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Vera_Pavlovna [14]
3 years ago
15

3 cm of water evaporated from a 200-hectare vertical walled reservoir during 24 hours. Storm water was added to the reservoir at

a constant rate of 3 m3/s during this period. Determine the volume in ha-cm of water released during the period (through the bottom of the reservoir) if the water level was the same at the beginning and the end of the day.
Engineering
1 answer:
ycow [4]3 years ago
7 0

Answer:

water released through the bottom of the reservoir in 24 hrs is 1992 ha-cm

Explanation:

Given the data in the question;

A = 200 hectare =  2 × 10⁶ m²

water evaporated Ve = 2 × 10⁶ m² × 3 × 10⁻² = 60000 m³ { in 24 hrs }

Water added by storm in 24hrs Vi = 3 × 24 × 3600 = 259200 m³

now let water released be Vr

ΔV = V_ini - V_final = 0

Vi - Ve - Vr = 0

Vr = Vi - Ve

Vr = 259200 m³ - 60000 m³

Vr = 199200 m³ = 19920000 m² - cm

Vr = 1992 ha-cm

Therefore, water released through the bottom of the reservoir in 24 hrs is 1992 ha-cm

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The design speed was used for the freeway exit ramp is 11 mph.

<h3>Design speed used in the exit ramp</h3>

The design speed used in the exit ramp is calculated as follows;

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<h3>Design speed</h3>

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f = (10)²/(15 x 19.4) - 0.01

f = 0.33

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Learn more about design speed here: brainly.com/question/22279858

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What is the magnitude of the maximum stress that exist at the tip of an internal crack having a radius of curvature of 1.9 x 10-
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Answer:

2800 [MPa]

Explanation:

In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

\sigma_{m} = 2 \sigma_{p} (\frac{l_{c}}{r_{c}})^{0.5}

Where \sigma_{m} is the magnitude of he maximum stress at the tip of the crack, \sigma_{p} is the magnitude of the tensile stress, l_{c} is 1/2 the length of the internal crack, and r_{c} is the radius of curvature of the crack.

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