A proposed embankment fill requires 7100 ft of compacted soil. The void ratio of the compacted fill is specified as 0.5. Four borrow pits are available as below, which list the respective void ratios of the soil and the cost per cubic feet for moving the soil to the proposed construction site. Make the necessary calculations to determine the total cost for each pit. Assume the specific gravity of soil solids, Gs to be the same at all pits.
1. Calculate the volume of solid (Vs) = ft3 Borrow Pit A: void ratio = 1.05 and the unit cost = $ 6 /ft3
2. Calculate the total cost to pay to Pit A = Borrow Pit B: void ratio = 0.61 and the unit cost = $ 7 /ft3
3. Calculate the total cost to pay to Pit B = Borrow Pit C: void ratio = 0.81 and the unit cost = $ 4 /ft3
4. Calculate the total cost to pay to Pit C = Borrow Pit D: void ratio = 0.73 and the unit cost = $6 /ft3
5. Calculate the total cost to pay to Pit D =

The value of floor(lg(n)) for n = 63 is found to be 5.

To determine the value of floor(lg(n)) for n = 63, we can calculate the base-2 logarithm of 63 and round down to the nearest integer.

The base-2 logarithm, denoted as lg(n), represents the exponent to which 2 must be raised to obtain the value of n. In this case, lg(63) is approximately 5.977, but since we need to round down, the result is 5.

Therefore, the value of floor(lg(n)) for n = 63 is 5.

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Answer:

ABCD

Explanation:

Couette flow is a type of fluid flow in which a viscous fluid is confined in a space between two surfaces, which move with respect to each other, producing a shearing motion within the fluid.

In this type of flow, the velocity profile is linear with zero velocity at the surface, which means there is zero fluid slip at the boundary or boundaries such as a sliding rigid plate, a cylinder sliding along a tube, or a shaft rotating in a supporting bearing.

Hence, the given statement is true, which means option (a) is correct.However, Couette flow is the flow of a fluid in a thin layer between two parallel surfaces, where the flow is driven by the motion of one of the surfaces relative to the other. As a result of this motion, there is a velocity gradient within the fluid, which is proportional to the distance from the stationary surface. Couette flow can be used to model fluid flow in many practical engineering applications, such as lubrication in bearings, flow through pipes, and the operation of turbines and pumps.

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Answer:

b

Explanation:

i did it yeater dayajsbs

The exit pressure will be 1.3273 bar = 0.1327 MPa and the exit specific internal energy 228.61 KJ/Kg .

1.) In graph,

1-2 = Compressor

3-4 = Throttle value

\(T_{1} \\\) = \(T_{4}\) = -20°C = 253 K

\(T_{3}\) = 25°C = 298 K

\(P_{3}\) = 0.8 MPa

Assuming saturated fluid is entering the value at point (3),

using table of R-134a properties

∵ \(h_{3} =h_{4}\) = 234.49 KJ/Kg (at 25°C,8 bar)

at \(T_{3}\) = 25° , \(P_{3}\) = 0.8 MPa =8 bar

At -20°C,   \(hf_{4}\) = 173.6 KJ/Kg

                 \(hg_{4}\) = 386.6 KJ/Kg

\(h_{4} = hf_{3} + n_{4} ( hg_{4} - hf_{4} )\) = 234.49

173.6 + \(n_{4}\) [386.6 - 173.6] = 234.49

             \(n_{4}\) = 0.285

\(P_{4} = P_{1} = P_{sat}\) = 1.3273 bar = 0.1327 MPa

2.)

Exit specific internal energy, using table R-134a

\(U_{4} = uf_{4} + n_{4} (ug_{4} - uf_{4} )\)

at -20°c, \(uf_{4}\\\) = 173.65, ufg = 192.85 KJ/ Kg

\(U_{4}\) = 173.65 + 0.285 (192.85)

\(U_{4}\) = 228.61 KJ/Kg

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A. Tracking your online activity and then selling that data without your knowledge.

Social networking can make users vulnerable to various criminal activities, including identity theft, cyberbullying, and scams. One of the most significant risks is the exposure of personal information, which can be used for targeted advertising, phishing, or even physical harm. Cybercriminals can use social media to track users' online activity and collect personal data without their knowledge or consent, which can be sold or used for identity theft. Moreover, criminals can use social media to spread malicious content or gain access to user accounts, posing as trustworthy friends or authority figures. Therefore, users should be aware of the risks and take measures to protect their privacy and security, such as reviewing privacy settings, avoiding sharing sensitive information, and being cautious about accepting friend requests from unknown users.

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c

it could not be d it cover about all the lights u would have to have thick dirt can't be a way to little dirt u would have to just throw dirt on ur lights and b same thing to little amount of dirt

The major types of interior walls and partitions in a larger building are:

It is different because it is more durable than the rest and it is also:

Conclusively, the coats of plaster that are used over expanded metal lath are:

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For a given water content, the application of lower compaction energy increases the dry unit weight of the soil. So, option A holds the correct answer choice.

Compaction refers to a process that brings about an increase in soil unit weight or density, accompanied by a decrease in air volume. There is usually no change in water content. The degree of compaction is measured by dry unit weight and depends on the water content and compaction effort such as weight of the roller, the number of passes, the number of impacts, and weight of the hammer. So when lower compaction energy is put in, it does not have any effect on the dry unit weight of the soil.

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The code you have provided is a for-loop that prints out the comma character ',' 11 times. The comma character has an ASCII value of 44, so the loop will output 44 11 times. This is why the answer is 44.

The for loop starts at 0 and will continue to run as long as the variable I is less than or equal to 10. Each time the loop runs, the variable I is incremented by 1.

The system.out.println() statement inside the for loop will print out a comma each time the loop runs.

So the output of this code will be 11 commas, like this:
```
,,,,,,,,,,,
```

There is no calculation happening in this code, so the answer is 44. It seems like there may be a mistake or typo in the question.

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In the radio communication system, multipath is the propagation phenomenon that causes diffusion or reflection in multiple different directions of a signal.

Multipath is a propagation mechanism that impacts the propagation of signals in radio communication. Multipath results in the transmission of data to the receiving antenna by two or more paths. Diffusion and reflection are the causes that create multiple paths for the signal to be delivered.

Diffraction occurs when a signal bends around sharp corners; while reflection occurs when a signal impinges on a smooth object. When a signal is received through more than one path because of the diffraction or reflection, it creates phase shifting and interference of the signal.

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For industrial machinery, straight exposed vertical risers of intermediate metal conduit with threaded couplings are permitted if supported at the top and bottom no more than 10 feet apart.

As long as they are supported at the top and bottom no more than 10 feet apart, straight exposed vertical risers of intermediate metal conduit with threaded couplings are acceptable for industrial machines. This requirement makes sure that the conduit is stable and well-supported in order to guard against any risks or damage. To ensure compliance and safety when installing industrial gear, it is crucial to follow the precise rules and specifications outlined by local electrical codes and standards.

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Answer: A leak in the system

Explanation:

If an engine fails during rollout or just before takeoff, immediately shut both throttles and land the aircraft safely. Before reaching a safe single engine speed right away after takeoff, drop your nose to increase velocity.

Closing the throttle, turning off the fuel pump, setting the mixture control to idle cutoff, and simply cranking the engine is the most reliable hot start method I've found.

If an engine fails during rollout or just before takeoff, immediately shut both throttles and land the aircraft safely. Before reaching a safe single engine speed right away after takeoff, drop your nose to increase velocity. If you are unable to climb, close both throttles and land straight ahead.

The typical practice for the majority of aircraft would be to abandon takeoff if an engine failed during takeoff. In small aircraft, the pilot should turn the throttles down to idle, activate the speed brakes (if provided), and apply the brakes as needed if the engine fails before VR (Rotation Speed).

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According to the sector model of North American city structure, members of low-income groups tend to live in the central business district (CBD) and the inner-city area.

The sector model of North American city structure is an urban land use model that was created by Homer Hoyt in 1939. It divides a city into a series of sectors that radiate outwards from the CBD. These sectors differ in terms of their socio-economic characteristics and are defined by their level of wealth, housing type, and level of social and economic status.

Land use patterns in the sector model are concentrated around the central business district and expand radially outwards, with the most desirable areas occupying the outermost ring. Members of low-income groups tend to live in the central business district and the inner-city area.

Hence, it can be concluded that according to the sector model of North American city structure, members of low-income groups tend to live in the central business district (CBD) and the inner-city area.

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You should search the zones to the left and right of your vehicle, as well as any areas directly in front of the intersection, such as crosswalks, bike lanes, and medians. Additionally, you should be aware of any oncoming traffic or pedestrians who may be crossing the intersection.

When approaching an intersection, it is important to pay attention to all of the potential hazards around you. First, you should search the zones to the left and right of your vehicle. This includes looking for oncoming traffic, pedestrians, or cyclists who may be crossing the intersection or entering the roadway. Additionally, you should be aware of any obstacles that may be directly in front of you, such as crosswalks, bicycle lanes, or medians. By taking the time to search all of these zones and be aware of your surroundings, you can increase your safety and the safety of others.

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The fundamental components of any digital system are logic gates. It is an electronic circuit with one or more inputs and just one output.

In any digital system, logic gates serve as the fundamental building elements. One or more inputs may be present, but there is only one output in this electronic circuit. An established logic governs the relationship between the input and the output. This is the basis behind the names AND, OR, NOT, etc. given to logic gates.

Using cables that stand for 1s and 0s, we transmit data through computers. In order for computers to someday perform more difficult tasks like computing the 50th digit of, they need to be able to manipulate those 1s and 0s. Input wire 1s and 0s are converted via logic gates in computers.

A K-map is a particular type of truth table that facilitates the mapping of parameter values and leads to a shortened Boolean expression. Functions with two to four variables are ideal candidates for K-maps.

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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 ...

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

Find the cover for the memory module slot. Insert the module by sliding it. The majority of socket alignments are roughly 30 degrees. Change the memory.

Memory is the ability to take in information from the surrounding environment, process it, store it, and then recall it at a later time—sometimes years later. A memory storage system or filing cabinet are frequently used as analogies for human memory.

Memory is a necessary part of every person's life. We are unable to function with in present or consider the future without a remembrance of the past. We wouldn't be able to recall our actions from the previous day, the present, or the future.

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Answer:

B. rotor

Explanation:

The correct answer Is rotor because the others are part of a cars drivetrain

The drive train system exists as a critical element of a vehicle and the transmission exists as an integral part of the drive train. B rotors NOT part of a car's drive train.

A drive train exists not really a single part of your car – it's a set of drive train components that interact with the engine to drive the wheels and different regions of the vehicle to thrust it into motion. These components often contain the transmission, differential, driveshaft, axles, CV joints, and wheels.

Therefore, the correct answer is option B rotors.

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Answer:

Bulk modulus: ß = - ∆p/(∆V/V)

∆p = (3039 - 1520)x10³ = 1519 kPa

∆V = 1231 - 1232 = -1 m³

V = 1232 m³

ß = - 1519/(-1/1232) = 1.87x10^6 kPa = 1.87 GPa

Explanation:

a.The average bulk modulus of elasticity of the liquid is 1.87 GPa

b. Coefficient of compressibility 0.5437 GPa-¹

c Velocity of sound 1.87 x 10^9P

a. Bulk modulus of elasticity

ß = - ∆p/(∆V/V)

First step  is to determine ∆p

∆p = (3039 kpa - 1520 kpa)x10³

∆p  = 1519 kPa

Second step is to determine ∆V

∆V = 1231 litres - 1232 litres

∆V = -1 m³

Now let determine the Bulk modulus of elasticity

Bulk modulus of elasticity= - 1519/(-1/1232)

Bulk modulus of elasticity= 1.87x10^6 kPa

Bulk modulus of elasticity= 1.87 GPa

b. The coefficient of compressibility

Coefficient of compressibility=β =1/K

Coefficient of compressibility=β =1/1.87

β =0.5437 GPa-¹

C. Velocity of sounds  in the medium with a density of 1593 kg/m3

V=√K/ρ

V=√1.87×10^9/ 1593

V=1083m/s

V = 1.87 x 10^9P

Inconclusion:

a.The average bulk modulus of elasticity of the liquid is 1.87 GPa

b. Coefficient of compressibility 0.5437 GPa-¹

c Velocity of sound 1.87 x 10^9P

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The questions in the pre-lab assignment focus on various aspects of the experiment, including physical separation techniques, calculation of sample mass, and determination of component percentages.

Step 1: The appropriate physical separation techniques to complete the flowchart could include methods such as filtration, evaporation, and magnetic separation, depending on the specific requirements of the experiment. These techniques are used to separate the different components of the mixture based on their physical properties.

Step 2: To calculate the mass of the sample, one typically weighs the entire mixture before performing any separation steps. This measurement provides the initial mass of the mixture, which serves as a reference for subsequent calculations.

Step 3: The mass of the rice can be determined by isolating the rice component through a suitable separation technique and weighing the collected rice. The difference between the initial mass of the mixture and the mass of the other components will represent the mass of the rice.

Step 4: Similarly, the mass of the sand can be calculated by separating the sand from the mixture and weighing it. The difference between the initial mass and the mass of the remaining components will correspond to the mass of the sand.

Step 5: To determine the mass of the salt, the salt component needs to be separated from the mixture and weighed. The difference between the initial mass and the masses of the other components will indicate the mass of the salt.

Step 6: The percentage of each component in the mixture can be calculated by dividing the mass of each component by the initial mass of the mixture and multiplying by 100. This calculation provides the proportion of each component relative to the total mass of the mixture.

By addressing these questions, students gain a better understanding of the experimental procedure, the use of physical separation techniques, and the importance of accurately calculating masses and percentages in the context of the experiment.

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The elastic properties of the brass specimen enables it to return to its

original length when stressed below the yield strength.

The correct responses are;

4. a. The final length is 90.0 m.

b. The final length is 97.2 mm.

Reasons:  

a. Diameter of the brass alloy = 7.5 mm

Length of the specimen = 90.0 mm

Force applied = 6000 N

The equation for the applied stress, σ, is presented as follows;

\(\sigma = \dfrac{Force \ applied}{Area \ of \ specimen} = \dfrac{6000 \, N}{\pi \cdot \left(\dfrac{7.5 \times 10^{-3}}{2} \, m} \right)^2 } \approx 135.81 \ \mathrm{MPA}\)

Depending on the cold working condition, 135.81 MPa is below the yield

strength, and the brass will return to its original condition when the force is

removed. The final length is remains as 90.0 m.

b. When the applied force is F = 16,500 N, we have;

\(\sigma = \dfrac{16,500\, N}{\pi \cdot \left(\dfrac{7.5 \times 10^{-3}}{2} \, m} \right)^2 } \approx 373.48\ \mathrm{MPA}\)

The stress found for the force of 16,500 N is above the yield stress of

brass, and it is therefore, in the plastic region.

From the stress strain curve, the strain can be estimated by drawing a line

from the point of the 373.48 MP on the stress strain curve, parallel to the

elastic region to intersect the strain axis, which gives a value of strain

approximately, ε = 0.08.

The length of the specimen is given by the formula; \(l_i = l_0 \cdot (1 + \epsilon)\)

Therefore;

\(l_i\) = 90 × (1 + 0.08) = 97.2

The final length of the specimen, \(l_i\) = 97.2 mm

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Answer:

Along the zero to measure an object on a ruler

The garbage collector can deallocate the Integer object with the value 28 since it no longer has any references pointing to it. Likewise replaced is the reference to the Integer object with the value 40.

However in Java, unlike C/C++, short int and long int are invalid data types. Moreover, Java does not distinguish between signed and unsigned in any way.

The five fundamental data types that are supported by the majority of modern computer languages are Integral, Floating Point, Character, Character String, and Composite types. A number of specific subtypes are also included in each general group.

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Answer:

Answer

have PPE on wash hands after

Explanation:

Answer:The magnetic field in a solenoid formula is given by, B = μoIN / L B = (1.26×10−6 × 15 × 360) / 0.8 B = 8.505 × 10−3 N/Amps m

Explanation:

The magnetic field generated by the solenoid is 8.505 × 10−4 N/Amps m.

A) Angle of departure The angle of departure of a root locus from a complex pole pair is given by: \($$\theta = 180^\circ - \phi_p - \phi_z$$\)$$\theta = 180^\circ - \phi_p - \phi_z$$

where θ is the angle of departure, φp is the phase angle at the pole, and φz is the phase angle at the zero. From the transfer function given, we can find the complex poles as follows:

$$(s+2)^2 + 1^2 = 0$$$$\

implies s = -2 \pm j$$At these poles, the phase angle is given by:

\($${\rm \phi_p} = tan^{-1}\left(\frac{-Im}{Re}\right) = tan^{-1}\left(\frac{-1}{-2}\right) = 26.6^\circ$$\)

The angle of departure is therefore:

\($$\theta = 180^\circ - 26.6^\circ - 0^\circ = 153.4^\circ$$B)\)

Break-in pointThe break-in point is the point where the root locus enters the real axis. It occurs when the real axis is a part of the locus and lies between two real poles or real zeros.

For a second-order system, the break-in point occurs when the angle of departure and the angle of arrival to the real axis are equal.In this case, we can see that there are no real zeros or real poles, so the locus cannot enter the real axis.

Therefore, there is no break-in point for this system.

\($$G(s) = \frac{K(s+1)}{s^2 + 4s + 5}$$a)\)The angle of departure of the root locus from the complex poles is \($$153.4^{\circ}$$b)\) The root locus does not enter the real axis, so there is no break-in point.

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