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3 years ago

Problem 2: Sieve Analysis and Soil Gradation

Engineering

1 answer:

Marina CMI [18]3 years ago

6 0

<u>Explanation:</u>

% Gravel = 100 - 72% = 28%

% Sand = 100 - 28 - 15 = 57%

% Fires = 100 - 0 - 28 - 57 = 15%

$\begin{aligned}&D_{n} \cdot 0.03 \quad ; \quad D_{x}=1.1 \quad ; \quad D_{0}=3.3\\&C_{u}=\frac{D_{60}}{D_{10}}=\frac{3.3}{0.03}=110\\&C_{c}=\frac{D_{30}^{2}}{D_{0} \cdot D_{0}}=\frac{1.1^{2}}{3.3 \times 0.03}=12.22\end{aligned}$

The soil is poorly graded soil.

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Vika [28.1K]

Answer:

Explanation:

its B

4 0

3 years ago

An automobile has a mass of 1200 kg. What is its kinetic energy, in ki, relative to the road when traveling at a velocity of 50

netineya [11]

Answer:

a)Ek=115759.26J

b)23.15kW

Explanation:

kinetic energy(Ek) is understood as that energy that has a body with mass when it travels with a certain speed, it is calculated by the following equation

$Ek=0.5mv^{2}$

for the first part

m=1200Kg

V=50km/h=13.89m/s

solving for Ek

$Ek=0.5(1200)(13.89)^2$

Ek=115759.26J

For the second part of the problem we calculate the kinetic energy, using the same formula, then in order to find the energy change we find the difference between the two, finally divide over time (15s) to find the power.

m=1200kg

V=100km/h=27.78m/s

$Ek=0.5(1200)(27.78)^2=462963J\\$

taking into account all of the above the following equation is inferred

ΔE= $\frac{Ek2-EK1}{T}=\frac{462963-115759.26}{15} =23146.916W=23.15kW$

3 0

3 years ago

A soil weighs 2,520 lbs/CY in its in situ condition, 1,970 lb/CY in its loose condition after excavation, and 3,025 lbs/CY in an

azamat

Answer:

load factor = 0.782

Shrink Factor = 0.833

no of truck is 62500

Explanation:

given data

soil weighs in situ condition = 2,520 lbs/CY

soil weighs in loose condition = 1,970 lb/CY

soil weighs in embanked state = 3,025 lbs/CY

average volume = 16 LCY

soil from a borrow pit = 1 million CCY

solution

first we get here Load Factor that is express as

load factor = $\frac{1,970}{2550}$

load factor = 0.782

and Shrink Factor will be as

Shrink Factor = $\frac{2520}{3025}$

Shrink Factor = 0.833

and

no of truck will be

no of truck = $\frac{1000000}{16}$

no of truck is 62500

6 0

4 years ago

A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 27 . It has b

Mamont248 [21]

Answer:

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

Explanation:

Given data;

Let,

critical stress required for initiating crack propagation Cc = 112MPa

plain strain fracture toughness = 27.0MPa

surface length of the crack = a

dimensionless parameter = Y.

Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m

Also for 6.2mm length of surface crack;

Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m

The dimensionless parameter

Cc = Kic/(Y*√pia*a)

Y = Kic/(Cc*√pia*a)

Y = 27/(112*√pia*4.4*10-³)

Y = 2.05

Now,

Cc = Kic/(Y*√pia*a)

Cc = 27/(2.05*√pia*3.1*10-³)

Cc = 135.78MPa

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

For more understanding, I have provided an attachment to the solution.

4 0

4 years ago

The explosion of a hydrogen bomb can be approximated by a fireball with a temperature of 7200 K, according to a report published

Iteru [2.4K]

Answer:

The rate of radiation of the energy is $E_r = 1.523747635*10^9 W/m^2$

The irradiation is $G =46.177\ kW/m^2$

The amount of energy absorbed is $E_B = 461.772 KJ$

The oak Tree would catch fire because the temperature of the blast(7200 K) is higher than the flammability limit (650 K) of the oak tree and secondly the thickness is very small

Explanation:

From the question we are told that

The temperature is $T = 7200K$