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sukhopar [10]
3 years ago
13

technician A says that in any circuit, electrical current takes the path of least resistance. technician B says that while this

is true in a series circuit, it's not entirely true in a parallel circuit. who is correct?
Engineering
1 answer:
ArbitrLikvidat [17]3 years ago
3 0

Answer:

  technician A is correct

Explanation:

Technician B has circuit topologies confused. In a series circuit, there is only one path for electrical current to take. In a parallel circuit, the current will divide between paths in proportion to the inverse of their resistance. The least resistance path will have the most current.

Technician A is mostly correct.

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Given a matrix, clockwise-rotate elements in it. Please add code to problem3.cpp and the makefile. Use the code in p3 to test yo
rusak2 [61]

Answer:

/* C Program to rotate matrix by 90 degrees */

#include<stdio.h>

int main()

{

int matrix[100][100];

int m,n,i,j;

printf("Enter row and columns of matrix: ");

scanf("%d%d",&m,&n);

 

/* Enter m*n array elements */

printf("Enter matrix elements: \n");

for(i=0;i<m;i++)

{

 for(j=0;j<n;j++)

 {

  scanf("%d",&matrix[i][j]);

 }

}

 

/* matrix after the 90 degrees rotation */

printf("Matrix after 90 degrees roration \n");

for(i=0;i<n;i++)

{

 for(j=m-1;j>=0;j--)

 {

  printf("%d  ",matrix[j][i]);

 }

 printf("\n");

}

 

return 0;

 

}

5 0
4 years ago
The best grade of hardwood lumber that is generally available is _____​
Vesnalui [34]

Answer:

FAS

Explanation:

first and second grade

5 0
2 years ago
What is the area enclosed by the cycle area of the Carnot cycle illustrating on a P-V diagram?
Inga [223]

Answer:

The work of the cycle.

Explanation:

The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.

The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.

7 0
3 years ago
Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o
jekas [21]

Complete Question

Complete Question is attached below.

Answer:

V'=5m/s

Explanation:

From the question we are told that:

Diameter d=0.10m

Power P=4.0kW

Head loss \mu=10m

 \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu

 \frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10

 H_m=(\frac{200*10^3}{1000*9.8}-10)

 H_m=10.39m

Generally the equation for Power is mathematically given by

 P=\rho gQH_m

Therefore

 Q=\frac{P}{\rho g H_m}

 Q=\frac{4*10^4}{1000*9.81*10.9}

 Q=0.03935m^3/sec

Since

 Q=AV'

Where

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 A=7.85*10^{-3}

Therefore

 V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}

 V'=5m/s

5 0
3 years ago
A resistivity meter is measured in
Bingel [31]
Ohms ..................
3 0
3 years ago
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