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2 years ago

Why might construction crews want to install pipes before the foundation is poured?

1 answer:

6 0

I’m a concrete mason myself and I can tell you it is a pain in the butt to Roto hammer a hole into the concrete to put the pipe in it’s a lot easier to just pour the concrete around it

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In particular, a system may or may not be (1) Memoryless, (2) Time invariant, (3)Linear, (4) Casual, (5) Stable.

egoroff_w [7]

Answer:

a. True

Explanation:

A system may be sometimes casual, time invariant, memoryless, stable and linear in particular.

Thus the answer is true.

A system is casual when the output of the system at any time depends on the input only at the present time and in the past.

A system is said to be memoryless when the output for each of the independent variable at some given time is fully dependent on the input only at that particular time.

A system is linear when it satisfies the additivity and the homogeneity properties.

A system is called time invariant when the time shift in the output signal will result in the identical time shift of the output signal.

Thus a system can be time invariant, memoryless, linear, casual and stable.

4 0

2 years ago

True or False; The Neutrons in an atom have a neutral charge.

yanalaym [24]

Answer:

true

Explanation:

if it is not true it is false

3 0

3 years ago

Read 2 more answers

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 19.636

luda_lava [24]

Answer:

The original length of the specimen is found to be 76.093 mm.

Explanation:

From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:

Initial Volume = Final Volume

πd1²L1/4 = πd2²L2/4

d1²L1 = d2²L2

L1 = d2²L2/d1²

where,

d1 = initial diameter = 19.636 mm

d2 = final diameter = 19.661 mm

L1 = Initial Length = Original Length = ?

L2 = Final Length = 75.9 mm

Therefore, using values:

L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²

5 0

3 years ago

Water flows in a pipeline. At a point in the line where the diameter is 7 in., the velocity is 12 fps and the pressure is 50 psi

PolarNik [594]

Answer:

a) P₂ = 3219.11 lbf / ft² , b) P₂ = 721.91 lbf / ft² , c) P₂ = 5707.31 lbf / ft²

Explanation:

For this exercise we can use the fluid mechanics equations, let's start with the continuity equation, index 1 is for the starting point and index 2 for the end point of the reduction

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

The area of a circle is

A = π r² = π/4 d²

v₂ = v₁ (d₁ / d₂)²

Let's calculate

v₂ = 12 (7/3)²

v₂ = 65 feet / s

Now let's use Bernoulli's equation

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

P₁ - P₂ = ρ g (y₂ –y₁) + ½ ρ (v₂² - v₁²)

Case 1. The pipe is horizontal, so

y₁ = y₂

P₁ - P₂ = ½ ρ (v₂² –v₁²)

P₂ = P₁ - ½ ρ (v₂² –v₁²)

ρ = 62.43 lbf / ft³

P₁ = 50 psi (144 lbf/ ft² / psi) = 7200 lbf / ft²

P₂ = 7200 - ½ 62.43 / 32 (65² -12²)

P₂ = 7200 - 3980.89

P₂ = 3219.11 lbf / ft²

Case 2 vertical pipe with water flow up

y₂ –y₁ = 40 ft

P₁ - P₂ = ρ g (y₂ –y₁) + ½ rho (v₂² - v₁²)

7200 - P₂ = 62.43 (40) + ½ 62.43 / 32 (65 2 - 12 2) =

P₂ = 7200 - 2497.2 - 3980.89

P₂ = 721.91 lbf / ft²

Case 3. Vertical water pipe flows down

y₂ –y₁ = -40

P₂ = 7200 + 2497.2 - 3980.89

P₂ = 5707.31 lbf / ft²

3 0

2 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

3 years ago