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garri49 [273]
3 years ago
11

A seawall with an opening is used to dampen the tidal influence in a coastal area (and limit erosion). The seawall is 2.5 m long

(in the direction perpendicular to the page), the mean sea level on the sea side is 2.2 m above the centroid of the slot. The mean sea level on the land side of the wall is 1.4 m above the centroid of the slot. The sea side tide fluctuates +0.6 m, and the landward tide fluctuates +0.4 m.
The slot extends the entire length of the wall and is estimated to have a discharge coefficient of 0.80.


If the volume of seawater moving from the sea side to the land side of the wall is to be no more than 6 000 m' in 18 hours, what is the maximum allowable height of the slot?


Note: the discharge coefficient is the ratio of the actual flow rate over the predicted flow rate. It mostly accounts for the fact that the average velocity is different from the velocity.
Engineering
1 answer:
raketka [301]3 years ago
3 0

Answer:

The maximum allowable Height of the slot is 11.685mm

Explanation:

The explanation is attached. The approach used is Bernoulli's equation

Download pdf
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A crankcase heater is often used to prevent refrigerant from mixing with compressor oil during periods of:
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Low ambient temperature

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Ultimate tensile strength is: (a) The stress at 0.2% strain (b) The stress at the onset of plastic deformation (c) The stress at
MissTica

Answer:

By definition the ultimate tensile strength is the maximum stress in the stress-strain deformation. The stress at 0.2% strain, the stress at the onset of plastic deformation, the stress at the end of the elastic deformation and the stress at the fracture correspond, by definition, to other points of the stress-strain curve.

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The most important element of green construction is that it is a(n) __________ approach to building. *
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Environmentally friendly


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3 years ago
You hang a heavy ball with a mass of 42 kg from a silver rod 2.7 m long by 1.9 mm by 2.6 mm. You measure the stretch of the rod,
nadezda [96]

Answer:

Explanation:

cross sectional area  A = 1.9 x 2.6 x 10⁻⁶ m²

= 4.94 x 10⁻⁶ m²

stress = 42 x 9.8 / 4.94 x 10⁻⁶

= 83.32 x 10⁶ N/m²

strain = .002902 / 2.7

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= 83.32 x 10⁶ / 1.075 x 10⁻³

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3 years ago
The time to half-maximum voltage is how long it takes the capacitor to charge halfway. Based on your experimental results, how l
satela [25.4K]

Answer:

Time taken for the capacitor to charge to 0.75 of its maximum capacity = 2 × (Time take for the capacitor to charge to half of its capacity)

Explanation:

The charging of a capacitor/the build up of its voltage follows an exponential progression and is given by

V(t) = V₀ [1 - e⁻ᵏᵗ]

where k = (1/time constant)

when V(t) = V₀/2

(1/2) = 1 - e⁻ᵏᵗ

e⁻ᵏᵗ = 0.5

In e⁻ᵏᵗ = In 0.5 = - 0.693

-kt = - 0.693

kt = 0.693

t = (0.693/k)

Recall that k = (1/time constant)

Time to charge to half of max voltage = T(1/2)

T(1/2) = 0.693 (Time constant)

when V(t) = 0.75

0.75 = 1 - e⁻ᵏᵗ

e⁻ᵏᵗ = 0.25

In e⁻ᵏᵗ = In 0.25 = -1.386

-kt = - 1.386

kt = 1.386

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t = 2 × T(1/2)

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