The coil polarity in a waste-spark system is determined by the direction in which the coil is wound (left-hand rule for conventional current flow)and can’t be changed. For example, if a V-8 engine has a firing order of 18436572 and the number 1 cylinder is on compression, which cylinder will be on the exhaust stroke?
Answer:
camshaft, in internal-combustion engines, rotating shaft with attached disks of irregular shape (the cams), which actuate the intake and exhaust valves of the cylinders.
Explanation:
I'm taking an engineering/tech class. I hope this helps! :)
Answer:
Codes for each of the problems are explained below
Explanation:
PROBLEM 1 IN C++:
#include<iostream>
using namespace std;
//fib function that calculate nth integer of the fibonacci sequence.
void fib(int n){
// l and r inital fibonacci values for n=1 and n=2;
int l=1,r=1,c;
//if n==1 or n==2 then print 1.
if(n==1 || n==2){
cout << 1;
return;
}
//for loop runs n-2 times and calculates nth integer of fibonacci sequence.
for(int i=0;i<n-2;i++){
c=l+r;
l=r;
r=c;
cout << "(" << i << "," << c << ") ";
}
//prints nth integer of the fibonacci sequence stored in c.
cout << "\n" << c;
}
int main(){
int n; //declared variable n
cin >> n; //inputs n to find nth integer of the fibonacci sequence.
fib(n);//calls function fib to calculate and print fibonacci number.
}
PROBLEM 2 IN PYTHON:
def fib(n):
print("fib({})".format(n), end=' ')
if n <= 1:
return n
else:
return fib(n - 1) + fib(n - 2)
if __name__ == '__main__':
n = int(input())
result = fib(n)
print()
print(result)
Answer:
the minimum shaft diameter is 35.026 mm
the maximum shaft diameter is 35.042mm
Explanation:
Given data;
D-maximum = 35.020mm and d-minimum = 35.000mm
we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6
so From table, Selection of International Trade Grades metric series
the grade tolerance are;
ΔD = IT7(0.025 mm)
Δd = IT6(0.016 mm)
Also from Table "Fundamental Deviations for Shafts" metric series
Sf = 0.026
so
D-maximum
Dmax = d + Sf + Δd
we substitute
Dmax = 35 + 0.026 + 0.016
Dmax = 35.042 mm
therefore the maximum diameter of shaft is 35.042mm
d-minimum
Dmin = d + Sf
Dmin = 35 + 0.026
Dmin = 35.026 mm
therefore the minimum diameter of shaft is 35.026 mm
Answer: B
Explanation:
One good way to improve your gas mileage is to accelerate smoothly and directly to a safe speed.
Hope this helps!
