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fenix001 [56]
3 years ago
13

A concrete mix design calls for 6.5 sacks of cement, a water/cement ratio of 0.45, and an air content of 2.5%. 1. Complete the m

ix design usage chart below by filling in the shaded cells. Material Type Description Design Quantity Specific Gravity Volume (ft3 ) Cement Type II lb 3.15 Coarse Aggregate ¾" x #4 1,190 lb 2.75 Coarse Aggregate ⅜" x #8 lb 2.72 Fine Aggregate Concrete Sand 1,453 lb 2.66 Water Mixing Water gal 1.00 Air % by Volume 2.5% - Yield lb - 27.00 2. What is the unit weight of this mix in pcf?
Engineering
2 answers:
RUDIKE [14]3 years ago
7 0

Answer:

28.6 kg

Explanation:

The final weight can be calculated from the mixing data and formulae which is given as follows:

cement content = \frac{water content}{water - cement ratio}

Computing the parameters and checking the tables gives 28.6 kg.

4vir4ik [10]3 years ago
5 0

Answer:

The answer is 28.6 kg

Explanation:

The final weight can be calculated from the mixing data and formulae which is given as follows:

cement content = water content divided by (water minus cement ratio)

Therefore:

Cement content = water content/(water - cement ratio)

After all the parameters had been imputed and calculated, we will arrive at a value of 28.6 kg.

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A metal shear can be used to cut flat stock , round stock , channel iron and which of the following?
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3 years ago
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

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v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

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3 years ago
Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters
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4 0
3 years ago
Diodes can be used to protect electronic components from certain flowing in another Direction describe another application outsi
Anvisha [2.4K]

Answer:

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Explanation:

Diodes are like check valves, keeping current from flowing both ways.  Used to create d.c. out of a.c by rectification.  Also to block flow if d.c. power like a battery is hooked up in reverse polarity.

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