Answer:
Both Techs A and B
Explanation:
Electronic braking systems are controlled by the electronic brake control module. It is a microprocessor that processes information from wheel-speed sensors and the hydraulic brake system to determine when to release braking pressure at a wheel that's about to lock up and start skidding and activates the anti lock braking system or traction system when it detects it is necessary.
Some electronic brake control modules can be programmed to the size of the vehicle's new tires to restore proper electronic brake control performance. While others may require replacing the module to match the module's programming to the installed tire size. So, both technicians A and B are correct.
Answer:

Explanation:
In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.
Lets take




All external matting gears will rotates in opposite direction with respect to each other.
So the speed of gear third can be given as follows


Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
Answer:
See explaination
Explanation:
#include <iostream>
#include<string.h>
using namespace std;
bool isPalindrome(string str, int lower, int upper){
if(str.length() == 0 || lower>=upper){
return true;
}
else{
if(str.at(lower) == str.at(upper)){
return isPalindrome(str,lower+1,upper-1);
}
else{
return false;
}
}
}
int main(){
string input;
cout<<"Enter string: ";
cin>>input;
if(isPalindrome(input,0,input.length()-1)){
cout<<input<<" is a palindrome"<<endl;
}
else{
cout<<input<<" is NOT a palindrome"<<endl;
}
return 0;
}
Answer:
insert (array[] , value , currentsize , maxsize )
{
if maxsize <=currentsize
{
return -1
}
index = currentsize-1
while (i>=0 && array[index] > value)
{
array[index+1]=array[index]
i=i-1
}
array[i+1]=value
return 0
}
Explanation:
1: Check if array is already full, if it's full then no component may be inserted.
2: if array isn't full:
- Check parts of the array ranging from last position of range towards initial range and determine position of that initial range that is smaller than the worth to be inserted.
- Right shift every component of the array once ranging from last position up to the position larger than the position at that smaller range was known.
- assign new worth to the position that is next to the known position of initial smaller component.