In order to properly construct this road you have determined that you need to bring in 135,000 CCY of fill material to construct the sub-base. The fill material that is available is dry earth, with a weight of 2800 lbs/BCY and a moisture content of 7%. The percent swell for this soil is 25%. The final compaction level will be 118 PCF dry density and the optimum moisture content is 6%. How many BCY of earth must be excavated to meet this requirement

The scenario with the terminology it represents should be matched as follows:

A ratio can be defined as a mathematical expression that's used to denote the proportion of two (2) or more quantities with respect to one another and the total quantities.

A proportion can be defined as an expression which is typically used to represent (indicate) the equality of two (2) ratios. This ultimately implies that, proportions can be used to establish that two (2) ratios are equivalent and solve for all unknown quantities.

A scale factor can be defined as the ratio of two (2) corresponding length of sides or diameter in two similar geometric figures such as polygons.

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Answer:

Ratio: Justin’s car can drive 10 miles per gallon, or 10:1. He wants to travel 40 miles, meaning he needs at least 4 gallons in his car (10:1 = 40:4).

Proportion: Jonathan designs a new car. The car can run 48 miles per gallon, or 48:1.

Scale: Kinsey’s sketches show the design of her lawn mower with its exact proportions but in a smaller size.

Explanation:

Took the quiz 100%

The property is that polyurethanes have a good tear resistance as well as tensile strength.

Polyurethane is an umbrella word for a group of polymers formed via the condensation of polyisocyanates and polyalcohols. Despite its xenobiotic origins, polyurethane has been discovered to be biodegradable by naturally occurring microbes.

Polyurethanes are synthesized by reacting a polyol (an alcohol having more than two reactive hydroxyl groups per molecule) with a diisocyanate or a polymeric isocyanate in the presence of appropriate catalysts and additives

To manage cell fate and tissue healing, biodegradable polyurethane platforms have been used. Cell bonding, spreading, multiplication, and separation are all influenced by framework features such as compound, mechanical, and primary. These are not fully fixed as the science of biodegradable polyurethanes, platform manufacturing technologies, and surface change approaches.

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The hardware that I am using to take this course is a Laptop computer running on windows operating system.

When you send a query, like your previous message, your laptop's operating system breaks it into smaller units called packets. These packets contain parts of your message, destination address, and other metadata.

They're sent over the internet through your Internet Service Provider (ISP), which directs them through routers and switches. These network devices ensure packets take the most efficient path.

They reach the server where I'm hosted. The server processes your query, generates a response, divides it into packets, sends them back through the same process, and your laptop reassembles them to display my answer.

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if you add a bit too much solvent when filling the volumetric flask to the mark then the solution must be discarded and a new solution should be prepared from the first step of the procedure.

A solvent is a substance that dissolves a solute and produces a solution. The word solvent is derived from the Latin word solvus, which means to "loosen, untie, solve."

A supercritical fluid, solid, gas, or liquid can also serve as a solvent in addition to the more common liquid form. All the ions and proteins in a cell are dissolved in water, which is the most frequent solvent used by living things. Water is a solvent for polar molecules.

With temperature, different amounts of solute can dissolve in different amounts of solvent. Paints, paint stripers, inks, and dry cleaning are among the major uses of solvents.

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A strain gauge should have a low stiffness in order to get high sensitive readings. This is because a strain gauge is designed to measure small changes in length or deformation in a material due to external forces, and a low stiffness allows for greater flexibility and sensitivity to these changes.

A strain gauge is a type of sensor that is used to measure the deformation or strain of a material. It consists of a thin wire or foil that is bonded to the surface of the material being measured. When the material is deformed due to an external force, the wire or foil also stretches or compresses, causing a change in resistance. This change in resistance can be measured and used to calculate the strain or deformation of the material. The sensitivity of a strain gauge is determined by its stiffness, which is the resistance of the wire or foil to deformation. A strain gauge with a high stiffness will be less sensitive to small changes in deformation, while a strain gauge with a low stiffness will be more sensitive to these changes. Therefore, in order to get high sensitive readings from a strain gauge, it should have a low stiffness. This will allow it to detect even small changes in deformation, which can be useful in applications such as structural monitoring or strain testing.

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a.Ellen is facing the eye condition of myopia since it is negative

b.Far point is -1m

What is myopia?

Both myopias, high and low. Low myopia is another name for mild nearsightedness (typically less than three diopters of myopia). Three to six diopters of myopia is considered to be moderate nearsightedness. High myopia is another name for severe nearsightedness (myopia of more than 6 diopters). In general, nearsighted children get more nearsighted as they age, but by the time they reach their 20s, their need for glasses has stabilized.

Power (P) =  -1.0D

f (negative) =  diverging lens

p = 1/f

So, far point is 1m

Hence to conclude ellen has myopia and the far point is 1 m

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Answer:

A. The force on the car was greater since it was moving

Explanation:

Hope this helps : )

Answer:  C. The forces exerted by the car and wall were equal

Explanation:  Isaac Newton’s third law states if an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A.

Answer:

C. 14.55

Explanation:

12 x 10 = 120

120 divded by 10 is 12

so now we do the left side

7 x 3 = 21 divded by 10 is 2

so now we have 14

and the remaning area is 0.55

so 14.55

The contact would be hardwired in series with the emergency stop button.

This configuration ensures that if the fault relay contact opens due to a processor malfunction, the emergency stop button will also open the circuit, halting the operation of the machinery. This provides an additional layer of safety in case the fault relay contact fails to function properly or is bypassed.

Hardwiring the contact in parallel with the emergency stop button could lead to unintended operation of the machinery, as the emergency stop button may not be able to override the malfunctioning processor.

Programming the fault relay contact as part of the master control reset or zone control last state instruction would not provide the necessary fail-safe mechanism for halting machinery in the event of a processor malfunction.

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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.

The absolute pressure of air at which it deviates from the continuum concept can be calculated using the given information.

To calculate the absolute pressure, we need to determine the number of molecules per cubic millimeter. Given that a gas is considered rarefied when it contains less than 10^12 molecules per cubic millimeter, and Avogadro's number is\(6.023 × 10^23\) molecules per mole, we can use these values to find the absolute pressure.

To convert from molecules per cubic millimeter to moles per cubic millimeter, we divide the given value by Avogadro's number:

\(10^12 molecules/mm^3 / (6.023 × 10^23 molecules/mol) = 1.66 × 10^-12 mol/mm^3\)

Now, we can convert the number of moles per cubic millimeter to pressure in Pascals (Pa) using the ideal gas law equation:

PV = nRT

Assuming the gas is at a temperature of 20 degrees Celsius, which is 293.15 Kelvin, and using the ideal gas constant R = 8.314 J/(mol·K), we can rearrange the equation to solve for pressure:

P = (nRT) / V

Given that the volume is 1 mm^3, the number of moles per cubic millimeter is 1.66 × 10^-12 mol/mm^3, and the temperature is 293.15 K, we can calculate the absolute pressure as:

\(P = (1.66 × 10^-12 mol/mm^3) × (8.314 J/(mol·K)) × (293.15 K) / (1 mm^3)P ≈ 3.89 × 10^-6 Pa\)

Therefore, the absolute pressure for air when it deviates from the continuum concept is approximately 3.89 × 10^-6 Pa.

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Answer:

I'm not 100%sure but I think it's the first one

The square feet coverage of a gallon of primer may vary depending on the quality of the primer and the porosity of the surface being primed. On average, a gallon of primer can cover approximately 200-300 square feet. However, it is always best to consult the manufacturer's instructions for the specific primer being used as it will provide a more accurate estimate of coverage.

Generally, when buying paint or primer, it is important to know the coverage area to ensure that you purchase enough to cover the entire surface adequately. Factors such as the number of coats, the thickness of the coat, and the method of application will all play a role in determining the coverage area. For example, using a roller instead of a brush may result in a more even application and, therefore, use less primer. Furthermore, it is important to keep in mind that using too little primer may result in uneven coverage and may not provide adequate protection against moisture and other environmental factors.

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Two technicians are discussing solder repair. Technician A says that solder should be applied to the soldering iron tip while soldering Technician B says that the solder should be applied to the wire joint while soldering. The correction technician is: "Both Technician A and B (Option C)

Soldering is a joining method that uses solder to bind several metals together. Solder is a metal alloy typically composed of tin and lead that is melted using a hot iron.

The iron is heated to a temperature above 600 degrees Fahrenheit before cooling to form a strong electrical connection. To put it another way, soldering is the process of uniting metal pieces to produce a mechanical or electrical link.

It commonly employs a low melting point metal alloy (solder), which is heated and applied to the metal pieces to be connected, and when the solder solidifies, it attaches to the metal parts and makes a connection.

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Full Question:

Two technicians are discussing solder repair. Technician A says that solder should be applied to the soldering iron tip while soldering Technician B says that the solder should be applied to the wire joint while soldering. Who is Correct?

Answer:

120

Explanation:

Answer:

40.7 m

Explanation:

Let point 1 represent the surface of the water, point 2 be the top of the water trajectory and the reference be the bottom of the tank. Hence:

\(z_1=16\ m,P_1=2\ atm,V_1=0(velocity\ at\ the \ surface \ of\ water\ is\ low)\\V_2=0,P_2=P\\\\Using\ Bernoulli\ equation:\\\\\frac{P_1}{\rho g} +\frac{V_1^2}{2g}+z_1= \frac{P_2}{\rho g} +\frac{V_2^2}{2g}+z_2\\\\Sine\ V_1=0,V_2=0:\\\\\frac{P_1}{\rho g} +z_1=\frac{P}{\rho g} +z_2\\\\z_2=\frac{P_1}{\rho g} -\frac{P}{\rho g} +z_1\\\\z_2=\frac{P_1-P}{\rho g} +z_1\\\\z_2=\frac{P_{1.gage}}{\rho g} +z_1\\\\\)

\(z_2=\frac{2\ atm}{1000\ kg/m^3*9.81\ m/s^2}(\frac{101325\ N/m^2}{1\ atm} )(\frac{1\ kg.m/s^2}{1 \ N} ) +20\\\\z_2=20.7+20\\\\z_2=40.7\ m\)

Explanation:

Ohm's law applies: V = IR, and variations. The current in a series circuit is the same through every element. That current is given as 1.1 A.

Rt = (8 V)/I3 = (8 V)/(1.1 A) = 80/11 Ω = 7 3/11 Ω

R3 = Rt -R1 -R2 = 7 3/11 -2 -5 = 3/11 Ω

It = I3 = 1.1 A

I1 = I2 = I3 = 1.1 A

V1 = (I1)(R1) = (1.1 A)(2 Ω) = 2.2 V

V2 = (I2)(R2) = (1.1 A)(5 Ω) = 5.5 V

V3 = (I3)(R3) = (1.1 A)(7/11 Ω) = 0.3 V

Electrical Engineering Textbooks: These textbooks provide comprehensive information on conductor applications and insulation types. They cover topics such as conductor materials, their properties, and various insulation materials used in different applications.

Online Resources: There are several websites dedicated to electrical engineering and related topics that offer information on conductor applications and insulation types. Some reliable sources include IEEE (Institute of Electrical and Electronics Engineers) Xplore, Electrical Engineering Stack Exchange, and All About Circuits. These platforms have forums, articles, and technical papers discussing conductor applications and insulation types.Manufacturers' Websites: Electrical component manufacturers often provide detailed information on conductor applications and insulation types.

For example, companies like General Cable, Southwire, and Prysmian Group have websites that describe their product offerings, including conductor applications and insulation types. You can explore their product catalogs or technical specifications for more specific details.Industry Standards and Codes: Various industry standards and codes outline conductor applications and insulation types. The National Electrical Code (NEC) and the International Electrotechnical Commission (IEC) standards are widely followed in electrical engineering. These standards often provide guidelines and requirements for conductor selection and insulation materials based on the intended application.Remember, it's essential to cross-reference information from multiple sources to ensure accuracy and a comprehensive understanding.

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A PMOS transistor conducts when the control input (gate) is LOW or at a logic level of 0.

In a PMOS transistor, the source is connected to VDD (the positive supply voltage), and the drain is connected to the output (load). When the gate of the PMOS transistor is at a logic level of 0 or LOW, it creates a channel between the source and drain, allowing current to flow from the source to the drain and turning the transistor ON. This results in a low resistance path between the output and VDD, allowing a high logic level to be present at the output.

Conversely, when the gate of the PMOS transistor is at a logic level of 1 or HIGH, it blocks the channel between the source and drain, preventing current from flowing and turning the transistor OFF. This results in a high resistance path between the output and VDD, causing the output to be pulled up to VDD through a resistor, which translates to a logic level of 0 at the output.

Therefore, a PMOS transistor conducts when the control input (gate) is at a logic level of 0, and it does not conduct when the control input (gate) is at a logic level of 1.

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To create a full Fraction class, implement instance variables, a constructor, and several methods such as getNumerator, getDenominator, setNumerator, setDenominator, add, subtract, multiply, and toString. Test the class using FractionTester to ensure proper functionality.

To create a full Fraction class, you need to implement several methods. Let's go through each method step by step:

1. Create your instance variables and constructor:
  - Instance variables are the properties or attributes of the Fraction class, such as numerator and denominator.
  - The constructor is a special method used to initialize the instance variables when a Fraction object is created.

2. Implement the following methods in the Fraction class:
  a. `public int getNumerator()`: This method should return the numerator of the fraction.
  b. `public int getDenominator()`: This method should return the denominator of the fraction.
  c. `public void setNumerator(int x)`: This method should set the numerator of the fraction to the given value, `x`.
  d. `public void setDenominator(int x)`: This method should set the denominator of the fraction to the given value, `x`.
  e. `public void add(Fraction other)`: This method should add the given `other` fraction to the current fraction.
  f. `public void subtract(Fraction other)`: This method should subtract the given `other` fraction from the current fraction.
  g. `public void multiply(Fraction other)`: This method should multiply the current fraction by the given `other` fraction.
  h. `public String toString()`: This method should return a string representation of the fraction.

3. Use the FractionTester file to test your Fraction class as you implement each method.

Make sure to pay attention to the correct implementation of each method, as they will be crucial for the functionality of the Fraction class.

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Answer:

Engineering law and ethics are the moral codes associated with the field in which engineers are meant to adhere strictly to.

These ethics help in the safety of the engineers as a result of wearing protective clothing being a part of the ethics . Accuracy should also be taken important in research and daily activities. Corrupt practices and activities which poses risk to the environment is also strictly prohibited.

Explanation:

The three-second rule is recommended for passenger vehicles during ideal road and weather conditions. Slow down and increase your following distance even more during adverse weather conditions or when visibility is reduced. Also increase your following distance if you are driving a larger vehicle or towing a trailer.

Answer:

Its specific resistance will be around 0.015

Answer:

12 mins

Explanation:

The summation of the forces in vertical direction

= Fb - Fd - w = 0 ∴ Fd = Fb - w  ----- ( 1 )

Fb ( buoyant force ) = Pair * g * Vballoon ------- ( 2 )

Pair = air density , Vballoon = volume of balloon

Vballoon = \(\frac{\pi D^3}{6}\) ,  where D = 4  ∴  Vballoon = 33.51 ft^3

g = 32.2 ft/s^2

From property tables

Pair = 2.33 * 10^-3 slug/ft^3

μ ( dynamic viscosity ) = 3.8 * 10^-7 slug/ft.s

Insert values into equation 2

Fb = ( 2.33 * 10^-3 ) * ( 32.2 ) *( 33.51 )  = 2.514 Ib

∴ Fd = 2.514 - 2.5 = 0.014 Ib ( equation 1 )

Assuming that flow is Laminar  and  RE < 1

Re = (Pair * vd) / μair   -------- ( 3 )

where:  Pair = 2.33 * 10^-3 slug/ft^3 ,  vd = ( 987 * 4 ) ft^2/s ,   μair = 3.8 * 10^-7 slug/ft.s

Insert values into equation 3

Re = 2.4 * 10^7 (  this means that the assumption above is wrong )

Hence we will use drag force law

Assume Cd = 0.5

Express  Fd using the relation below

Fd = 1/2* Cd * Pair * AV^2

therefore V = 1.39 ft/s

Recalculate  Reynold's number using  v = 1.39 ft/s

Re = 34091

from the figure Cd ≈ 0.5 at Re = 34091  

Finally calculate the rise time ( time taken to reach an altitude of 1000 ft )

t = h/v

  = 1000 / 1.39 = 719 seconds ≈ 12 mins

Answer:

sorry im answering questions for the points cuz im built dfferent

Explanation:

Answer: That happens to me too sometimes. I hate when people give the wrong answers...like if u don't know the answer, then don't respond

70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

(as per my thinking)

Answer:

70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

(a) The active part of the inflow expands its cross-section as it passes the tidal turbine due to conservation of mass and energy.

(b) Several reasons contribute to the theoretical maximum efficiency not being achieved in practical tidal turbine deployments, including turbulence, non-uniform flow, and mechanical losses.

(c) Estimating the total fluid loading on the Alstom 1.4 MW OceadeTM turbine requires considering the flow velocity, rotor and tower dimensions, and making assumptions about the flow characteristics and structural properties.

(d) The estimates of forces obtained from the fluid loading calculations are essential for designing the turbine and tower structure by ensuring that they can withstand the anticipated loads and stresses.

(a) The active part of the inflow expands its cross-section as it passes the tidal turbine due to the principle of conservation of mass and energy. As the tidal flow encounters the turbine rotor, some of the kinetic energy of the flow is converted into mechanical energy to drive the turbine. To satisfy the conservation of mass, the cross-sectional area of the flow must increase to compensate for the reduction in flow velocity caused by energy extraction.

(b) Achieving the theoretical maximum efficiency in practical tidal turbine deployments is challenging due to several reasons. First, tidal flows are often characterized by turbulence, which disrupts the uniformity of the flow and reduces overall efficiency. Second, tidal flow itself is not uniformly distributed, and the flow characteristics vary with tidal cycles, further impacting efficiency. Lastly, mechanical losses in the turbine's components, such as friction and resistance, reduce the efficiency of energy conversion.

(c) Estimating the total fluid loading on the Alstom 1.4 MW OceadeTM turbine involves considering the flow velocity, rotor diameter, and tower dimensions. Assuming a tidal flow velocity of 3 m/s, the fluid loading can be estimated by considering the momentum change and forces acting on the rotor and tower surfaces. Assumptions may include a simplified flow model, neglecting factors such as turbulence and non-uniform flow, and assuming a stationary tower. These assumptions simplify the calculation while providing a reasonable estimate of the fluid loading.

(d) The estimates of forces obtained from the fluid loading calculations are crucial for designing the turbine and tower structure. These estimates help engineers determine the required structural strength, material selection, and design considerations to ensure that the turbine and tower can withstand the anticipated fluid forces and mechanical stresses. By considering the estimated forces, designers can optimize the structural integrity, stability, and reliability of the turbine and tower, ensuring safe and efficient operation in tidal environments.

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