Given the specific gravity of soil solids 2.75, calculate the zero-air-void unit weight in kN/m
3
for a soil with moisture content 5%,8%,10%,12% and 15%. For this soil determine the variation of dry density (kg/m
3
) at 10% and 15% for degree of saturation at 80%,90% and 100%

Answer:

are you wht

didn't understand the question

Answer:

Yes/No

Explanation:

NANI?!

Answer:

Explanation:

In the steady state total volume going out will be equal to total volume coming in , and

total salt coming in = total salt going out

Total volume coming in per minute = 2000 + 2 x 10³ x 60

= 122000 L .

Total salt coming in per minute = 1200 x 2000 + 20  x 2 x 10³ x 60

= 2400000 + 2400000 mg

= 4800000 mg

volume of water going out per minute = 122000 L

Total salt going out  per minute = 4800000 mg

concentration of water going out = 4800000 / 122000

= 39.344 mg / L

Pearson Math Lab is an online platform developed by Pearson Education, a leading educational publisher. The platform is designed to help students learn and practice math concepts at their own pace and in a personalized way.

It offers interactive learning tools, assessments, and homework resources for K-12 students, as well as college and university students.

Pearson Math Lab provides access to a range of digital resources, including eTextbooks, video lectures, animations, and online platform. These resources are designed to help students build foundational knowledge and practice problem-solving skills in math. The platform also provides immediate feedback and personalized recommendations based on students' performance, helping them to identify areas where they need more practice and support.

In addition to providing resources for students, Pearson Math Lab also includes features for instructors and educators. These features allow instructors to track student progress, monitor their performance, and customize learning paths to meet the needs of individual students. The platform also integrates with learning management systems (LMS) commonly used in educational institutions, making it easy for instructors to integrate Pearson Math Lab into their teaching and assessment practices.

Overall, Pearson Math Lab is a comprehensive platform that provides students with the tools and resources they need to learn and practice math in a flexible and personalized way.

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When a taxpayer sells their personal property, like a house, for less than what they paid for it, it is known as a capital loss. Herman sold his personal home at a capital loss. The IRS does not allow taxpayers to take a tax deduction for the sale of personal property, such as a primary residence.

The IRS does not allow taxpayers to take a tax deduction for the sale of personal property, such as a primary residence. As a result, Herman cannot use the loss as a tax deduction. However, there are a few exceptions to this rule: If Herman used the home for business purposes and it was a part of a business asset, he could be able to use the loss as a tax deduction.

If the home was converted to a rental property before it was sold, Herman may be able to use the loss as a tax deduction. If Herman is moving and the sale of the house qualifies as a qualified moving expense, he may be able to use the loss as a tax deduction.

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Answer Designed by Elijah E. Myers, it was constructed in the 1890s from Colorado white granite, and opened for use in November 1894. The distinctive gold dome consists of real gold leaf, first added in 1908, commemorating the Colorado Gold Rush.

Location: 200 East Colfax Avenue; Denver, Co...

Part of: Denver Civic Center (ID12001017)

Built: 1886–1901

Added to NRHP: October 16, 2012

Explanation:

Kerberos consists of two services: the Authentication Service (AS) and the Ticket-Granting Service (TGS).

The Authentication Service (AS) is responsible for verifying the identity of users or clients requesting access to a network or system.

When a user wants to authenticate, they send a request to the AS, typically by providing their username or principal. The AS responds by issuing a Ticket Granting Ticket (TGT) that contains a session key encrypted with the user's password or other authentication credentials. The TGT serves as proof of successful authentication and is used to request service tickets from the TGS. The Ticket-Granting Service (TGS) is responsible for granting service tickets to authenticated users. To obtain a service ticket, the user presents their TGT to the TGS, along with the requested service identifier. The TGS verifies the TGT, generates a Service Ticket (ST), and returns it to the user. The ST contains a session key encrypted with the service's secret key and can be used by the user to authenticate themselves to the requested service.

In summary, the Authentication Service (AS) verifies user identities and issues Ticket Granting Tickets (TGTs), while the Ticket-Granting Service (TGS) grants Service Tickets (STs) for accessing specific services. Together, these two services form the core components of the Kerberos authentication protocol, enabling secure authentication and access control within a network or system.

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Answer:

I think it is false!

Explanation:

Answer: I think it's true

Explanation:

Because if you were part of a play, you would have a part but if you work on props, you don't have a part onstage.

Answer:

The landscape architect uses knowledge of legal requirements to inform design choices.

Explanation:

The landscape architect doesn’t only draw on knowledge of law and government to design grounds for public buildings, but will draw on that knowledge quite often on many projects. The landscape architect’s knowledge of design is not purely intuitive but is a product of study and might involve research. The landscape architect’s knowledge of law and government, however, does come into play in making design choices that are consistent with the law and government regulations.

The statement "For the typical design of a municipal piping system, both the prandtl and reynolds numbers are usually very critical." is false.

This is False.

While both Prandtl and Reynolds numbers are important parameters in fluid mechanics, the criticality of their values in municipal piping system design varies depending on the application.

The Prandtl number represents the ratio of momentum diffusivity to thermal diffusivity. It is used to determine the thermal boundary layer thickness and heat transfer characteristics in fluids. In municipal piping systems, the Prandtl number may not be as critical as other parameters such as flow rate and pressure drop.

The Reynolds number, on the other hand, represents the ratio of inertial forces to viscous forces. It is used to predict the onset of turbulent flow and to determine pressure drop and pumping requirements. In municipal piping systems, the Reynolds number is an important parameter in designing for optimal flow rates and minimizing energy consumption.

In conclusion, while both Prandtl and Reynolds numbers are important in fluid mechanics, their criticality depends on the specific application and design requirements of municipal piping systems.

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Because of their fascinating properties, such as low elastic modulus, high stiffness-to-weight ratio, low coefficient of thermal expansion, and large specific surface area, lattice structures, also known as architected cellular structures, have been used in a variety of industrial industries.

Lattice structures can be employed in biomedical applications to lower the stiffness of metallic medical implants to that of bone, minimizing stress shielding while permitting fluid flow due to their porosity with a significant surface area-to-volume ratio to assist osseointegration.

The capacity of these structures to provide impact protection, a large surface area, and shock absorptions promotes their appeal. Lattice-based products can overcome standard production limits and help build high-performing items at a lower cost.

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Full Question:

Additive manufacturing of metallic lattice structures: unconstrained design, accurate fabrication, fascinated performances, and challenges.

Describe Lattice Structures.

To determine the vertical displacement of joint A using the method of virtual work, we need to first find the forces in each member. We can do this by using the method of sections and analyzing the left and right sides separately.

Starting with the left side, we can sum the vertical forces to get:
3k - F1*sin(45) - F2*sin(45) = 0
where F1 and F2 are the forces in members AB and AC, respectively.

Similarly, summing the moments about point A gives:
-8*F1*sin(45) - 12*F2*sin(45) = 0

Solving these two equations, we get F1 = 4.5 kips and F2 = 3 kips.

Now we can use the method of virtual work to find the displacement of joint A. We will apply a virtual load of 1 kip in the upward direction at joint A and find the corresponding change in potential energy of the structure.

The virtual work done by this load is:
Wv = 1*(3/12) = 0.25 in-kips

The change in potential energy is given by:
ΔU = F1*δ1 + F2*δ2
where δ1 and δ2 are the displacements in members AB and AC, respectively.

Using the principle of virtual work, we can equate the virtual work done to the change in potential energy:
Wv = ΔU

Substituting in the values we found earlier, we get:
0.25 = 4.5*δ1 + 3*δ2

We also know that the total vertical displacement at joint A is given by:
δA = δ1 + δ2

To solve for δA, we can use the fact that the displacement of any point in a structure is equal to the partial derivative of the total potential energy with respect to the force at that point. This is known as Castigliano's theorem.

Taking the partial derivative of ΔU with respect to F1, we get:
δ1 = -∂ΔU/∂F1 = -4.5*(12/29)/sin(45) = -1.33 in

Taking the partial derivative of ΔU with respect to F2, we get:
δ2 = -∂ΔU/∂F2 = -3*(12/29)/sin(45) = -0.89 in

Substituting these values into the equation for δA, we get:
δA = -1.33 - 0.89 = -2.22 in

Therefore, the vertical displacement of joint A is -2.22 in.

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To ensure that a Zener diode remains in the reverse breakdown region and maintains a constant voltage drop, the Zener current (IZ) must be greater than the Zener knee current (IZK) specified in its datasheet. However, I'm unsure about the relevance of "rl" in your question. Could you please provide more context or clarify what "rl" refers to?

Answer:

Explanation:

Complements are used in digital circuits, because it is faster to subtract by adding complements than by performing true subtraction. The binary complement of a number is created by reversing all bits and adding 1. The carry from the high-order position is eliminated.

Answer:

Given :- the volume of the pyramid is 36 cubic cm , find the volume of the prism on same base and same height as pyramid .

Answer :-

we know that,

Volume of pyramid = (1/3) * Base area * height .

Volume of prism = Base area * height .

so,

→ Volume of pyramid = 36 cm³

→ (1/3) * Base area * height = 36

→ Base area * height = 36 * 3

→ Base area * height = 108 cm³.

then,

→ Volume of prism = Base area * height .

→ Volume of prism = 108 cm³ (Ans.)

Explanation:

The passage argues that computer games are not as harmful as portrayed by the media and that their benefits are comparable to classical art or music training, challenging popular notions

The passage argues that computer games are not as harmful as portrayed by the media. It acknowledges that modern computer games may contain more sexual and violent content, but it contends that the complexity and aesthetics of electronic graphics provide a similar benefit to what classical art or music training provided for past generations.

The passage further implies that the benefits of classical art or music training are widely accepted and not questioned. Therefore, it suggests that the benefits of electronic graphics and computer games are also worth considering.

Overall, the passage seems to suggest that computer games, despite their controversial content, can provide benefits that are comparable to classical art or music training. It challenges the popular notion that computer games are harmful and suggests that their potential benefits should be acknowledged.

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Answer:

Following are the solution to the given question:

Explanation:

Line voltage:

\(V_L=\sqrt{3}V_{ph}=\sqrt{3}(120) \ v\)

Power supplied to the load:

\(P_{L}=\sqrt{3}V_{L}I_{L} \cos \phi\)

\(10\times 10^3=\sqrt{3}(120 \sqrt{3}) I_{L}\ (0.85)\\\\I_{L}= 32.68\ A\)

Check wye-connection, for the phase current:

\(I_{ph}=I_L= 32.68\ A\)

Therefore,

Phasor currents: \(32.68 \angle 0^{\circ} \ A \ ,\ 32.68 \angle 120^{\circ} \ A\ ,\ and\ 32.68 -\angle 120^{\circ} \ A\)  

Magnitude of the per-phase load impedance:

\(Z_{ph}=\frac{V_{ph}}{I_{ph}}=\frac{120}{32.68}=3.672 \ \Omega\)

Phase angle:

\(\phi = \cos^{-1} \ (0.85) =31.79^{\circ}\)

Please find the phasor diagram in the attached file.

Answer:

21/2 we get remainder 1

10/2 we get remainder 0

5/2 we get remainder 1

2/2 we get remainder 0

1. Now take the 1

The answer to the question is

(21)10=(10101)2

Answer:

SECTION LEARNING OBJECTIVES

By the end of this section, you will be able to do the following:

Distinguish between static friction and kinetic friction

Solve problems involving inclined planes

Section Key Terms

kinetic friction static friction

Static Friction and Kinetic Friction

Recall from the previous chapter that friction is a force that opposes motion, and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice.

There are different types of friction—kinetic and static. Kinetic friction acts on an object in motion, while static friction acts on an object or system at rest. The maximum static friction is usually greater than the kinetic friction between the objects.

Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going.

Figure 5.33 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them.

This function takes a sorted StaticArray `inputArray` as a parameter and returns a new StaticArray `resultArray` with duplicate values removed. It initializes the first element of `resultArray` with the first element of `inputArray` and then iterates through the remaining elements of `inputArray`. If an element is different from the previous element, it is added to `resultArray`. The resulting array is then resized to contain only the unique elements. The original `inputArray` is not modified.

Here's an example implementation of a function that removes duplicates from a sorted StaticArray:

```cpp

template<typename T, size_t N>

StaticArray<T, N> removeDuplicates(const StaticArray<T, N>& inputArray) {

   StaticArray<T, N> resultArray;

   // Initialize the first element of the result array

   resultArray[0] = inputArray[0];

   size_t uniqueIndex = 1;

   // Iterate through the input array starting from the second element

   for (size_t i = 1; i < N; i++) {

       // Check if the current element is different from the previous element

       if (inputArray[i] != inputArray[i - 1]) {

           resultArray[uniqueIndex] = inputArray[i];

           uniqueIndex++;

       }

   }

   // Resize the result array to the actual number of unique elements

   resultArray.resize(uniqueIndex);

   return resultArray;

}

```

Note: This implementation assumes that the `StaticArray` class provides the `operator[]` for accessing elements and the `resize` method for resizing the array. You may need to adapt the code depending on the specific implementation of your `StaticArray` class.

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a. The input-output equation for the given SSR is:
y = [10]x
Where y is the output and x is the input.
b. The transfer function for this system can be obtained by taking the Laplace transform of the input-output equation:
Y(s) = [10]X(s)
Dividing both sides by X(s), we get:
G(s) = Y(s)/X(s) = 10
Therefore, the transfer function of the system is G(s) = 10.


SSR stands for Solid State Relay which is an electronic device used for switching a load on or off in response to a control signal. In this case, the given SSR has an output response (y) that is directly proportional to the input control signal (x) with a gain of 10. The input-output equation represents the relationship between the input and output of the system, while the transfer function gives a mathematical representation of the system's behavior in the Laplace domain. In this case, the transfer function is a constant value of 10, indicating that the output of the system is always ten times the input signal.

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The following are consequences of burning coal for energy increased levels of CO2 in the atmosphere, increased heavy metals (lead and mercury) released into the air and Acidified rain. The correct options are a, b and c.

Burning coal produces a number of main emissions: Sulphur dioxide (SO2), which causes respiratory conditions and acid rain, Nitrogen oxides (NOx), which cause respiratory diseases and smog, Particulates that cause lung sickness, respiratory diseases, fog, and haze, The main greenhouse gas produced by burning fossil fuels (coal, oil, and natural gas) is carbon dioxide (CO2).

Mercury and other heavy metals have been associated with developmental and neurological harm in both humans and other animals. Power stations produce bottom ash and fly ash as byproducts of burning coal.

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(a) Confidence Interval ≈ (11.72, 12.08). (b) Confidence Interval ≈ (11.64, 12.16) and (c) The confidence level of the interval is at least 95%.

To find the confidence intervals for the mean sugar content, we can use the formula:

Confidence Interval = Sample Mean ± (Critical Value * Standard Error)

where the critical value is based on the desired confidence level and the standard error is calculated as the standard deviation divided by the square root of the sample size.

a. 95% confidence interval:

The critical value for a 95% confidence level is approximately 1.96.

Standard Error = 1.1 / sqrt(140) ≈ 0.093

Confidence Interval = 11.9 ± (1.96 * 0.093)

Confidence Interval ≈ (11.72, 12.08)

b. 99% confidence interval:

The critical value for a 99% confidence level is approximately 2.58.

Standard Error = 1.1 / sqrt(140) ≈ 0.093

Confidence Interval = 11.9 ± (2.58 * 0.093)

Confidence Interval ≈ (11.64, 12.16)

c. The given interval is (11.81, 11.99). This interval falls entirely within the 95% confidence interval calculated in part a. Therefore, the confidence level of the interval is at least 95%.

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(a) To calculate the distance above the hiker when the rock becomes visible, we can use the equation of motion:

d = ut + (1/2)gt^2

where d is the vertical distance, u is the initial vertical velocity (which is zero as the rock breaks loose), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Given that the height of the cliff is 113 m and the time is 1.48 s:

113 = 0 + (1/2) * 9.8 * (1.48)^2

Solving this equation, we find:

d ≈ 10.9 m

Therefore, the rock is approximately 10.9 meters above the hiker when he can see it.

(b) To calculate the time the hiker has to move before the rock hits his head, we can use the equation:

d = ut + (1/2)gt^2

where d is the vertical distance, u is the initial vertical velocity (which is zero), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Given that the total vertical distance is 113 m:

113 = 0 + (1/2) * 9.8 * t^2

Solving this equation, we find:

t ≈ 5.38 s

Therefore, the hiker has approximately 5.38 seconds to move before the rock hits his head.

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Answer:

540 W

Explanation:

Answer:

30that is the answer . . . . . . . . . .. . . . .

The gage pressure of a water column 30 m high is 294 kPa.

The gage pressure of a water column is directly proportional to its height and the density of water. In this case, the density of water is assumed to be 1000 kg/m3.

To find the gage pressure of a water column 30 m high, we can use the following equation:

P = ρ * g * h

Where:

P is the gage pressure,

ρ is the density of water,

g is the acceleration due to gravity, and

h is the height of the water column.

First, let's calculate the gage pressure of the water column 10 m high using the given information:

P1 = ρ * g * h1

= 1000 kg/m3 * 9.8 m/s2 * 10 m

= 98,000 N/m2

Since 1 N/m² is equal to 1 Pascal (Pa), the gage pressure of the water column 10 m high is 98,000 Pa or 98 kPa.

Now, let's find the gage pressure of the water column 30 m high using the same formula:

P2 = ρ * g * h2

= 1000 kg/m3 * 9.8 m/s2 * 30 m

= 294,000 N/m2

Converting the gage pressure to kPa:

P2 = 294,000 N/m2

= 294 kPa

Therefore, the gage pressure of a water column 30 m high is 294 kPa.

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Answer: Ethanol

Explanation:

The correct SI Form of the following combinations of Units are given as follows:

A) kN/μs = GN/s

B) Mg/mN; = Gg/N

C) MN/(kg.ms)  = GN/(kg.s)

The International System of Units, abbreviated SI in all languages and often pleonastically as the SI system, is the current version of the metric system and the world's most extensively used measuring system.

The System of Units, often known as the metric system, is frequently shortened as SI, which originates from the original French word, Système international d'unités.

A) kN/μs  = (10) ³N/ (10) ⁻6s

= (10)⁹ N/s

= GN/s

B) Mg/mN = (10⁶)g/10⁻³/N

= Gg/N

C) MN/ (kg.ms) = 10⁶N/kg * (10⁻³)s

= 10⁹ (N/kg · s)

= GN/Kg · S)

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Full Question:

Represent Each Of The Following Combinations Of Units In The Correct Si Form Using An Appropriate Prefix:

A) kN/μs

B) Mg/mN; and

MN/(kg.ms)

The received power if the effective aperture area of the receiving antenna is 20 m2 is 177.77 m2.

In physics, power is referred to as the rate of energy conversion or transfer over time. The unit of power in the SI system, often known as the International System of Units, is the Watt (W). A single joule per second is one watt.

Power was formerly referred to as activity in some research. A scalar quantity is power. As power is always a function of labor done, it follows that if a person's output varies during the day depending on the time of day, so will his power.

A measure of the pace at which energy is transferred, power is a physical quantity. As a result, it can be described as the pace of job completion relative to time.

Therefore, The received power if the effective aperture area of the receiving antenna is 20 m2 is 177.77 m2.

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The correct answer to this open question is the following.

The text structure of an article that discusses pharaohs and gives examples and explains how they look is a description.

The text structure called description allows the reader to fully know the characteristics of the people it is referring to, including some important details. That is why the author of a description text adds words like "such as" and "for example."

When describing something, the write is giving structure to the text and sequence. What comes first., what is followed, and so on.

That is why The text structure of an article that discusses pharaohs and gives examples and explains how they look is a description. It includes cause and effect sentences, and some comparisons in order to contrast an idea.

28.9% of the incident power is reflected back from the load.

(a) To indicate the position on the Smith chart with an "A", follow the steps mentioned below:

Step 1: Normalize the load impedance, zL

Step 2: Locate the normalized load impedance on the Smith Chart.

Step 3: Mark the position on the Smith Chart as "A".

Given, Transmission line is terminated in a normalized load impedance of ZLN = 2.0 - j(1.5).

To normalize the load impedance, we can use the following formula;zL = ZL/Z0

Where Z0 is the characteristic impedance of the transmission line.

zL = (2.0 - j(1.5))/1 = 2.0 - j1.5Locate this normalized impedance on the Smith Chart and mark it with "A". The figure of the Smith chart is given below:

Figure: Smith ChartWe have marked the position "A" on the Smith Chart.

Now, to find the normalized load admittance (yL), follow the steps mentioned below:

Step 1: Find the conjugate of the normalized load impedance, zL*.

Step 2: Use the following formula to find the admittance;yL = 1/zL*Where zL* is the conjugate of the normalized load impedance.

Given zL = 2.0 - j1.5, then;zL* = 2.0 + j1.5yL = 1/zL* = 0.32 + j0.24

Therefore, the normalized load admittance is yL = 0.32 + j0.24. We mark it as "B" on the Smith chart

.(b) To find the reflection coefficient at the load (both magnitude and phase), follow the steps mentioned below:

Step 1: Draw a line from the normalized load impedance (point A) to the center of the Smith Chart.

Step 2: Determine the intersection of this line with the unity circle.

Step 3: Draw a line from the center of the Smith Chart to the intersection of the line from step 2.

Step 4: The reflection coefficient at the load is the point where the line from step 1 intersects the line from step 3.

The figure of the Smith chart is given below:

Figure: Smith ChartWe have marked the normalized load impedance (point A) and the normalized load admittance (point B) on the Smith Chart. The line from point A intersects the unity circle at point C. The line from the center of the Smith Chart intersects point C at point D.

Therefore, the reflection coefficient at the load is point D. The magnitude and phase of the reflection coefficient are indicated on the Smith Chart as 0.53 30 °.

The percentage of incident power that is reflected back from the load is given by;ρL = |ΓL|^2Where ΓL is the reflection coefficient at the load.Then,ρL = (0.53)^2 = 0.28

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