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:
if you are speaking of the acronym then Engineering uses science and mathematics to solve everyday problems in society
Answer:
fracture will occur as the value is less than E/10 (= 22.5)
Explanation:
If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.
![\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}](https://tex.z-dn.net/?f=%5Csigma_m%20%3D%202%5Csigma_o%20%5B%5Cfrac%7Ba%7D%7B%5Crho_t%7D%5D%5E%7B1%2F2%7D)
= 15 GPa
fracture will occur as the value is less than E/10 = 22.5
Answer:
Explained
Explanation:
Cold working: It is plastic deformation of material at temperature below recrystallization temperature. whereas hot working is deforming material above the recrystallization temperature.
Given melting point temp of lead is 327° C and lead recrystallizes at about
0.3 to 0.5 times melting temperature which will be higher that 20°C. Hence we can conclude that at 20°C lead will under go cold working only.
Answer:
(a) the rate of heat transfer to the coolant is Q = 139.71W
(b) the surface temperature of the shaft T = 40.97°C
(c) the mechanical power wasted by the viscous dissipation in oil 22.2kW
Explanation:
See explanation in the attached files
