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2 years ago

In which of the following states would homes most likely have the deepest foundation?

Engineering

1 answer:

Evgen [1.6K]2 years ago

5 0

Answer:

D) Louisiana, to prevent homes from blowing away in hurricanes.

Explanation:

Louisiana is famous for hurricanes i think that if they were well prepared and had deeper foundation the damage may lower by a good lot.

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g A 30-m-diameter sedimentation basin has an average water depth of 3.0 m. It is treating 0.3 m3/s wastewater flow. Compute over

stepladder [879]

Answer:

The overflow rate is 4.24×10^-4 m/s.

The detention time is 7069.5 s

Explanation:

Overflow rate is given as volumetric flow rate ÷ area

volumetric flow rate = 0.3 m^3/s

area = πd^2/4 = 3.142×30^2/4 = 706.95 m^2

Overflow rate = 0.3 m^3/s ÷ 706.95 m^2 = 4.24×10^-4 m/s

Detention time = volume ÷ volumetric flow rate

volume = area × depth = 706.95 m^2 × 3 m = 2120.85 m^3

Detention time = 2120.85 m^3 ÷ 0.3 m^3/s = 7069.5 s

6 0

2 years ago

6. Question

valkas [14]

Answer:

Check the 2nd, 3rd and 4th statements.

Explanation:

4 0

2 years ago

In sleep, what does REM stand for?

sertanlavr [38]

Answer:

Rapid eye maneuver

Explanation:

during slumber, your eyes move quickly on different directions.

7 0

3 years ago

A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twi

Leviafan [203]

Answer:

F₁ = 1500 N

F₂ = 750 N

$F_{e}$ = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

= 7.5 x 1000 W

= 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 = F₁ - F₂

750 = 2 F₂ - F₂ ( ∵F₁ = 2 F₂ )

∴F₂ = 750 N

Now F₁ = 2 F₂

F₁ = 2 x F₂

F₁ = 2 x 750

F₁ = 1500 N , this is the maximum force.

Therefore we know,

$F_{max}$ = 3 x $F_{e}$

where $F_{e}$ is centrifugal force

$F_{e}$ = $F_{max}$ / 3

= 1500 / 3

= 500 N

8 0

3 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in

Sonja [21]

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:

σ = FL / π $R^{3}$

= 3000 $\times$ (40 $\times$ 10^-3) / π (5 $\times$ 10^-3)^3

σ = 305 $\times$ 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

= 2(305 $\times$ 10^6) (15 $\times$ 10^-3)^3 / 3(40 $\times$ 10^-3)

F = 17156 N.

7 0

3 years ago