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2 years ago

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The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:

1 answer:

Serggg [28]2 years ago

7 0

Answer:

a) 2622.903 N/m^3

b) 1.38233

c)4.878811765

Explanation:

Find the void ratio using the formula:

$y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1$

Here;

$G_{s}$ is specific gravity of soil solids

$y_{w}$ is unit weight of water = 998 kg/m^3

$w$ is the moisture content = 0.17

$e$ is the void ratio

$y$ is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

$w*G_{s} = S*e ..... Eq2$

S is saturation rate

Substitute Eq 2 into Eq 1

$y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w} }{1+e}$

$14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233$

Specific gravity of soil solids

$G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765$

Saturated Unit Weight

$y_{s} = \frac{(G_{s} + e)*y_{w} }{1+e} \\=\frac{(4.878811765 + 1.38233)*998 }{1+1.38233}\\\\= 2622.902571 N/m^3$

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Answer:

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Explanation:

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2 years ago

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2 years ago

1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by

zvonat [6]

Answer:

299.36 feet

Explanation:

$To \ find \ the \ distance \ of \ the \ ball \ from \ the \ home \ plate. \\ \\ From \ the \ given \ information:$

$Height \ h = 4 \ ft$

$Initial \ speed \ V_o = 98 \ ft/s ec$

$The \ angle \ \theta = 45^0$

$Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s$

$U_x = V_o \ cos 45 = \dfrac{98}{\sqrt{2}}$

$U_y = V_o \ sin 45 = \dfrac{98}{\sqrt{2}}$

So;

$S_y = u_y t - \dfrac{1}{2}gt^2$

$-1 =\dfrac{98}{\sqrt{2}}t - \dfrac{1}{2}*32*1.85t^2$

By solving:

$t_1 = 4.32 \ sec$

Thus;

$horizontal \ distance = U_x t$

$= \dfrac{98}{\sqrt{2}}\times 4.32$

$\mathbf{=299.36 \ feet}$

$\mathbf{Thus \ , the \ distance \ from \ the \ home \ plate \ = \ 299.36 \ feet}$

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2 years ago

2. Determine the surface area of a primary settling tank sized to handle a maximum hourly flow of 0.570 m3/s at an overflow rate

Hitman42 [59]

Answer:

The surface area of the primary settling tank is 0.0095 m^2.

The effective theoretical detention time is 0.05 s.

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The surface area of the tank is calculated by dividing the volumetric flow rate by the overflow rate.

Volumetric flow rate = 0.570 m^3/s

Overflow rate = 60 m/s

Surface area = 0.570 m^3/s ÷ 60 m/s = 0.0095 m^2

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Volume of the tank = surface area × depth = 0.0095 m^2 × 3 m = 0.0285 m^3

Detention time = 0.0285 m^3 ÷ 0.570 m^3/s = 0.05 s

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