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mixas84 [53]
2 years ago
13

technician a says that use of firmer, urethane bushings in the suspension improves the vehicle's road-holding ability and handli

ng. Technician B says that firmer bushings help eliminate torque steer in some FWD vehicles. Who is correct?
Engineering
1 answer:
tensa zangetsu [6.8K]2 years ago
5 0

it is technician B i belive im not sure the question tho

You might be interested in
Air at p=1 atm enters a long tube of length 2.5 m and diameter of 12 mm at an inlet temperature of Tm,i=100oC and mass flowrate
Annette [7]

Answer:

The heat transfer q = 18.32W

Explanation:

In this question, we are asked to calculate the heat entering the tube and also evaluate properties at T =400K

Please check attachment for complete solution and step by step explanation

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
Reference Parameters (returning multiple values): Write a C++ function that converts standard time to military time. Inputs incl
valkas [14]

Answer:

Code is given as below:

Explanation:

#include <iostream>

using namespace std;

//function prototype declaration

void MilitaryTime(int, int, char, int &, int &);

int main()

{

    //declare required variables

    int SHour, SMin, MHour, MMin;

    char AorP;

    //promt and read the hours from the user

    cout<<"Enter hours in standard time : ";

    cin>>SHour;

    //check the hours are valid are not

    while(SHour<0 || SHour>12)

    {

         cout<<"Invalid hours for standard time. "

             <<"Try again..."<<endl;

         cout<<"Enter hours in standard time : ";

         cin>>SHour;

    }

    //promt and read the minutes from the user

    cout<<"Enter minutes in standard time : ";

    cin>>SMin;

    //check the minutes are valid are not

    while(SMin<0 || SMin>59)

    {

         cout<<"Invalid minutes for standard time. "

             <<"Try again..."<<endl;

         cout<<"Enter minutes in standard time : ";

         cin>>SMin;

    }

    //promt and read the am or pm from the user

    cout<<"Enter standard time meridiem (a for AM p for PM): ";

    cin>>AorP;

    //check the meridiem is valid are not

    while(!(AorP=='a' || AorP=='p' || AorP=='A' || AorP=='P'))

    {

         cout<<"Invalid meridiem for standard time. "

             <<"Try again..."<<endl;

         cout<<"Enter standard time meridiem (a for AM p for PM): ";

         cin>>AorP;

    }

    //call function to calculate the military time

    MilitaryTime(SHour, SMin, AorP, MHour, MMin);

    //fill zeros and display standard time

    cout.width(2);

    cout.fill('0');

    cout<<SHour<<":";

    cout.width(2);

    cout.fill('0');

    cout<<SMin;

    if(AorP=='a' || AorP=='A')

         cout<<" am = ";

    else

         cout<<" pm = ";

    //fill zeros and display military time

    cout.width(2);

    cout.fill('0');

    cout<<MHour;

    cout.width(2);

    cout.fill('0');

    cout<<MMin<<endl;

    system("PAUSE");

    return 0;

}

//function to calculate the military time with reference parameters

void MilitaryTime(int SHour, int SMin, char AorP, int &MHour, int &MMin)

{

    //check the meredium is am or pm

    //and calculate hours

    if(AorP=='a' || AorP=='A')

    {

         if(SHour==12)

             MHour = 0;

         else

             MHour = SHour;

    }

    else

         MHour = SHour+12;

    MMin = SMin;

5 0
3 years ago
Think of the differences between circuit-switching and packet-switching paradigms in the Internet core design. Assume an Interne
dem82 [27]

Answer:

0.264 ; 0.079

Explanation:

Given that:

Sample size, n = 100

Probability of being active, p = 1% = 1/100 = 0.01

Using the binomial probability relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Probability that more than 1 user will be active

P(x > 1) = 1 - [p(x=0) + p(x = 1)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x > 1) = 1 - [0.366 + 0.370]

P(x > 1) = 0.264

2.)

Probability that more than 2 user will be active

P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185

P(x > 1) = 1 - [0.366 + 0.370 + 0.185]

P(x > 1) = 0.079

7 0
2 years ago
Given vectors A = xˆ2−yˆ +zˆ3 and B = xˆ3−zˆ2, find a vector C whose magnitude is 9 and whose direction is perpendicular to both
natka813 [3]

Answer:

Vector C = 1.334i + 8.671j + 2k or 1.334x + 8.671y + 2z

Explanation:

The concept applied to solve the question is cross product of vector, AXB since vector C is perpendicular to vector A and B and this is solved by applying the 3X3 determinant method.

A detailed step by step explanation is attached below.

7 0
3 years ago
Read 2 more answers
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