The department recently purchased a new 3-phase lathe, and he is required to wire the power supply. The nameplate of the motor on the lathe indicated that it is delta connected with an equivalent impedance of (5 +j15) 2 per phase. The workshop has a balanced star connected supply and you measured the voltage in phase A to be 230 Ɖ0⁰ V. (a) Discuss three (3) advantage of using a three phase supply as opposed to a single phase supply (b) Draw a diagram showing a star-connected source supplying a delta-connected load. Show clearly labelled phase voltages, line voltages, phase currents and line currents. (c) If this balanced, star-connected source is connected to the delta-connected load, calculate: i) The phase voltages of the load ii) The phase currents in the load iii) The line currents iv) The total apparent power supplied

Answer:

  12 volts

Explanation:

The voltages across parallel-connected items are identical. (In fact, that's why you can measure the voltage by connecting the voltmeter in parallel with the circuit element.)

The voltage drop across each bulb is 12 volts.

Answer:

116,090 Btus

Explanation:

Answer:

sorry di ko alam

Explanation:

Answer:

Explained below

Explanation:

Perfect Elastic Plastic in steel design is simply a method whereby the structural members are selected using the criteria of the overall ultimate capacity of the system. However, when safety is considered, the applied loads are usually increased by factors of safety as prescribed in the relevant steel design codes. Therefore, this model of design is just based on the yield capacity of the steel.

Answer:

In I-Q modulation technique, many symbols or signals with different amplitude and phase are generated. What is the map of all of the symbols called:____.

d. constellation diagram.

Explanation:

A constellation diagram is a representation of all the possible symbols that a system can transmit.  It shows these as a collection of points on the map.  For a signal modulated by a digital modulation scheme, the constellation diagram is used to display both the ideal (reference) signal and the actual measured signal on the same plot.  Modulation simply means the process of converting data into electrical signals so that they are optimized for transmission.

Proteus simulation is  a software tool. It is used for the   designing and testing  of electronic circuits. So thetool will be conneted to the LCD as well as the cips on the left.

This tool is popular among engineers and students as it allows them to test their designs virtually with  ease, eliminating the need to physically build and test the circuit.

Through robust circuitry simulations under different conditions, designers can identify faults in functionality promptly and make necessary changes before building the physical prototype.

The utilization of Proteus simulation equips individuals with the ability to devise and assess electronic circuits in a highly efficient and economical manner, specifically targeting those found in mobile phones.

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Correct option is d.estimate landing loading rate.The anteroposterior ground reaction force could be used to estimate landing loading rate.

The anteroposterior ground reaction force is a measure of the force exerted by the body on the ground during movement. It represents the component of the force that acts in the forward-backward direction. By analyzing the anteroposterior ground reaction force, it is possible to estimate the landing loading rate, which refers to the rate at which force is applied to the body upon landing.

During activities such as jumping, the landing loading rate is an important parameter to consider as it can affect the risk of injury. A higher landing loading rate indicates a rapid increase in force upon landing, which may result in greater stress on the joints and tissues of the body.

Conversely, a lower landing loading rate suggests a more gradual increase in force, which can be less detrimental to the body.

By using the anteroposterior ground reaction force, researchers and practitioners can assess the landing loading rate and make informed decisions regarding training, rehabilitation, and injury prevention strategies.

Monitoring and analyzing this parameter can help identify individuals who may be at a higher risk of injury due to excessive loading rates and enable the implementation of targeted interventions to reduce injury risk.

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The answer response are:

1. In a steel beam, the designation w21x68 refers to its size and shape. The "W" stands for wide flange and the number "21" indicates the depth of the beam in inches. The weight per foot of the beam is approximately 68 pounds.

2. Shear connections are designed to resist shear forces in a beam, while moment connections are designed to resist bending moments. Shear connections typically have bolts or welds connecting the beam to the supporting column, while moment connections have additional plates and bolts to provide extra strength against bending moments.

3. Composite construction is a type of building construction that combines two materials, usually steel and concrete, to create a more efficient and cost-effective structure. The advantages of composite construction include increased strength and stiffness, improved fire resistance, and reduced construction time and costs.

4. Steel structures have the advantage of being strong and durable, but they can also be vulnerable to fire due to their high melting point. To overcome this disadvantage, steel structures can be coated with fire-resistant materials or designed with additional fire protection measures, such as sprinkler systems or fire-resistant insulation. An example of a steel structure that has been designed for fire resistance is the World Trade Center in New York City.

5. An open web steel joist is a type of steel beam that has a web consisting of a series of smaller steel bars or rods. The lightweight steel decking is attached to the joist using special clips or fasteners that are designed to hold the decking securely in place.

A steel beam is a structural element made of steel that is used to support heavy loads in construction and engineering projects. Steel beams come in a variety of shapes and sizes, including I-beams, H-beams, and wide flange beams.

Therefore, They are designed to resist bending and shearing forces, and are often used to span long distances or support the weight of other structural components, such as floors, roofs, and walls. Steel beams are popular in construction due to their strength, durability, and versatility, and can be found in a wide range of buildings, from skyscrapers to residential homes.

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False. When voltage levels lag (experience a momentary decrease), the extra voltage can severely damage or destroy equipment. Voltage levels that increase can cause damage as well, but not as severely as when they decrease.

Voltage is a measure of electrical potential energy per unit charge. It is the difference in electrical potential between two points in an electrical circuit, and the amount of work required to move a unit charge from one point to the other. Voltage is measured in units of volts (V). Voltage is an important concept in electricity since it is responsible for the flow of current in a circuit. Voltage is also used to measure the power of a source, such as a battery. Voltage is used to power a variety of electrical devices, including motors and lights. In addition, the voltage of a circuit can be adjusted to regulate the current, or the flow of electrons.

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Answer:

  (√6)/3 ≈ 0.8165

Explanation:

The RMS value is the square root of the mean of the square of the waveform over one period. It will be ...

  \(\displaystyle\sqrt{\frac{1}{T}\left(\int_{\frac{T}{4}}^{\frac{3T}{4}}{1^2}\,dt+\int_{\frac{3t}{4}}^{\frac{5t}{4}}{(\frac{-4}{T}}(t-T))^2\,dt\right)}=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\left.\frac{16}{T^2}\cdot\frac{1}{3}(t-T)^3\right|_{\frac{3t}{4}}^{\frac{5t}{4}}\right)}\\\\=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\frac{T}{6}\right)}=\sqrt{\frac{2}{3}}=\boxed{\frac{\sqrt{6}}{3}\approx0.8165}\)

ANSWER:

Aerospace Engineering. ...

Chemical Engineering. ...

Biomedical Engineering.

EXPLANATION:

This is all i know but ... I hope this helps~

(a) For noncolloidal silt, the width of the channel necessary to limit shear and prevent erosion is approximately [insert numerical value] meters.

(b) For fine gravel, the width of the channel necessary to limit shear and prevent erosion is approximately [insert numerical value] meters.

To determine the necessary width of the channel, we can utilize the Manning's equation, which relates the channel flow rate, slope, roughness coefficient, and hydraulic radius. Given that the channel needs to carry 5 m3/s of clear water and has a slope of 0.08%, we can calculate the required width for different bottom conditions.

(a) For a bottom of noncolloidal silt, which typically has a Manning's roughness coefficient (n) of around 0.025, we can use the equation Q = (1.49/n) * (A * R^(2/3)) * S^(1/2), where Q is the flow rate, A is the cross-sectional area of the channel, R is the hydraulic radius, and S is the slope. Rearranging the equation to solve for the width (A = width * depth), we can substitute the given values and solve for the width:

5 = (1.49/0.025) * (width * depth * (width * depth / (2 * (width + depth))))^(2/3) * (0.0008)^(1/2)

Simplifying the equation and solving numerically, we find that the necessary width of the channel to limit shear and prevent erosion is approximately 6.14 meters.

(b) For a bottom of fine gravel, which typically has a higher Manning's roughness coefficient (n) of around 0.035, we can repeat the same calculation with the updated roughness coefficient. Substituting the values and solving for the width:

5 = (1.49/0.035) * (width * depth * (width * depth / (2 * (width + depth))))^(2/3) * (0.0008)^(1/2)

Simplifying and solving numerically, we find that the necessary width of the channel in this case is approximately 0.47 meters.

Practicality of the results:

The calculated widths provide estimates for preventing channel erosion based on the given parameters. However, it is essential to consider other factors and practical constraints in real-world scenarios. These factors may include local regulations, site-specific conditions, and potential variations in flow rates. Engineering expertise and detailed site assessments are crucial to ensure the practicality and safety of any channel design or modification.

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angular velocity of the reel after 3 seconds is 5.11 rad/s. We cannot calculate the frictional force without knowing the coefficient of kinetic friction.

To solve this problem, we need to use the principle of conservation of energy and the equation of rotational motion.
First, let's calculate the moment of inertia of the reel. The moment of inertia is given by the formula:
I = Mk^2
where M is the mass of the reel and k is the radius of gyration about its center of gravity. We are given that the weight of the reel is 150 Ib, so we can convert this to mass using the formula:
M = W/g
where W is the weight of the reel and g is the acceleration due to gravity. Substituting the given values, we get:
M = 150/32.2 = 4.66 slugs
Now we can calculate the moment of inertia:
I = Mk^2 = 4.66 (1.25)^2 = 7.3125 slug-ft^2
Next, let's find the work done by the torque on the reel. The work done is given by the formula:
W = MΔθ
where M is the torque and Δθ is the angular displacement. We are given that the torque is 25 Ib ft and the reel starts from rest, so initially Δθ = 0. At the end of 3 seconds, the angular displacement is given by:
Δθ = ωt + (1/2)αt^2
where ω is the final angular velocity, α is the angular acceleration, and t is the time. We are asked to find the final angular velocity after 3 seconds, so we rearrange the equation and substitute the given values:
ω = (Δθ - (1/2)αt^2)/t = (0.5)(α)(t) = (0.5)(τ/I)(t)
where τ is the torque and I is the moment of inertia. Substituting the given values, we get:
ω = (0.5)(25/7.3125)(3) = 5.11 rad/s
Finally, let's find the frictional force acting on the reel. The frictional force is given by the formula:
f = μN
where μ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the reel, which we already calculated to be 150 Ib. We are not given the value of μ, so we cannot calculate the frictional force.

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The net work done by the Nitrogen in lifting the mass and the pressure is; W_n = 0.6257 kJ

We are given;

Mass of argon; m_a = 0.1 kg

mass of piston; m_p = 5 kg

Pressure 1; p₁ = 100 kPa

Temperature 1; T₁ = 300 K

Temperature 2; T₂ = 320 K

Ideal gas constant; R = 0.208 kJ/kg.K

γ = 1.67

Since the reaction is adiabatic and slowly reversible, then we will use the formula; PV^(γ) = C

Thus, work done by argon is;

W_ar = [m_a * R(T₁ - T₂)]/(γ - 1)

W_ar = [0.1 * 0.208(300 - 320)]/(1.67 - 1)

W_ar = -0.6028 kJ

Work done by nitrogen will be;

W_n = |W_ar| + (m_p * g * m_a)/1000

W_n = 0.6028 + (5 * 9.81 * 0.1)/1000

W_n = 0.6257 kJ

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Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature \(T_1 = -16^0\ C\)

Quality \(x_1 = 0.2\)

Outlet:

Temperature \(T_2 = -16^0 C\)

Quality  \(x_2 = 1\)

The following data were obtained at saturation properties of R134a at the temperature of -16° C

\(v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg\)

\(v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg\)

\(m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}\)

\(\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}\)

A limit switch is a device that limits the angle of the boom at its highest position.

The limit switch will automatically shut down the machinery if the boom is rotated too far, helping to prevent potential damage. When the limit switch is reached, a mechanical or electrical signal is sent to the control unit of the crane, and the machine will shut off. The angle of the boom can be set manually or with a remote control, and the limit switch ensures that the boom does not go past its set angle. By limiting the boom's angle, the limit switch helps to protect the machinery and personnel operating the crane.

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There are three main methods of asphalt pavement recycling: hot in-place recycling, cold in-place recycling, and full-depth reclamation. The predominant method is full-depth reclamation, which involves pulverizing the existing pavement and mixing it with a stabilizing agent before compacting and overlaying with new asphalt.

This method not only recycles the existing materials, but also strengthens the base and subbase layers, leading to a more durable and longer-lasting pavement.

1. Cold in-place recycling (CIR)
2. Hot in-place recycling (HIR)
3. Full-depth reclamation (FDR)

Among these methods, Hot in-place recycling (HIR) is the predominant method. Here's a brief summary of this method:

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Answer:

Strap wrench (I think)

Explanation:

The strap doesnt have any rough edges so it shouldn't mark the workpiece (assuming that is what you meant when you said 'marring'). I would recommend to wait for some other answers though since i haven't studied this yet :)

The first thing to do if your vehicle leaves the roadway is to take your foot off the accelerator.

It's crucial to maintain composure and refrain from making quick or hostile motions when your car leaves the road.

Stepping off the gas pedal allows you to gently slow down the car and regain control. It enables you to concentrate on analysing the circumstance and selecting the most appropriate line of action.

You should retain a tight grip on the steering wheel and steer softly after removing your foot from the pedal to keep the car under control.

Thus, if it is safe to do so, it is crucial to maintain your composure and move the car back onto the road with calm, controlled motions.

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Control points are areas of the vehicle structure that are used to monitor the precision of the vehicle body's dimensions as they are being assembled. Thus, option B is correct.

In a full frame car, the body is independent of the frame and fastened by bushings and bolts. The unibody is a structure made up of integrated panels that are welded, bonded, and riveted together.

Therefore, In a full frame car, the body is independent of the frame and fastened by bushings and bolts. The unibody is a structure made up of integrated panels that are welded, bonded, and riveted together.

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The option that is not a derived units is known to be option  E) kg.

The units that are known to be used for any form of derived quantities are said to be called the derived units.

They are:

Note that this unit are derived because they are known to be derived because they have to be solved for using different ways. The others such as kg m-3 is one that need to be derived to arrive at it.

Therefore, based on the above, one can say that The option that is not a derived units is known to be option  E) kg because it is one that cannot be derived,

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The following are derived units EXCEPT

Options

A) kg m-3

B)N

C)Ns

D)m3

E)kg

Answer:

a) 4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.

b) 21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.

c) 0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.

d) 0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.  

e) 23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.

Explanation:

a) 4 hectómetros cuadrados a decámetros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un hectómetro cuadrado equivale a 100 decámetros cuadradps. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

\(x = 4\,Hm^{2}\times\frac{100\,Dm^{2}}{1\,Hm^{2}}\)

\(x = 400\,Dm^{2}\)

4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.

b) 21345 centímetros cuadrados a metros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

\(x = 21345\,cm^{2}\times \frac{1\,m^{2}}{10000\,cm^{2}}\)

\(x = 2,135\,m^{2}\)

21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.

c) 0,592 kilómetros cuadrados a metros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 1000000 metros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

\(x = 0,592\,km^{2}\times \frac{1000000\,m^{2}}{1\,km^{2}}\)

\(x = 592000\,m^{2}\)

0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.

d) 0,102 metros cuadrados a centímetros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

\(x = 0,102\,m^{2}\times \frac{10000\,cm^{2}}{1\,m^{2}}\)

\(x = 1020\,cm^{2}\)

0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.

e) 23911 kilómetros cuadrados a hectómetros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 100 hectómetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

\(x = 23911\,km^{2}\times \frac{100\,Hm^{2}}{1\,km^{2}}\)

\(x = 2391100\,Hm^{2}\)

23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.

Answer:

What’s the most important reason we should need a mask?

we were because it protects us from germs and viruses.

Is the ban on removal of the mask requirement being implemented too fast in an area without general consensus (moral, political, cultural, technological)?

Face coverings limit the volume and travel distance of expiratory droplets dispersed when talking, breathing, and coughing.

The fact that the dataset includes people who all live in the same zip code might get in the way of accuracy and fairness.

When a dataset consists of people from the same zip code, it can introduce biases and limitations that hinder the accuracy and fairness of the analysis. Zip codes often correspond to specific geographical areas, which means that the dataset might lack diversity in terms of demographics, socioeconomic status, and other important variables. This lack of representation can lead to skewed results and inaccurate conclusions when trying to make generalizations or predictions about a broader population.

In terms of accuracy, relying on a dataset with a homogeneous zip code can result in misleading findings. The insights derived from such a limited sample may not be applicable to the entire population or other regions. It could lead to overgeneralizations or assumptions that may not hold true when considering a more diverse population. Additionally, certain zip codes may have unique characteristics or circumstances that do not reflect the broader population, further distorting the accuracy of the analysis.

Furthermore, the fairness of the analysis may be compromised when using a dataset limited to a single zip code. If important variables such as income, race, or education level are not adequately represented in the dataset due to the lack of diversity within the zip code, any conclusions drawn from the analysis may not be equitable or inclusive. This can have implications for decision-making processes, policy development, or resource allocation, as it may perpetuate existing inequalities or overlook specific needs and challenges faced by different groups within the larger population.

Therefore, it is crucial to ensure that datasets are representative and diverse, including individuals from various zip codes and geographic areas, to enhance both the accuracy and fairness of the analysis.

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Answer:

#include <iostream>

using namespace std;

// named constant to give mnemonic name to "magic number"

const float SqFtPerGal = 350.0f;

// the function main() is always the entry point of the application

int main()

{

float length, width, area, paint; // to hold user values and results

// prompt user for length and width of wall

cout << "Enter length of wall : " << flush;

cin >> length;

cout << "Enter width of wall : " << flush;

cin >> width;

// calculate area and amount of paint needed

area = length * width;

paint = area / SqFtPerGal;

// output results with reasonable text

cout << "You need " << paint << " gallons of paint to cover "

<< area << " square feet of wall." << endl;

// program stops executing when it returns from main(), 0 means O.K.

return 0;

}

Explanation:

Answer:

Alternating Current (a.c) can be reversed since it relates to sinusoidal curves while Direct Current (d.c) can not be reversed.

Explanation:

\(.\)

Difference between AC and DC

Electric current flows in two ways as an alternating current (AC) or direct current (DC). In alternating current, current keeps switching directions periodically – forward and backward. While in the direct current it flows in a single direction steadily. The main difference between AC and DC lies in the direction in which the electrons flow. In DC, the electrons flow steadily in a single direction while electrons keep switching directions, going forward and then backwards in AC. Let us learn more differences between them in the next few sections.

To compute the life of a bearing, we need to consider the number of revolutions it makes. Given that the bearing rotates at 1750 rpm continuously for 24 hours a day, we can calculate the number of revolutions in a day.

Number of revolutions in a day = rpm * 60 minutes * 24 hours
Number of revolutions in a day = 1750 * 60 * 24 = 2,520,000 revolutions

Next, we need to calculate the number of revolutions in five years, taking into account leap years.

Number of revolutions in five years = number of revolutions in a day * 365 days * 5 years
Number of revolutions in five years = 2,520,000 * 365 * 5 = 4,608,000,000 revolutions

Therefore, the life of the bearing that rotates at 1750 rpm continuously for 24 hours a day for five years is approximately 4,608,000,000 revolutions.

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