Which of the followong parts does not rotate during starter operation? A. Commutator segments B. Armature windings c. Field wind
1 answer:
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Answer:
Glass: Low-Loss dielectric
α = 8.42*10^-11 Np/m
β = 468.3 rad/m
λ = 1.34 cm
up = 1.34*10^8 m/s
ηc = 168.5 Ω
Tissue: Quasi-Conductor
α = 9.75 Np/m
β = 12.16 rad/m
λ = 51.69 cm
up = 0.52*10^8 m/s
ηc = 39.54 + j 31.72 Ω
Wood: Good conductor
α = 6.3*10^-4 Np/m
β = 6.3*10^-4 Np/m
λ = 10 km
up = 0.1*10^8 m/s
ηc = 6.28*( 1 + j )
Explanation:
Given:
Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz
Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.
Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz
Find:
Determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate α, β, λ, up, and ηc:
Solution:
- We need to determine the loss tangent to determine category of the medium as follows:
σ / w*εr*εo
Where, w is the angular speed of wave
εo is the permittivity of free space = 10^-9 / 36*pi
- Now we classify as follows:
- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.
Glass: Low-Loss dielectric
Tissue: Quasi-Conductor
Wood: Good conductor
- Now we will use categorized material base equations from Table 17-1 as follows:
Glass: Low-Loss dielectric
α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)
α = 8.42*10^-11 Np/m
β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)
β = 468.3 rad/m
λ = 2*pi / β = 2*pi / 468.3
λ = 1.34 cm
up = λ*f = 0.0134*10^10
up = 1.34*10^8 m/s
ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)
ηc = 168.5 Ω
Tissue: Quasi-Conductor
α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)
α = 9.75 Np/m
β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)
β = 12.16 rad/m
λ = 2*pi / β = 2*pi / 12.16
λ = 51.69 cm
up = λ*f = 0.5169*100*10^6
up = 0.52*10^8 m/s
ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5
ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5
ηc = 39.54 + j 31.72 Ω
Wood: Good conductor
α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )
β = α = 6.3*10^-4 Np/m
λ = 2*pi / β = 2*pi / 6.3*10^-4
λ = 10 km
up = λ*f = 10,000*1*10^3
up = 0.1*10^8 m/s
ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4
ηc = 6.28*( 1 + j )
Answer:
Check the explanation
Explanation:
#include <iostream>
using namespace std;
void insert(int* arr, int* size, int value, int position){
if(position<0 || position>=*size){
cout<<"position is greater than size of the array"<<endl;
return ;
}
*size = *size + 1 ;
for(int i=*size;i>position;i--){
arr[i] = arr[i-1];
}
arr[position] = value ;
}
void print(int arr[], int size){
for(int i=0;i<size;i++){
cout<< arr[i] <<" ";
}
cout<<" "<<endl;
}
void remove(int* arr, int* size, int position){
* size = * size - 1 ;
for(int i=position;i<*size;i++){
arr[i] = arr[i+1];
}
}
int count(int arr[], int size, int target){
int total = 0 ;
for(int i=0;i<size;i++){
if(arr[i] == target)
total += 1 ;
}
return total ;
}
int main()
{
int size;
cout<<"Enter the initial size of the array:";
cin>>size;
int arr[size],val;
cout<<"Enter the values to fill the array:"<<endl;
for(int i=0;i<size;i++){
cin>>val;
arr[i] = val ;
}
int choice = 5,value,position,target ;
do{
cout<<"Make a selection:"<<endl;
cout<<"1) Insert"<<endl;
cout<<"2) Remove"<<endl;
cout<<"3) Count"<<endl;
cout<<"4) Print"<<endl;
cout<<"5) Exit"<<endl;
cout<<"Choice:";
cin>>choice;
switch(choice){
case 1:
cout << "Enter the value:";
cin>>value;
cout << "Enter the position:";
cin>>position;
insert(arr,&size,value,position);
break;
case 2:
cout << "Enter the position:";
cin>>position;
remove(arr,&size,position);
break;
case 3:
cout<<"Enter the target value:";
cin>>target;
cout <<"The number of times "<<target<<" occured in your array is:" <<count(arr,size,target)<<endl;
break;
case 4:
print(arr,size);
break;
case 5:
cout <<"Thank you..."<<endl;
break;
default:
cout << "Invalid choice..."<<endl;
}
}while(choice!=5);
return 0;
}
Kindly check the attached images below for the code output.
