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

Engineering

1 answer:

Ghella [55]2 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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A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th

Helga [31]

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at

Replacing by the values already known for vo, vf and a, and solving for t, we get:

t = vf-vo /a = (29 m/s – 11.2 m/s) / 2.2 m/s = 8 sec

5 0

2 years ago

An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi

Fittoniya [83]

Answer:

a.) -147V

b.) -120V

c.) 51V

Explanation:

a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.

c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.

Honestly, these things take practice to get used to. It's really hard to explain this.