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Yuki888 [10]
4 years ago
12

Consider a drainage basin having 60% soil group A and 40% soil group B. Five years ago the land use pattern in the basin was ½ w

ooded area with poor cover and ½ cultivated land (row crops/contoured and terraces) with good conservation treatment. Now the land use has been changed to 1/3 wooded area with poor cover, 1/3 cultivated land (row crops/contoured and terraces) with good conservation treatment, and 1/3 commercial and business area.
(a) Estimate the increased runoff volume during the dormant season due to the land use change over the past 5-year period for a storm of 35 cm total depth under the dry antecedent moisture condition (AMC I). This storm depth corresponds to a duration of 6-hr and 100-year return period. The total 5-day antecedent rainfall amount is 30 mm. (Note: 1 in = 25.4 mm.)
(b) Under the present watershed land use pattern, find the effective rainfall hyetograph (in cm/hr) for the following storm event using SCS method under the dry antecedent moisture condition (AMC I).
Engineering
1 answer:
victus00 [196]4 years ago
3 0

Answer:

Please see the attached file for the complete answer.

Explanation:

Download pdf
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Determine the minimum radius of a horizontal circular curve for a route having a 70 mph design speed, super-elevation (e) 3%, an
patriot [66]

Answer:

a) Rmin ≈ 52 m

b) D = 110.184°

c) Lc ≈ 59 m

d) PT Station = 2827+12.37

PT Station = 2827+78.63

Explanation:

a) Given

v = 70 mph = (70 mph)(1,609 m/1 mile)(1 h/3600 s) = 31.286 m/s

e = 3% = 0.03

f = 0.12

a) We can use the equation

Rmin = v²/(127*(e + f))

⇒ Rmin = (31.286)²/(127*(0.03 + 0.12))

⇒ Rmin = 51.38 m ≈ 52 m

b) We can use the equation

D = 5729.578/R  ⇒  D = 5729.578/52

⇒  D = 110.184°

c) We apply the formula

Lc = R*Δ/57.3

If  Δ = 65°  we have

Lc = 52*65/57.3

⇒ Lc = 58.98 m ≈ 59 m

d) If the PI is station 2827+45.50 we get the tangent length T as follows:

T = R*tan(Δ/2)

⇒ T = 52*tan(65/2) = 33.13 m

then, the station of the PC will be

PC Station = PI - T

⇒ PC Station = (2827+45.50) - (0+33.13) = 2827+12.37

and the station of the PT will be

PT Station = PI + T

⇒ PT Station = (2827+45.50) + (0+33.13) = 2827+78.63

3 0
4 years ago
In one study the critical stress intensity factor for human bone was calculated to be 4.05 MN/m3/2. If the value of Y in Eq. (2.
Diano4ka-milaya [45]
Where is Eq.(28) ?? You should show it to find the result
6 0
3 years ago
Ordan has _ 5 8 can of green paint and _ 3 6 can of blue paint. If the cans are the same size, does Jordan have more green paint
Morgarella [4.7K]

Answer:

Jordan has more green paints

Explanation:

Given

Green = \frac{5}{8}

Blue = \frac{3}{6}

Required

Which paint does he have more?

For better understanding, it's better to convert both measurements to decimal.

For the green paint:

Green = \frac{5}{8}

Green = 0.625

For the blue paint:

Blue = \frac{3}{6}

Blue = 0.5

By comparison:

0.625 > 0.5

<em>This means that Jordan has more green paints</em>

3 0
3 years ago
Develop an E-R model (E-R Diagram) for the application belowDatabase RequirementsLogin
nikklg [1K]

Answer:

As the ER diagram is to be drawn in some tool so it is attached as snapshot i have taken from my tool.

Explanation:

Please find snapshot below for detailed ER diagram.

4 0
3 years ago
Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,
Lyrx [107]

Answer:

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0

Explanation:

v_0 = Initial velocity of ball A

v_A=v_0\cos45^{\circ}

v_B = Initial velocity of ball B = 0

(v_A)_n' = Final velocity of ball A

v_B' = Final velocity of ball B

e = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

Coefficient of restitution is given by

e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

Adding the above two equations we get

2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0

\boldsymbol{\therefore v_B'=0.6364v_0}

(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0

From the conservation of momentum along the plane of contact we have

(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}

v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0.

5 0
3 years ago
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