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marin [14]
3 years ago
15

Modify the one-dimensional continuity equation to include rainfall over the free surface

Engineering
1 answer:
KonstantinChe [14]3 years ago
3 0

The rainfall run off model  HEC-HMS is combined with river routing model. They are used for simulating the rainfall process.

Explanation:

The HEC - HMS rainfall model is used for simulating the rainfall runoff process. In this study the soil conservation service and curve number method is used to calculate the sub basin loss in basin module.

It provides various options for providing the rainfall distributions in the basin. It has the control specification module used to control the time interval for the simulations.

The one dimensional continuity equation is

бA / бT + бQ / бx= 0

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Remember from Lab 3C that Mad Libs are activities that have a person provide various words, which are then used to complete a sh
Harlamova29_29 [7]

Answer:

Python code is explained below

Explanation:

CODE(TEXT): -

9A.py:-

def sentence(): #function to read input and display

  name = input("Enter a name: ") #input name

  place = input("Enter a place: ") #input place

  number = int(input("Enter a number: ")) #input number typecasted to int

  noun = input("Enter a plural noun: ") #input plural noun

  adjective = input("Enter an adjective: ") #input adjective

  print("\n{} went to {} to buy {} different types of {}".format(name, place, number, noun)) #format is used to display o/p

  print("but unfortunately, the {} were all {} so {} went back to {} to return them.\n".format(noun, adjective,name, place))

op = "n" #initially op = n i.e. user not wanting to quit

while op != "y" : #while user does not input y

  sentence() #input words from user

  op = input("Do you want to quit [y/n]? ") #prompt to continue or quit

  if op == "y": #if yes then print goodbye and exit

      print("Goodbye")

      break

  else: #else print newline and take input

      print("")

9B.py: -

#inputs are strings by default applying a split() function to a string splits it into a list of strings

#then with the help of map() function we can convert all the strings into integers

numbers = list(map(int, input("Enter the input: ").split()))

sum = 0 #sum is initially 0

for i in range(len(numbers)): #loops for all the elements in the list

  sum = sum + numbers[i] #add the content of current element to sum

avg = sum/ len(numbers) #average = (sum of all elements)/ number of elements in the list

max = numbers[0] #initially max = first number

for i in range(1, len(numbers)): #loops for all remaining elements

  if numbers[i] > max : #if their content is greater than max

      max = numbers[i] #update max

print("The average and max are: {:.2f} {}".format(avg, max)) #format is used to print in formatted form

#.2f is used to print only 2 digits after decimals

9C.py: -

#inputs are strings by default applying a split() function to a string splits it into a list of strings

#then with the help of map() function we can convert all the strings into integers

numbers = list(map(int, input("Enter a list of numbers: ").split()))

numbers = [elem for elem in numbers if elem >= 0] #using list comprehension

#loops for all elements of the list and keep only those who are >= 0

numbers.sort() #list inbuilt function to sort items

print("The non-negative and sorted numbers are: ", end = "")

for num in numbers: #for all numbers in the list print them

  print("{} ".format(num), end = "")

7 0
3 years ago
How do you make a 3d print
yulyashka [42]

Answer:you need a 3d printer

Explanation:

5 0
2 years ago
A water skier leaves the end of an 8 foot tall ski ramp with a speed of 20 mi/hr and at an angle of 250. He lets go of the tow r
klemol [59]

Answer:

At highest point:

y1 = 10.4 ft

v1 = (26.5*i + 0*j) ft/s

When he lands:

x2 = 31.5 ft (distance he travels)

t2 = 1.19 s

V2 = (26.5*i - 25.9*j) ft/s

a2 = -44.3°

Explanation:

Since he let go of the tow rope upon leaving the ramp he is in free fall from that moment on. In free fall he is affected only by the acceleration of gravity. Gravity has a vertical component only, so the movement will be at constant acceleration in the vertical component and at constant speed in the horizontal component.

20 mi / h = 29.3 ft/s

If the ramp has an angle of 25 degrees, the speed is

v0 = (29.3 * cos(25) * i + 29.3 * sin(25) * j) ft/s

v0 = (26.5*i + 12.4*j) ft/s

I set up the coordinate system with the origin at the base of the ramp under its end, so:

R0 = (0*i + 8*j) ft

The equation for the horizontal position is:

X(t) = X0 + Vx0 * t

The equation for horizontal speed is:

Vx(t) = Vx0

The equation for vertical position is:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

The equation for vertical speed is:

Vy(t) = Vy0 + a * t

In this frame of reference a is the acceleration of gravity and its values is -32.2 ft/s^2.

In the heighest point of the trajectory the vertical speed will be zero because that is the point where it transitions form going upwards (positive vertical speed) to going down (negative vertical speed), and it crosses zero.

0 = Vy0 + a * t1

a * t1 = -Vy0

t1 = -Vy0 / a

t1 = -12.4 / -32.2 = 0.38 s

y1 = y(0.38) = 8 + 12.4 * 0.38 + 1/2 * (-32.2) * (0.38)^2 = 10.4 ft

The velocity at that moment will be:

v1 = (26.5*i + 0*j) ft/s

When he lands in the water his height is zero.

0 = 8 + 12.4 * t2 + 1/2 * (-32.2) * t2^2

-16.1 * t2^2 + 12.4 * t2 + 8 = 0

Solving this equation electronically:

t2 = 1.19 s

Replacing this time on the position equation:

X(1.19) = 26.5 * 1.19 = 31.5 ft

The speed is:

Vx2 = 26.5 ft/s

Vy2 = 12.4 - 32.2 * 1.19 = -25.9 ft/s

V2 = (26.5*i - 25.9*j) ft/s

a2 = arctg(-25.9 / 26.5) = -44.3

3 0
3 years ago
Someone please please help me and explain!! I will give brainliest if right!!!
Mariulka [41]

R=10+15+30

55 is the answer to the question

5 0
3 years ago
Read 2 more answers
Consider a system with two tasks, Task1 and Task2. Task1 has a period of 200 ms, and Task2 has a period of 300 ms. All tasks ini
Murrr4er [49]

<u>Explanation:</u>

Task 1 time period = 200ms, Task 2 time period = 300ms

Task ticked = \frac{1000ms}{200ms}= 5  →  5 times

Task 2 ticked =\frac{1000ms}{300ms} = 3.33 → 3 times

At 600 ms → 200ms 200ms 200ms

                     300ms → \frac{30ms}{60ms}

Largest time period = H.C.M of (200ms, 300ms)

                                 = 600ms

4 0
3 years ago
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