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Naddika [18.5K]

3 years ago

15

A shrinkage limit test is performed on a soil. The initial mass and volume of the soil are: V1=20.2cm^3 , while the final mass a

nd volume are M2=24g and V2=14.3cm^3 . Note that in the initial state the soil is saturated, whereas in the final state the soil is completely dry.
Calculate:
a. the shrinkage limit SL of the soil.
b. the void ratio at the SL.
c. Gs of the soil solids.
d. the initial void ratio.

1 answer:

love history [14]3 years ago

3 0

C. Gs of the soil solids

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Answer:

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Explanation:

We are given that

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Initial speed, u=110 km/h= $110\times \frac{5}{18}=30.56 m/s$

Where $1 km/h=5/18 m/s$

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$g=9.81 m/s^2$

Final speed, v=0

Change in kinetic energy, $\Delta K.E=\frac{1}{2}m(v^2-u^2)$

$\Delta K.E=\frac{1}{2}(1100)(0-(30.56)^2)=-513652.48 J$

$\Delta K.E=-\frac{513652.48}{1000}=-513.652 KJ$

Where 1 KJ=1000 J

Change in potential energy, $\Delta P.E=mgh(h_2-h_1)$

Initially height, h1=0

Using the formula

$\Delta P.E=1100\times 9.81(40-0)$

$\Delta P.E=431640J$

$\Delta P.E=431.64KJ$

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3 years ago

Assume a steel pipe of inner radius r1= 20 mm and outer radius r2= 25 mm, which is exposed to natural convection at h = 50 W/m2.

Mekhanik [1.2K]

Answer:

98,614.82 W/m²

Explanation:

$Q = 2\pi hL(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})$

Where;

Q = the amount of heat loss from the pipe

h = the heat transfer coefficient of the pipe = 50 W/m².K

T₁ = the ambient temperature of the pipe = 30⁰C

T₂ = the outside temperature of the pipe = 100⁰C

L= the length of pipe

r₁ = inner radius of the pipe = 20mm

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To determine the amount of heat loss from the pipe per unit length

From the equation above

$\frac{Q}{L} = 2\pi h(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})$

$\frac{Q}{L} = 2\pi* 50(\frac{100-30}{Ln\frac{25}{20}})$

$\frac{Q}{L} = 314.159(\frac{70}{0.223})$

$\frac{Q}{L} = 314.159(313.901)$ = 98,614.82 W/m²

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4 years ago

The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface

slamgirl [31]

Answer:

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

Explanation:

Given that

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So now putting the values

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We know that

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q=39.15 W/m²

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