệm:

Đo dòng điện là sử dụng các dụng cụ như ôm kê, vôn kế, ampe kế, tần số kế… để xác định các đại lượng vật lý của dòng điện

-Đo lường điện để làm gì?

Phát hiện hư hỏng sự cố trong mạch điện và các thiết bị vi mạch

Xác định các giá trị cần đo

Đánh giá chất lượng của các thiết bị sau sản xuất

Xác định thông số kỹ thuật của thiết bị

-Phân loại dụng cụ đo điện: Hiên nay có 2 phương pháp phân loại chính

a. Theo nguyên lý làm việc

Dụng cụ đo kiểu điện từ

Dụng cụ đo kiểu điện động

Dụng cụ đo kiểu cảm ứng

Dụng cụ đo kiểu từ điện

b. Theo đai lượng, giá trị cần đo

+Đo điện năng: Ví dụ công tơ điện

+Đo điện áp: Ví dụ: Vôn kế

+Đo dòng điện: Ví dụ: Ampe kế

+Đo công suất: Oát kế

+Đo điện trở: Ôm kế

Sai số khi đo: Khi đo lường luôn xảy ra các sai số

Sai số tương đối: Là tỷ lệ % của sự chênh lệch giữa giá trị đo được và giá trị thự

Answer:

The plotted diagram of line distance versus the transmission speed is uploaded below.

Explanation:

Given the data in the question;

Bit length B = 1000 bits

propagation velocity V = 2 × 10⁸ m/s

now, we know that the bit length of a link is expressed as;

B = R × d/V

where V is propagation velocity

d is the distance

R is the transmission speed

B is bit length

so we substitute

1000 = R × d/(2 × 10⁸)

1000 = Rd/(2 × 10⁸)

2 × 10¹¹ = Rd

R = 2 × 10¹¹ / d

R = 2E+11

Hence, we plot the transmission speed versus line distance; as shown in the image BELOW.

From the plot, if the transmission speed increases, the distances between stations decreases and vise versa.

Hence, both are inversely proportional.

The graph plot of line distance versus the transmission speed for a bit length of 1000 bits is; plotted below with Lν = 2 × 10¹¹ m/s

We are given;

Bit length; B = 1000 bits

Propagation velocity; V = 2 × 10⁸ m/s

Formula for bit length of a link is expressed as;

B = ν × L/V

where;

V is propagation velocity

L is the distance

ν is the transmission speed

B is bit length

Thus, plugging in the relevant values gives;

1000 = ν × L/(2 × 10⁸)

Thus;

Lν = 1000 × 2 × 10⁸

Lν = 2 × 10¹¹ m/s

Thus, find the attached image of a graph showing the line distance versus the transmission speed for a bit length of 1000 bits.

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Answer:

Approximately 60% of all collisions at railroad crossings occur when active warning devices are present and functioning.

The percentage of railroad accidents that occur at properly functioning automatic warning devices can vary depending on various factors such as the location, type of railroad, and overall safety measures in place.

Here,

These warning devices, such as gates, lights, and bells, are designed to alert drivers and pedestrians to the presence of a train at a railroad crossing. When they are working correctly, they provide an effective warning system to prevent accidents.

It's important to note that while automatic warning devices greatly reduce the risk of accidents, human error or negligence can still lead to incidents.

It is essential for all road users to obey the signals and exercise caution when approaching and crossing railroad tracks, even when warning devices are functioning properly.

However, it is generally expected that the occurrence of accidents at properly functioning automatic warning devices is relatively low.

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Answer:

See below

Explanation:

Enter-key also called the "Return key," it is the keyboard key that is pressed to signal the computer to input the line of data or the command that has just been typed.It Was the Return KeyThe Enter key was originally the "Return key" on a typewriter, which caused the carriage to return to the beginning of the next line on the paper. In a word processing or text editing application, pressing Enter ends a paragraph. A character code for return/end-of-line, which is different in Windows than it is in the Mac, Linux or Unix, is inserted into the text at that point.

Answer:

True

Explanation:

Once there are two yellow lines having inner broken lines on the two sides of a center traffic lane, what this is trying to tell you is that you can use those lanes to start a left hand turn, or a U-turn from the both directions of traffic. However you cannot use it for passing. This is sometimes misunderstood by road users and drivers.

Answer:

Beam of 25" depth and 12" width is sufficient.

I've attached a detailed section of the beam.

Explanation:

We are given;

Beam Span; L = 20 ft

Dead load; DL = 0.50 k/ft

Live load; LL = 0.65 k/ft.

Beam width; b = 12 inches

From ACI code, ultimate load is given as;

W_u = 1.2DL + 1.6LL

Thus;

W_u = 1.2(0.5) + 1.6(0.65)

W_u = 1.64 k/ft

Now, ultimate moment is given by the formula;

M_u = (W_u × L²)/8

M_u = (1.64 × 20²)/8

M_u = 82 k-ft

Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.

Effective depth of a beam is given by the formula;

d_eff = d - clear cover - stirrup diameter - ½Main bar diameter

Now, let's adopt the following;

Clear cover = 1.5"

Stirrup diameter = 0.5"

Main bar diameter = 1"

Thus;

d_eff = 25" - 1.5" - 0.5" - ½(1")

d_eff = 22.5"

Now, let's find steel ratio(ρ) ;

ρ = Total A_s/(b × d_eff)

Now, A_s = ½ × area of main diameter bar

Thus, A_s = ½ × π × 1² = 0.785 in²

Let's use Nominal number of 3 bars as our main diameter bars.

Thus, total A_s = 3 × 0.785

Total A_s = 2.355 in²

Hence;

ρ = 2.355/(22.5 × 12)

ρ = 0.008722

Design moment Capacity is given;

M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12

Φ is 0.9

f’c = 4,000 psi = 4 kpsi

fy = 60,000 psi = 60 kpsi

M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12

M_n = 220.03 k-ft

Thus: M_n > M_u

Thus, the beam of 25" depth and 12" width is sufficient.