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siniylev [52]
3 years ago
9

Which of the following is not a conductor of electricity.

Engineering
2 answers:
Usimov [2.4K]3 years ago
4 0

Answer:

Rubber

Explanation:

Rubber is not a conductor of electricity

Nadusha1986 [10]3 years ago
4 0
3. Rubber because electricity can’t travel through it
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Answer:

Power required to overcome aerodynamic drag is 50.971 KW

Explanation:

For explanation see the picture attached

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Create two arrays with 5 elements each: one will hold Strings and the second will hold integers. Write a program to ask the user
MrMuchimi

Answer:

#include <iostream>

#include <iomanip>

#include <string>

using namespace std;

int main() {

   string name[5];  

   int age[5];  

   int i,j;  

   for ( i = 0; i<=4; i++ ) {  

       cout << "Please enter student's name:";  

       cin >> name[i];  

       cout << "Please enter student's age:";  

       cin >> age[i];          

   }  

for (i=0;i<=4;i++){

   cout<<"Age of  "<< name[i]<<"  is  "<<age[i]<<endl;  

}

}

Output of above program is displayed in figure attached.

5 0
3 years ago
1. A _____ is applied to a wall or roof rafters to add strength and keep the building straight and plumb.
PSYCHO15rus [73]

Answer:

A force must s applied to a wall or roof rafters to add strength and keep the building straight and plumb

7 0
3 years ago
Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 10mA. The zener has an incremental resistance of rZ = 30
hram777 [196]

Answer:

The answer is given in the explanation.

Explanation:

The circuit is as indicated in the attached figure.

From the analytical description the zener voltage is given as

V_z=V_z_o+I_zr_z

Here

Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.

The equivalent model is shown in the attached figure.

From the above equation, Vzo is calculated as

V_z_o=V_z-I_zr_z

Here Vz is given as 7.5 V

Iz is given as 10 mA

rz is given as 30 Ω

Thus the Vzo is given as

V_z_o=V_z-I_zr_z\\V_z_o=7.4-30*10*10^{-3}\\V_z_o=7.5-0.3\\V_z_o=7.2 V

The value of I_L is given as 5 mA

Now the expression of current is as

I=I_z+I_L\\I=10mA+5mA\\I=15 mA

Now the resistance is calculated as

R=\dfrac{V-Vo}{I}\\R=\dfrac{10-7.2}{15*10^{-3}}\\R=186.66

So the value of resistance is 186.66 Ω.

Considering the supply voltage is increased by 10%

V is 10-10%*10=10+1=11 so the

R=\dfrac{V-Vo}{I}\\186.66=\dfrac{11-V_o}{15*10^{-3}}\\V_o=8.2 V

Considering the supply voltage is decreased by 10%

V is 10-10%*10=10-1=9 so the

R=\dfrac{V-Vo}{I}\\186.66=\dfrac{9-V_o}{15*10^{-3}}\\V_o=6.2 V

Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA

so

R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{11-V_o}{10*10^{-3}}\\V_o=9.13 V

Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus

V_z_o=V_z-I_zr_z\\V_z_o=7.5-30*0.5*10^{-3}\\V_z_o=7.5-0.015\\V_z_o=7.485 V

R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{9-7.485}{I}\\I=10.71 mA

The load voltage is 7.485 V

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