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3 years ago

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A fill covering a wide area is to be placed at the surface of this profile. The fill has a total unit weight of 20 kN/m^3 and is

3 m thick. Assume that the data for the sample at 7.0 m are representative of the entire clay profile. Also assume that the clay is heavily over consolidated and that the danse sands at the surface of the profile are so stiff that they do not contribute to the settlement. Find the settlement of the surface due to compression of the clay layer

1 answer:

goblinko [34]3 years ago

8 0

Answer:

hello your question lacks some information attached below is the complete question with the required information

answer : 81.63 mm

Explanation:

settlement of the surface due to compression of the clay ( new consolidated )

= 81.63 mm

attached below is a detailed solution to the given problem

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The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing

Viktor [21]

Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that $\mu = 0.8$

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

$P(X \leq 4) + P(X > 4) = 1$

We want P(X > 4). Then

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

$P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595$

$P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438$

$P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383$

$P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014$

0.14% probability of observing more than 4 errors in the carpet

5 0

3 years ago

A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container

kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

$delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}$

In an isobaric process heat (Q) is defined as:

$Q= m*Cp*dT$

Replacing in the equation for entropy

$delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}$

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:

$delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}$

Solving the integral we get the expression to estimate the entropy change in the system

$delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})$

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is $0.85\frac{kJ}{Kg*K}$

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

$Q= m*Cp*dT$

With $dt=T_{2}-T_{1}$ clearing for T2 we get:

$T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K$

Now we can estimate the entropy change in the system

$delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}$

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.

b) see picture.

3 0

3 years ago

Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 240°C. The water is slowly heated at constan

Digiron [165]

Answer:

The final temperature of water is 381.39 °C.

Explanation:

Given that

Mass of water = 5 kg

Heat transfer at constant pressure Q = 2960 KJ

Initial temperature = 240 °C

We know that heat transfer at constant pressure given as follows

$Q=mC_p\Delta T$

We know that for water

$C_p=4.187\ \frac{KJ}{kg.K}$

Lets take final temperature of water is T

So

$Q=mC_p\Delta T$

$2960=5\times 4.187(T-240)$

T=381.39 °C

So the final temperature of water is 381.39 °C.

7 0

3 years ago

A batch of 1000 is split into 10 smaller batches of equal size 100. The processing time of each unit is 2

Vika [28.1K]

The lead time of the actual batch will be in

2950 in minutes

<h3>What is Processing Time?</h3>

This refers to the amount of time which is taken for a processor to run a procedure and return a result.

We can see that a batch of 1000 is split so that they each have 10 smaller batches which has an equal size of 100 each, then if the processing time is 2 mins per machine and the set up time is 30 mins.

Hence, when this batch is processed over a serial line of 5 machines, then the lead time of the actual batch would be 2950 in minutes

Read more about processing time here:

brainly.com/question/18444145

4 0

2 years ago

What is the measurement below?

Bess [88]

Explanation:

इसिसिसिसैस्स्स्स्स्स्स्स्स्स्सूस्सोस्स्स्स्स्स

8 0

3 years ago