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Mekhanik [1.2K]
3 years ago
6

The lab technician you recently hired tells you the following: Boss, an undisturbed sample of saturated clayey soil was brought

to me from the Mission Valley site. I measured the mass of the sample to be 600 grams. I then measured the mass of the sample after placing it in the oven for 24 hrs. I found this mass to be 200 grams. Can you help me determine the water content?" Determine the water content.
Engineering
1 answer:
harkovskaia [24]3 years ago
6 0

Answer:

The water of the saturated clayed soil is 66.67 %.

Explanation:

Given;

mass of saturated clayed soil, M_s = 600 g

mass of dry soil sample, M_d = 200 g

mass of water content, M_w = M_s - M_d = 600 g - 200 g = 400 g

The water content is determined as;

M_w(\%)  = \frac{M_s - M_d}{M_s} *100\%\\\\M_w(\%)  = \frac{600-200}{600} *100 \% \\\\M_w(\%)  = 66.67 \%

Therefore, the water of the saturated clayed soil is 66.67 %.

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An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi
Fittoniya [83]

Answer:

a.) -147V

b.) -120V

c.) 51V

Explanation:

a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.

c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.

Honestly, these things take practice to get used to. It's really hard to explain this.

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3 years ago
2.) A fluid moves in a steady manner between two sections in a flow
Talja [164]

Answer:

250\ \text{lbm/min}

625\ \text{ft/min}

Explanation:

A_1 = Area of section 1 = 10\ \text{ft}^2

V_1 = Velocity of water at section 1 = 100 ft/min

v_1 = Specific volume at section 1 = 4\ \text{ft}^3/\text{lbm}

\rho = Density of fluid = 0.2\ \text{lb/ft}^3

A_2 = Area of section 2 = 2\ \text{ft}^2

Mass flow rate is given by

m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}

The mass flow rate through the pipe is 250\ \text{lbm/min}

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}

The speed at section 2 is 625\ \text{ft/min}.

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3 years ago
Determine whether or not each of the following four transaction execution histories is serializable. If a history is serializabl
ludmilkaskok [199]

Answer:

Option D. w1[x] w2[u] w2[y] w1[y] w3[x] w3[u] w1[z]

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The execution in the option D is correct. This is because there is more than one reasonable criterion.

8 0
2 years ago
The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam
lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

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The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

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The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy

T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

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SG > SGw * 4/3* sin(30) * (cos(30))^2

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For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0
2 years ago
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Answer:

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From the diagram of P-v And T-s we can understand so easily.

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