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3 years ago

Q.17) A 50-acre catchment containing cropland is converted ot a Qatar mail

Engineering

1 answer:

Ghella [55]3 years ago

3 0

Answer:

Option D

Explanation:

A post development hydrograph will have lower concentration time and lower infiltration losses and hence sooner peak and higher peak and more runoff or higher area under graph. Therefore, all the answers are correct hence option D

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Given a force of 72 lbs at a distance of 15 ft, calculate the moment produced.

Elis [28]

Answer:

1425.78 N.m

Explanation:

Moments of force is calculated as ;

Moments= Force * distance

M= F*d

The S.I unit for moment of force is Newton-meter (N.m)

Given in the question;

Force = 72 lbs

1 pound = 4.45 N

72 lbs = 4.45 * 72=320.4 N

Distance= 15 ft

1ft= 0.3048 m

15 ft = 15*0.3048 = 4.57 m

d= 4.57 m

M= F*d

M=320.4*4.57 =1425.78 N.m

5 0

3 years ago

Thanks for the help!

Dafna11 [192]

Answer:

the 1st one i think

Explanation:

6 0

3 years ago

Ang ______ ay pagiging totoo o matuwid at hindi mahilig magsinungaling

IceJOKER [234]

Answer:

$\purple{\rule{26pt}{5555555pt}}$

6 0

3 years ago

A(n) _____ is done to test each individual component (often a program) to ensure that it is as defect-free as possible.

lyudmila [28]

Answer:

Unit test

Explanation:

Defintion: A test of each individual component (often a program) to ensure that it is as defect-free as possible.

6 0

2 years ago

A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the t

Minchanka [31]

Answer:

$Q=7.3\times 10^{-3} m^3/s$

Explanation:

Given that

At top $d_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm$

$\rho =900\dfrac{Kg}{m^3}$

We know that

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2$

$A_1V_1=A_2V_2$

$\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2$

$\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2$

$V_2=8.02V_1$

$Z_2=12 sin60^{\circ}$

$\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}$

So $V_1=1.30$ m/s

We know that flow rate Q=AV

$Q=A_1V_1$

By putting the values

$A_1=\dfrac{\pi}{4}d^2$

$Q=7.3\times 10^{-3} m^3/s$

To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.

4 0

3 years ago