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Verizon [17]
3 years ago
8

Travel Time Problem: Compute the time of concentration using the Velocity, Sheet Flow Method for Non-Mountainous Orange County a

nd SCS method at a 25 year storm evert.
Location Slope (%) Length (ft) Land Use
1 4.5 1000 Forest light underbrush with herbaceous fair cover.
2 2.5 750 Alluvial Fans (eg. Natural desert landscaping)
3 1.5 500 Open Space with short grasses and good cover
4 0.5 250 Paved Areas (1/4 acre urban lots)
Engineering
1 answer:
Vaselesa [24]3 years ago
5 0

Answer:

Total time taken = 0.769 hour

Explanation:

using the velocity method

for sheet flow ;

Tt = \frac{0.007(nl)^{0.8} }{(Pl)^{5}s^{0.4}  }  

Tt = travel time

n = manning CaH

Pl = 25years

L = how length ( ft )

s = slope

For Location ( 1 )

s = 0.045

L = 1000 ft

n = 0.06 ( from manning's coefficient table )

Tt1 = 0.128 hour

For Location ( 2 )

s = 2.5 %

L= 750

n = 0.13

Tt2 = 0.239 hour

For Location ( 3 )

s = 1.5%

L = 500 ft

n = 0.15

Tt3 = 0.237  hour

For Location (4)

s = 0.5 %

L = 250 ft

n = 0.011

Tt4 = 0.165 hour

hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4

                                              = 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour

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Race car is accelerating and has a velocity of 10 m/s @ t=0. It completes a lap on a circular track of 400 m in 14 seconds. Calc
wariber [46]

Answer:

component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²

magnitude of acceleration is  22.98 m/s²

Explanation:

given data

velocity = 10 m/s

initial time to = 0

distance s = 400 m

time t = 14 s

to find out

components and magnitude of acceleration after the car has travelled 200 m

solution

first we find the radius of circular track that is

we know  distance S = 2πR

400 = 2πR

R = 63.66 m

and tangential acceleration is

S = ut + 0.5 ×at²

here u is initial speed and t is time and S is distance

400 = 10 × 14  + 0.5 ×a (14)²

a = 3.37 m/s²

and here tangential acceleration is constant

so  velocity at distance 200 m

v² - u² = 2 a S

v² = 10² + 2 ( 3.37) 200

v = 38.05 m/s

so radial acceleration at distance 200 m

ar = \frac{v^2}{R}

ar = \frac{38.05^2}{63.66}

ar = 22.74 m/s²

so magnitude of total acceleration is

A = \sqrt{a^2 + ar^2}

A = \sqrt{3.37^2 + 22.74^2}

A = 22.98 m/s²

so magnitude of acceleration is  22.98 m/s²

8 0
3 years ago
What is made in heaven?​
kramer

Answer:

Babies come from heaven didn't you know?

3 0
2 years ago
Find the following for an input of 120 VAC(RMS), 60 hertz, given a 10:1 stepdown transformer, and a full-wave bridge rectifier.
atroni [7]

Answer:

(i) 169.68 volt

(ii) 16.90 volt

(iii) 16.90 volt

(iv) 108.07 volt

(v) 2.161 A

Explanation:

Turn ratio is given as 10:1

We have given that input voltage v_p=120volt

(i) We know that peak voltage is give by v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt

(ii) We know that for transformer \frac{v_p}{v_s}=\frac{n_p}{n_s}

So \frac{169.08}{v_s}=\frac{10}{1}

v_s=16.90volt

So peak voltage in secondary will be 16.90 volt

(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary

So peak voltage of the rectifier will be 16.90 volt

(iv) Dc voltage of the rectifier is given by v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt

(v) Now dc current is given by i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A

4 0
2 years ago
A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The plane strain fracture toughness of the co
murzikaleks [220]

Since the applied stress required for failure due to crack propagation is still higher than 550 MPa, the ceramic is expected to fail due to overload and not because of the flaws

Explanation:

<u>Plane -Strain Fracture toughness is calculated as</u>

k_{IC}=fб\sqrt{\pi a}

F=geometry factor of the flaw

б=Stress applied

k_{IC}=Fracture toughness

a=Flaw size

<u>Given that </u>

Internal Flaw,a=0.001cm

Fracture Toughness k_{IC}=45MPa\sqrt{m}

Tensile Strength б=550 MPa

Geometry Factor,f=1

<u>Calculation</u>

An internal Flaw i s 0.001 cm

2a=0.001cm

a=0

6 0
3 years ago
You will create an array manipulation program that allows the user to do pretty much whatever they want to an array. When launch
enyata [817]

Answer:

Check the explanation

Explanation:

#include <iostream>

using namespace std;

void insert(int* arr, int* size, int value, int position){

if(position<0 || position>=*size){

cout<<"position is greater than size of the array"<<endl;

return ;

}

*size = *size + 1 ;

for(int i=*size;i>position;i--){

arr[i] = arr[i-1];

}

arr[position] = value ;

}

void print(int arr[], int size){

for(int i=0;i<size;i++){

cout<< arr[i] <<" ";

}

cout<<" "<<endl;

}

void remove(int* arr, int* size, int position){

* size = * size - 1 ;

for(int i=position;i<*size;i++){

arr[i] = arr[i+1];

}

}

int count(int arr[], int size, int target){

int total = 0 ;

for(int i=0;i<size;i++){

if(arr[i] == target)

total += 1 ;

}

return total ;

}

int main()

{

int size;

cout<<"Enter the initial size of the array:";

cin>>size;

int arr[size],val;

cout<<"Enter the values to fill the array:"<<endl;

for(int i=0;i<size;i++){

cin>>val;

arr[i] = val ;

}

int choice = 5,value,position,target ;

do{

cout<<"Make a selection:"<<endl;

cout<<"1) Insert"<<endl;

cout<<"2) Remove"<<endl;

cout<<"3) Count"<<endl;

cout<<"4) Print"<<endl;

cout<<"5) Exit"<<endl;

cout<<"Choice:";

cin>>choice;

switch(choice){

case 1:

cout << "Enter the value:";

cin>>value;

cout << "Enter the position:";

cin>>position;

insert(arr,&size,value,position);

break;

case 2:

cout << "Enter the position:";

cin>>position;

remove(arr,&size,position);

break;

case 3:

cout<<"Enter the target value:";

cin>>target;

cout <<"The number of times "<<target<<" occured in your array is:" <<count(arr,size,target)<<endl;

break;

case 4:

print(arr,size);

break;

case 5:

cout <<"Thank you..."<<endl;

break;

default:

cout << "Invalid choice..."<<endl;

}

}while(choice!=5);

return 0;

}

Kindly check the attached images below for the code output.

3 0
3 years ago
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