a cylindrical metal specimen has a diameter of 15.0 mm and a length of 150 mm. it is deformed elastically with a tensile stress of 50 mpa. (i) if the elongation must be less than 0.072 mm, which of the metals in the table below are suitable candidates? why? (ii) if, additionally, the maximum permissible diameter decrease is 0.0023 mm, which (if any) of the metals that satisfy the criterion in part (i) are suitable candidates? why?

The container that traps the most heat is a shoebox covered in aluminum foil.

An aluminum foil is used for food storage, and to wrap items, such as meats, to minimize moisture loss while cooking.

Aluminum foil is also an excellent insulator. This is due to the fact that it reduces heat emission by reflecting it back.

Therefore, in a shoebox container, the aluminum foil helps to trap most heat better than a plastic wrap or wax paper.

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The rating of the service disconnecting means for a single-family home must be at least 100 amperes, 3-wire.Therefore, 100 amperes is the required minimum rating of service size for a one-family home.

What is the service's required minimum ampere rating?

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Answer:

Explanation:

To determine which launcher was more accurate, we can compare the average distance of the bean bags from the target center.

We can use the distance formula to calculate the distance between two points:

d = sqrt((x2-x1)^2 + (y2-y1)^2)

where d is the distance, (x1, y1) is the first point, and (x2, y2) is the second point.

For Nachie's bean bags, the average distance from the target center is:

d = sqrt((0.05-0)^2 + (0.08-1)^2) = sqrt(0.0025 + 0.0184) = sqrt(0.02)

For Ihwari's bean bags, the average distance from the target center is:

d = sqrt((0.08-0)^2 + (1-1)^2) = sqrt(0.0064 + 0) = sqrt(0.0064) = 0.08

Since Ihwari's bean bags were closer to the target center on average, it can be concluded that Ihwari's launcher is more accurate than Nachie's.

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:

\(\mathbf{F_{AB} = 13785.06 N }\)

\(\mathbf{F_{AC} = -5062.38 N }\)

\(\mathbf{F_{AD} = -5062.38 N }\)

Explanation:

From the given information:

Let calculate the position vector of AB, AC, and AD

To start with AB; in order to calculate the position vector of AB ; we have:

\(r_{AB}^{\to} = r _{OA}^{\to} - r_{OB}^{\to} \\ \\ r_{AB}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( 1.5 \ \hat i + \ 1. 5 \hat j ) \\ \\ r_{AB}^{\to} = ( 2.6 \ \hat i - 1.5 \ \hat i + 2.6 \ \hat j - 1.5 \ \hat j - 2.2 \ \hat k) \\ \\ r_{AB}^{\to} = (1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) m\)

To calculate the position vector of AC; we have:

\(r_{AC}^{\to} = r _{OA}^{\to} - r_{OC}^{\to} \\ \\ r_{AC}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( 2\ \hat i + \ \hat j - 1.2 \ \hat k) \\ \\ r_{AC}^{\to} = ( 2.6 \ \hat i - 2\ \hat i + 2.6 \ \hat j - \ \hat j - 2.2 \ \hat k + 1.2 \ \hat k) \\ \\ r_{AC}^{\to} = (0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) m\)

To calculate the position vector of AD ; we have:

\(r_{AD}^{\to} = r _{OA}^{\to} - r_{OD}^{\to} \\ \\ r_{AC}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( \hat i + \ 2 \hat j - 1.2 \ \hat k) \\ \\ r_{AD}^{\to} = ( 2.6 \ \hat i - \hat i + 2.6 \ \hat j - 2 \ \hat j - 2.2 \ \hat k + 1.2 \ \hat k) \\ \\ r_{AD}^{\to} = (1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) m\)

However; let's calculate the force in AB, AC and AD in their respective unit vector form;

To start with unit vector AB by using the following expression; we have:

\(F_{AB}^{\to} = F_{AB} \dfrac{ r _{AB}^{\to} }{|r_{AB}^{\to}} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{\sqrt{ (1.1)^2 + (1.1)^2 + (-2.2 )^2 }} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{ \sqrt{7.26}} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{ 2.6944} \\ \\ \\ F_{AB}^{\to} = F_{AB} (0.408 \ \hat i+ 0.408 \ \hat j - 0.8165 \ \hat k ) N\\\)

The force AC in unit vector form is ;

\(F_{AC}^{\to} = F_{AC} \dfrac{ r _{AC}^{\to} }{|r_{AC}^{\to}} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{\sqrt{ (0.6)^2 + (1.6)^2 + (-1 )^2 }} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{ \sqrt{3.92}} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{1.9798} \\ \\ \\ F_{AC}^{\to} = F_{AC} (0.303 \ \hat i+ 0.808 \ \hat j - 0.505 \ \hat k ) N\\\)

The force AD in unit vector form is ;

\(F_{AD}^{\to} = F_{AD} \dfrac{ r _{AD}^{\to} }{|r_{AD}^{\to}|} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{\sqrt{ (1.6)^2 + (0.6)^2 + (-1 )^2 }} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{ \sqrt{3.92}} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{1.9798} \\ \\ \\ F_{AD}^{\to} = F_{AD} (0.808 \ \hat i+ 0.303 \ \hat j - 0.505 \ \hat k ) N\\\)

Similarly ; the weight of the lunar Module is:

W = mg

where;

mass = 13500 kg

acceleration due to gravity=  1.82 m/s²

W = 13500 × 1.82

W = 24,570 N

Also. we known that the load is shared by four landing gears; Thus, the vertical reaction force exerted by the ground on each landing gear can be expressed as:

\(R =\dfrac{W}{4}\)

\(R =\dfrac{24,570}{4}\)

R = 6142.5 N

Now; the reaction force at point A in unit vector form is :

\(R^{\to} = Rk^{\to} \\ \\ R^{\to} = (6142.5 \ k ^{\to}) \ N\)

Using the force equilibrium at the meeting point of the coordinates at A.

\(\sum F^{\to} = 0\)

\(F_{AB}^{\to} +F_{AC}^{\to} + F_{AD}^{\to} + R^{\to} =0\)

\([F_{AB} (0.408 \ \hat i + 0.408 \ \hat j - 0.8165 \ \hat k ) N + F_{AC} (0.303 \ \hat i + 0.808 \ \hat j - 0.505 \ \hat k ) N + F_{AD} (0.808 \ \hat i + 0.303 \ \hat j - 0.505 \ \hat k) N + (6142.5 \ k^ \to ) ]\)

\(= [ ( 0.408 F_{AB} +0.303 F_{AC} + 0.808F_{AD}) \hat i + (0.408 F_{AB}+0.808F_{AC}+0.303F_{AD}) \hat j + (-0.8165 F_{AB} -0.505F_{AC} -0.505 F_{AD} +6142.5 ) k ^ \to ] = 0\)

From above; we need to relate and equate each coefficients i.e i ,j, and \(k ^ \to\) on both sides ; so, we can re-write that above as;

\(0.408 F_{AB} +0.303 F_{AC} + 0.808F_{AD}) =0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ --- (1) \\ \\ 0.408 F_{AB}+0.808F_{AC}+0.303F_{AD}) =0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ --- (2) \\ \\ -0.8165 F_{AB} -0.505F_{AC} -0.505 F_{AD} +6142.5 = 0 --- (3)\)

Making rearrangement and solving by elimination method;

\(\mathbf{F_{AB} = 13785.06 N }\)

\(\mathbf{F_{AC} = -5062.38 N }\)

\(\mathbf{F_{AD} = -5062.38 N }\)

The force vector of each member, depends on the magnitude of the

force and the unit vector of the member.

Responses:

The force supported by the members are;

Resolving the given members into unit vectors gives;

\(\hat u_{AB} = \mathbf{\dfrac{(2.6 - 1.5) \cdot \hat i + (2.6 - 1.5)\cdot \hat j + (-2.2)\cdot \hat k }{\sqrt{(2.6- 1.5)^2 + (2.6 - 1.5)^2 + (-2.2)^2}}}\)\(\dfrac{(2.6 - 1.5) \cdot \hat i + (2.6 - 1.5)\cdot \hat j + (-2.2)\cdot \hat k }{\sqrt{(2.6- 1.5)^2 + (2.6 - 1.5)^2 + (-2.2)^2}}= 0.40825 \cdot \hat i + 0.40825\cdot \hat j - 0.81625\cdot \hat k\)

Similarly, we have;

\(\hat u_{AC} =\mathbf{ \dfrac{(2.6 - 2) \cdot \hat i + (2.6 - 1)\cdot \hat j + (-2.2+1.2)\cdot \hat k }{\sqrt{(2.6- 2)^2 + (2.6 - 1)^2 + (-2.2+1.2)^2}}}\)

\(\dfrac{(2.6 - 2) \cdot \hat i + (2.6 - 1)\cdot \hat j + (-2.2+1.2)\cdot \hat k }{\sqrt{(2.6- 2)^2 + (2.6 - 1)^2 + (-2.2+1.2)^2}} =\dfrac{0.6\cdot \hat i +1.6\cdot \hat j -1\cdot \hat k }{\sqrt{0.6^2 + 1.6^2 + (-1.)^2}}\)

\(\dfrac{0.6\cdot \hat i +1.6\cdot \hat j -1\cdot \hat k }{\sqrt{0.6^2 + 1.6^2 + (-1.)^2}}= 0.303046\cdot \hat i + 0.80812\cdot \hat j - 0.50508\cdot \hat k\)

\(\hat u_{AD} =\mathbf{ \dfrac{(2.6 - 1) \cdot \hat i + (2.6 -2)\cdot \hat j + (-2.2 + 1.2)\cdot \hat k }{\sqrt{(2.6-1)^2 + (2.6 -2))^2 + (-2.2 + 1.2)^2}}}\)

\(\hat u_{AD} =\mathbf{0.80812\cdot \hat i+ 0.303046\cdot \hat j - 0.50508\cdot \hat k}\)

The forces are therefore;

\(\vec F_{AB} =\mathbf{ F_{AB} \cdot \left ( 0.40825 \cdot \hat i + 0.40825\cdot \hat j - 0.81625\cdot \hat k \right)}\)

\(\vec F_{AC} =\mathbf{ F_{AC} \cdot \left (0.303046\cdot \hat i + 0.80812\cdot \hat j - 0.50508\cdot \hat k\right)}\)

\(\vec F_{AD} = \mathbf{F_{AD} \cdot \left (0.80812\cdot \hat i+ 0.303046\cdot \hat j - 0.50508\cdot \hat k\right)}\)

\(Weight \ on \ the \ assembly = \dfrac{13,500 \, kg \times 1.82 \, m/s^2}{4} = 6,142.5 \, \hat k N\)

Which gives;

\(\mathbf{0.40825 \cdot \hat i \cdot F_{AB}}\) + \(0.303046\cdot \hat i \cdot F_{AC}\) + \(0.80812\cdot \hat i \cdot F_{AD}\) = 0

\(0.40825 \cdot \hat j \cdot F_{AB}\) + \(0.80812\cdot \hat j \cdot F_{AC}\) + \(0.303046 \cdot \hat j \cdot F_{AD}\left\) = 0

\(-0.81625\cdot \hat k \cdot F_{AB}\) - \(0.50508\cdot \hat k \cdot F_{AC}\) - \(0.50508\cdot \hat k \cdot F_{AD}\) +  \(\mathbf{6,142.5 \, \hat k}\) = 0

Which gives;

\(-0.81625\cdot \hat k \cdot F_{AB}\) - \(0.50508\cdot \hat k \cdot F_{AC}\) - \(0.50508\cdot \hat k \cdot F_{AD}\) =  \(-6,142.5 \, \hat k\)

Solving gives;

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Answer: It is  a nominal rate per year

Answer:

yes

Explanation:

an idea could be a jack o lantern monster since it Halloween tomorrow  

If you want to take 5 measurements per decade for frequencies between 1 kHz and 100 kHz, then the following frequencies should be chosen: 1 kHz, 10 kHz, 100 kHz, 1 MHz, and 10 MHz.

Taking multiple measurements per decade in the frequency range from 1 kHz to 100 kHz can be done in order to gain a better understanding of the behavior of a signal in a given frequency range. By taking multiple measurements, you can gain insight into the signal's spectral characteristics such as its frequency content, spectral magnitude, and spectral phase.

Additionally, taking multiple measurements per decade can provide more precise information about the signal's power spectral density, and is often used in applications such as signal processing and communications to accurately measure system performance.

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Integrated circuits have revolutionized modern electronics by combining multiple electronic components into a single chip, enhancing performance, compactness, and efficiency.

Integrated circuits, also known as microchips or ICs, have played a pivotal role in revolutionizing modern electronics. These tiny wonders have fundamentally transformed the way electronic devices are designed and manufactured, enabling unprecedented advancements in various industries.

By integrating multiple electronic components onto a single chip, integrated circuits have brought about remarkable improvements in performance, compactness, and efficiency.

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The javascript code that satisfies the given question that throws a range error exception if any of the numbers is greater than 50.  is given below

function processNumbers(numList) {

   var result = 0;

   if (numList.length < 4)

       throw new Error();

   for (var index = 0; index < numList.length; index++) {

       if (numList[index] > 75)

           throw new RangeError();

       result += numList[index] * 1.3 * index;

   }

   return result;

}

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The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ \(\longrightarrow\) 2Cl₂ + 2H₂O

\(Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}\)

Which gives;

\(80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}\)

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

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False. Data scientists do not necessarily need much business domain knowledge is a false statement. In fact, having a strong understanding of the business domain is crucial for data scientists to effectively analyze and interpret data in a meaningful way.

Business domain knowledge allows data scientists to understand the context and nuances of the data they are working with. They can identify important trends, patterns, and relationships that are relevant to the business and its goals. Without this knowledge, data scientists may struggle to interpret data accurately and make informed decisions. Additionally, having business domain knowledge enables data scientists to communicate effectively with stakeholders and provide actionable insights that can drive business growth and success.

In summary, business domain knowledge is a vital component of data science. Data scientists must possess a strong understanding of the business domain to effectively analyze and interpret data in a meaningful way. It is important for data scientists to continuously learn and expand their knowledge of the business domain to ensure they provide valuable insights that can inform strategic decisions.

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Solution :

import \($\text{java}.$\)util.*;

public \($class$\) currency{

  public static \($\text{void}$\) main(String\($[]$\) args) {

      Scanner input \($=$\) new Scanner(System\($\text{.in}$\));

      System\($\text{.out.}$\)print("Enter \($\text{number of}$\) quarters:");

      int quarters = input.nextInt();

      System.out.print("Enter number of dimes:");

      int \($\text{dimes =}$\) input.nextInt();

      System\($\text{.out.}$\)print("Enter number of nickels:");

      int nickels = input.nextInt();

      System\($\text{.out.}$\)print("Enter number of pennies:");

      int \($\text{pennies = }$\) input.nextInt();

      // computing dollors

      double dollars = (double) ((quarters*0.25)+(dimes*0.10)+(nickels*0.05)+(pennies*0.01));

      System\($\text{.out.}$\)format("You have : $%.2f",dollars);

  }

}

In the context of buildings, structures designed to mitigate the effects of seismic activity or strong winds are commonly referred to as seismic or base isolators.

These isolators act as shock absorbers to reduce the transmission of vibrations and energy from the ground to the building. Seismic or base isolators typically consist of various materials or devices that can absorb and dissipate energy. Some commonly used types of isolators include:

1. Rubber or Neoprene Bearings: These are made of rubber or neoprene pads that provide flexibility and damping to absorb and dissipate energy during seismic events.

2. Sliding Bearings: Sliding bearings allow the building to slide horizontally on a low-friction surface, such as PTFE (polytetrafluoroethylene). This sliding motion helps to reduce the transfer of seismic forces to the building.

3. Fluid Dampers: Fluid dampers utilize the properties of viscous fluid to absorb and dissipate energy. They consist of cylinders filled with a viscous fluid that flows through small orifices, generating resistance against the motion caused by seismic forces.

4. Steel Springs: Steel springs can be used as shock absorbers to reduce the impact of seismic or wind loads. They provide flexibility and absorb energy by deforming under the applied forces.

These seismic or base isolators are strategically placed at the base or foundation of the building to reduce the transfer of vibrations and energy. By incorporating these shock-absorbing mechanisms, buildings can better withstand seismic events and minimize structural damage.

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Answer:

Efficiency is defined as any performance that uses the fewest number of inputs to produce the greatest number of outputs. Simply put, you're efficient if you get more out of less.

Explanation:

Answer:

Option D. is correct

Explanation:

Joe uses one of his CAT5 patch cables to connect two hubs to add a new segment to his local network. As he can only connect to it from a workstation within that segment,  he is not able to reach the new network segment from his workstation.

The most problem is that the technician used a straight-through cable.

Option D. is correct.

Answer:

Se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.

Explanation:

Dado que para conseguir jugo de naranja concentrada, se parte de un extracto con 7% en peso de sólidos el cual se mete a un evaporador al vacío, y en el evaporador se elimina el agua necesaria para que el jugo salga con una concentración del 60% de peso de sólidos, si se introducen al proceso 1000 kg/h de jugo diluido, para calcular la cantidad de agua evaporada y de jugo concentrado saliente se debe realizar el siguiente cálculo;

1000 x 0.07 = 70

60 = 70

100 = X

100 x 70 / 60 = X

7000 / 60 = X

116.66 = X

Por lo tanto, se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.

Soil is a vital component of any construction process. It is used to provide a strong and stable foundation that can support the weight of the structure built on top of it. Soil needs to be compacted to its optimum bulk density of 2 gm/cc at 15% moisture content (mass basis) for an earthen dam structure.

If the soils are initially at 10% moisture content (mass basis) with 1.2 gm/cc bulk density, how much water volume needs to be added or removed before you start compaction for the 15 m long, 3 m high compacted dam with a bottom width of 5 m and top width of 2 m (trapezoidal section).The bulk density of the soil needs to be increased to 2 gm/cc, and it needs to be done at 15% moisture content.

Therefore, the volume of soil that needs to be added or removed = 140 - 84 = 56 m³At 10% moisture content, the soil density is 1.2 gm/cc, and it needs to be increased to 2 gm/cc by adding or removing water. The soil moisture content can be increased by adding water, and it can be reduced by removing water. In this case, water needs to be added to the soil.

The amount of water required to achieve the optimum moisture content can be calculated using the following formula: Water volume = (Final soil volume x Final soil density x Final moisture content) - (Initial soil volume x Initial soil density x Initial moisture content) = (84 x 2 x 0.15) - (140 x 1.2 x 0.10) = 4.02 m³Therefore, 4.02 m³ of water needs to be added to the soil to achieve the desired moisture content and bulk density.

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Here's a function called Close Enough which takes a const reference to a vector of strings and an int. This function is used to check if there are two adjacent strings in a list that are very similar to each other by determining the number of characters that can be different between any two strings next to each other.

The function starts by iterating through the vector of strings from the first to the second to last string. For each string, it then counts the number of differences between it and the next string by comparing the characters at each position.

If the number of differences is less than or equal to the given integer n, the function returns the index of the current string. If there is no such index, the function returns -1.

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Answer:

Explanation:

In this geometry, the force in each string is proportional to the cosine of the angle the opposite string makes with the horizontal, and is inversely proportional to the sine of the sum of the angles.

  AC = (15 N)cos(45°)/sin(60° +45°) = (15 N)cos(45°)/sin(105°) ≈ 10.98 N

  BC = (15 N)cos(60°)/sin(105°) ≈ 7.76 N

_____

The relation described above derives from the fact that the sum of vertical forces and the sum of horizontal forces are both zero when the system is in equilibrium.

Answer:

Explanation:

A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.

Answer:

a. True

Explanation:

The study of fluids in a state of rest or in motion and the forces involved in it is called fluid mechanics. Fluid mechanics has a wide range of applications in the field of mechanical engineering as well as civil engineering.

When we study the flow of fluid between any two flat plates that is indefinitely flat and is parallel, the flow of the fluid is known as plane Poiseulle flow. The profile of a plane Poiseulle flow is parabolic.

The velocity profile of a plane Poiseulle flow is :

\($\frac{u(y)}{U_{max}}=1-\left(\frac{2y}{h}\right)^2$\)

Thus the answer is TRUE.

Answer:

Wmax =  750 kw  < power developed ( 1000kw ) for a reversible    the cycle is Impossible

Explanation:

Hot reservoir Temperature = 1000 K

Cold reservoir Temperature = 500 K

Heat transfer ( energy received by Hot reservoir ) ( Q ) = 1500 kW

Heat transfer ( energy received by Cold reservoir via Hot reservoir ) = 1000 Kw

Calculate the rate of entropy production

The higher the entropy production the less efficient the system

Δs = Cp In ( T2 / T1 )

power developed = 1000 kW

considering that the cycle is reversible and the constant volume or constant pressure of the substance in the thermodynamic cycle is not given we will use the efficiency to determine if the cycle is possible or not

Л = efficiency

∴Л = 1 - T2 / T1 = 1 - ( 500 / 1000 ) = 0.5

note as well that;  Л = work output / work input = Wmax / Q

                                  = 0.5 = Wmax / Q

∴ rate of entropy production = Q ( 0.5 ) = 1500 * 0.5 = 750 kw

Given that Wmax =  750 kw  < power developed ( 1000kw ) for a reversible    the cycle is Impossible

Answer:

The answer is below

Explanation:

\(\theta_1=cos^{-1}0.28=73.74^o\ lagging\\\\S_1=125\angle 73.74^o=35\ kW+j120\ kVAR\\\\S_2=10\ kW-j40\ kVAR\\\\S_3=15\ kW\)

Total power = P = 35 kW + 10 kW + 15 kW = 60 kW

Total kVAR = 120 kVAR - 40 kVAR = 80 kVAR

\(Total\ apparent \ power =S= S_1+S_2+S_3=(35+j120)+(10-j40)+(15)\\\\S=60\ kW+j80\ kVAR=100\angle 53.13^o\\\\Current(I)=\frac{S^*}{V^*} \frac{100000\angle -53.13^o}{1400\angle0}=71.43\angle-53.13^o\\ \\Power\ factor (PF)=cos(53.13)=0.6\ lagging\\\\The \ new\ power\ factor\ is\ to \ be\ 0.8[cos^{-1}0.8=36.87^o], hence\ since\ the\ total\ \\real\ power(P)= 60\ kW, the\ capacitor\ kVAR(Q_c)\ is:\\\\Q_c=60tan(53.13)-60tan(36.87)=80-45=35\ kVAR\\\\\)

\(C=\frac{Q_c}{wV^2} =\frac{35000\ VAR}{2\pi*60\ Hz*1400\ V}=47.38\ \mu f\)

New current (I') = \(\frac{S'^*}{V^*}=\frac{60000-j45000}{1400}=53.57\angle-36.87^o\)

Current reduce from 71.43 A to 53.57 A

During normal Windows operations, you receive a BSOD (Blue Screen Of Death), and your computer shuts down. "Hard drive" might cause this BSOD error.

This BSOD error could cause the system to crash without warning following a pause error.

It implies that there may have been a disk file system corruption, or that your hard drive segments are most likely to blame. This has to do with the NTFS file system.

BSODs can be caused by HDDs (hard drives). However, since the computer is brand new, there is a chance even though it is extremely unlikely.

Hard drive errors are frequently caused by faulty hardware, poor system maintenance, viruses, or human error. To fix disk issues, utilize the Chkdsk software found in Windows operating systems.

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The creative process that helps you overcome writer's block is called Reset.

Thus, could be the answer.

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is called ecoines

Explanation:

(a) Using the switching scheme given we have determined the value of ∝ to produce an output with an amplitude of 90V at the fundamental frequency i.e. ∝ = 36.96°

(b) The THD of the load current is = 1.6919

The maximum displacement or distance made by a point on a vibrating body or wave relative to its equilibrium position is known as the amplitude in physics. It is equivalent to the length of the vibration path divided in half. A pendulum's amplitude is therefore equal to half the distance that the bob travels when swinging from one side to the other. Vibrating sources produce waves, whose amplitude is inversely proportional to the source's amplitude.

The maximum deviation of any point on the string from its position when the string is at rest is used to determine the amplitude of a transverse wave, such as the wave on a plucked string. The maximum displacement of a particle from its equilibrium position is used to determine the amplitude of a longitudinal wave, such as a sound wave.

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Loading coils on loop lines are bad for ADSL because loading coils in combination with naturally occurring capacitance in UTP II creates a low pass filter. So, the correct option is d.

A "loading coil" is an inductive coil that is placed along the circuit of a telephone line. This minimizes the attenuation of signals within 300 to 3500 Hertz i.e. voice band frequency. This assists in saving loss of signals especially for large distances.

A "local loop" refers to the two-wired connection between the user's house and telephone company's central office.

Asymmetric Digital Subscriber Line (ADSL) is a broadband connection that assists in faster transfer of data through the same copper telephonic lines. This is suitable for homes and small business setups.

When a loading coil is placed in local loop then there is high attenuation for signals above 4000 Hertz. Thus, low pass filter is formed due to the combination of the coils with Unshielded twisted pair (UTP) capacitance. This is why if a loop line is intended to carry ADSL then the loading coil must be disconnected.

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