What is not required for current to flow through a conductor

What is a combination circuit? A combination circuit:

Anon25 [30]

Answer:

Combination circuit; The basic strategy for the analysis of combination circuits involves using the meaning of equivalent resistance for parallel branches to transform the combination circuit into a series circuit.

Example:

The use of both series and parallel connections within the same circuit. In this case, light bulbs A and B are connected by parallel connections and light bulbs C and D are connected by series connections. This is an example of a combination circuit.

7 0

3 years ago

Write a iterative function that finds the n-th integer of the Fibonacci sequence. Then build a minimal program (main function) t

Natasha2012 [34]

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)

7 0

3 years ago

Read 2 more answers

Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1

babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0

4 years ago

When checking for a no-star concern, you notice that an engine has no spark Technician A says to turn on the ignition engine (en

lbvjy [14]

Answer:

Technician B

Explanation:

Technician B is correct in his argument. This is because according to what he said, as the computer pulses stimuli the coil will turn on and off, promoting an increase in the voltage that will cause the fluctuation. Technician A is incorrect because the procedure he indicated imposes that the voltage is checked at the negative terminal and not at the positive.

5 0

3 years ago

Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin

Alborosie

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2 - $\sqrt{}$ ( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy = tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2 - $\sqrt{}$ ( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

6 0

3 years ago