Answer:
baking soda and vinegar dish soap
Explanation:
it will create a bubbles and let it sit for 3 hours and it will go away
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 rate of heat transfer decreases.
Explanation:
Fin effectiveness is defined as the ratio of heat transfer rate from a finned surface to the heat transfer rate from the same surface if there were no fins. Its value is expected to be greater than 1.
Having effectiveness value of 0.9 which is less than 1 indicates that the heat transfer rate will decrease since a fin effectiveness smaller than 1 shows that the fin acts as insulation (thermal insulation).
Answer:
2.379m
Explanation:
The width = 23m
The depth = 3m
The radius is denoted as R
The wetted area is = A
The perimeter perimeter = P
Hydraulic radius
R = A/P
The area of a rectangular channel
= Width multiplied by Depth
A = 23x3
A = 69m²
Perimeter = (2x3)+23
P = 6+23
P= 29
Hydraulic radius R = 69/29
= 2.379m
This answers the question
Thank you!
Answer:
Step 1 of 3
Case A:
AISI 1018 CD steel,
Fillet radius at wall=0.1 in,
Diameter of bar
From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel
Tensile strength
Yield strength
The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.
At the critical stress elements on the top and bottom surfaces transverse shear is zero
Explanation:
See the next steps in the attached image
