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iren [92.7K]
3 years ago
9

What is differences Between hard shoulder & soft shoulder in civil Engineerin?

Engineering
1 answer:
r-ruslan [8.4K]3 years ago
5 0

Answer:

<em><u>The 'shoulder' of a road is the land to the edge of the road. On most roads without pavements, the shoulder is a strip of grass or a hedgerow. This is known as a 'soft shoulder'. On a motorway, this strip of land is hardstanding, hence the name 'hard shoulder.'</u></em>

<em><u>Mark</u></em><em><u> </u></em><em><u>as</u></em><em><u> brilliant</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u> </u></em>

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Which of these are not referenced in an assembly?
Otrada [13]
E. Parts they don’t resemble
7 0
3 years ago
The cubic capacity of a four-stroke over-square spark-ignition engine is 245cc. The over-square ratio is (1.1) The clearance vol
pashok25 [27]

Answer:

Bore = 7 cm

stroke = 6.36 cm

compression ratio = 10.007

Explanation:

Given data:

Cubic capacity of the engine, V = 245 cc

Clearance volume, v = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

V = \frac{\pi}{4}D^2L

or

V = \frac{\pi}{4}\frac{D^3}{1.1}

on substituting the values, we have

245 =  \frac{\pi}{4}\frac{D^3}{1.1}

or

D = 7.00 cm

Now,

we have

D/L = 1.1

thus,

L = D/1.1

L = 7/1.1

or

L= 6.36 cm

Now,

the compression ratio is given as:

\textup{compression ratio}=\frac{V+v}{v}

on substituting the values, we get

\textup{compression ratio}=\frac{245+27.2}{27.2}

or

Compression ratio = 10.007

4 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
what is the transfer function of the loaded filter? express your answer in terms of the variables r , l , rl , and s .
NISA [10]

Loaded, H_{Loaded}(s) = \frac{RR_{L} }{R+R_{L} } /(\frac{RR_{L} }{R+R_{L} }+SL) = \frac{RR_{L}/L }{R+R_{L} } /(\frac{RR_{L} /L}{R+R_{L} }+S) is the loaded filter's transfer function.

A graded filter that, by virtue of its weight and permeability, stabilises the foot of an earth dam or other construction when it is installed at the base of that structure.

Air filters with depth loaded are made to achieve precisely that. They add particles gradually to create air passageways, reducing constriction. You may save time and money by using filters that last longer thanks to them. The bigger particles are caught at the filter's beginning, while the smaller particles are caught as it gets closer. This is intended to avoid rapid surface loading, hence facilitating more airflow. This enables longer-lasting filtration as well.

On the other hand, surface loading filters catch every particle that is on its surface. No matter how big or little the particles are, it doesn't care.

Learn more about Loaded here:

brainly.com/question/20039214

#SPJ4

3 0
1 year ago
What are the de Broglie frequencies and wavelengths of (a) an electron accelerated to 50 eV (b) a proton accelerated to 100 eV
DaniilM [7]

Answer:

(a) De-Brogie wavelength is 0.173 nm and frequency is 2.42 x 10^16 Hz

(b) De-Brogie wavelength is 2.875 pm and frequency is 4.8 x 10^16 Hz

Explanation:

(a)

First, we need to find velocity of electron. Since, it is accelerated by electric potential. Therefore,

K.E of electron = (1/2)mv² = (50 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 8 x 10^(-18) J

Mass of electron = m = 9.1 x 10^(-31) kg

Therefore,

v² = [8 x 10^(-18) J](2)/(9.1 x 10^(-31) kg)

v = √1.75 x 10^13

v = 4.2 x 10^6 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(9.1 x 10^(-31) kg)(4.2 x 10^6 m/s)

<u>λ = 0.173 x 10^(-9) m = 0.173 nm</u>

The frequency is given as:

Frequency = f = v/λ

f = (4.2 x 10^6 m/s)/(0.173 x 10^(-9) m)

<u>f = 2.42 x 10^16 Hz</u>

(b)

First, we need to find velocity of proton. Since, it is accelerated by electric potential. Therefore,

K.E of proton = (1/2)mv² = (100 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 1.6 x 10^(-17) J

Mass of proton = m = 1.67 x 10^(-27) kg

Therefore,

v² = [1.6 x 10^(-17) J](2)/(1.67 x 10^(-27) kg)

v = √1.916 x 10^10

v = 1.38 x 10^5 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(1.67 x 10^(-27) kg)(1.38 x 10^5 m/s)

<u>λ = 2.875 x 10^(-12) m = 2.875 pm</u>

The frequency is given as:

Frequency = f = v/λ

f = (1.38 x 10^5 m/s)/(2.875 x 10^(-12) m)

<u>f = 4.8 x 10^16 Hz</u>

6 0
3 years ago
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