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igomit [66]
3 years ago
15

Explain the differences among sand, silt, and clay, both in their physical characteristics and their behavior in relation to bui

lding foundations.
Engineering
1 answer:
3241004551 [841]3 years ago
6 0

Answer:PHYSICALLY- sand is spherical in shape with large particles up to 0.18g

Silt also contains more of spherical particles with weights around 0.0029g.

Clay contain plate like particles,which are flat or layered. It's particles are generally lower that silt and sand below 0.0029g.

BEHAVIORS Sand particles are coarse and very porous,water can easily penetrate with no particles cohesion,does not retain water, difficult to expand,it is IDEAL FOR BUILDINGS.

Silt particles are also porous with some coarseness and no particles cohesion. retains water and can expand. NOT IDEAL FOR BUILDINGS.

Clay contain particles that are not coarse,they have high cohesiveness,they are not porous as it is difficult for water to penetrate. NOT IDEAL FOR BUILDINGS.

Explanation: Sand particles are coarse,very porous,contains spherical particles with large particles sizes and IDEAL FOR BUILDINGS

Silt particles are also porous, with some coarseness and with little or no particles cohesion, NOT IDEAL FOR BUILDINGS.

Clay soils are not porous water can not easily flow thought it,it is hard when dry and soft when wet. IT IS NOT A GOOD OPTION FOR BUILDINGS.

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A mass weighing 22 lb stretches a spring 4.5 in. The mass is also attached to a damper with Y coefficient . Determine the value
Dominik [7]

Answer:

Cc= 12.7 lb.sec/ft

Explanation:

Given that

m = 22 lb

g= 32 ft/s²

m = \dfrac{22}{32}=0.6875\ s^2/ft

x= 4.5 in

1 in = 0.083 ft

x= 0.375 ft

Spring constant ,K

K=\dfrac{m}{x}=\dfrac{22}{0.375}

K= 58.66  lb/ft

The damper coefficient for critically damped system

C_c=2\sqrt{mK}

C_c=2\sqrt{0.6875\times 58.66}

Cc= 12.7 lb.sec/ft

5 0
3 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in
Sonja [21]

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:  

σ = FL / πR^{3}

   = 3000 \times (40 \times 10^-3) / π (5 \times 10^-3)^3

σ = 305 \times 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

  = 2(305 \times 10^6) (15 \times 10^-3)^3 / 3(40 \times 10^-3)

F = 17156 N.  

7 0
3 years ago
Which basic principle influences how all HVACR systems work?
bezimeni [28]

Answer:

B) An increase in pressure can lower the boiling point of a liquid and change the temperature at which it turns to a gas.

Explanation:

B) An increase in pressure can lower the boiling point of a liquid and change the temperature at which it turns to a gas.

6 0
3 years ago
Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u
Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × 10^{-2} N/m

so average radius = \frac{diameter}{2} =  100 µm = 100 ×10^{-6} m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r²    .....................1

here r is radius and  Δp pressure difference and σ surface tension

Δp = \frac{2 \sigma }{r}    

put here value

Δp = \frac{2*2.69*10^{-2}}{100*10^{-6}}  

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

7 0
3 years ago
Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of
Alexxandr [17]

Answer:

P_m_i_n = 442KPA

Explanation:

We are given:

m = 1.06Kg

T_H = 1.2T_L

T = 22kj

Therefore we need to find coefficient performance or the cycle

COP_R = \frac {1}{(T_R/T_l) -1}

= \frac {1 }{1.2-1}

= 5

For the amount of heat absorbed:

Q_l = COP_R Wm

= 5 × 22 = 110KJ

For the amount of heat rejected:

Q_H = Q_L + W_m

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= = \frac{132}{1.06}

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

T_L = \frac{T_H}{1.2};

T_L = \frac{342.5}{1.2}

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at T_L = 12.4°c we have 442KPa

6 0
3 years ago
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