Ask question

Ask question

All categories

3 years ago

15

Name eight safety electrical devices including their functions and effects if not present.

1 answer:

madam [21]3 years ago

3 0

Answer:

Electricity has two hazards. A thermal hazard occurs when there is electrical overheating. A shock hazard occurs when an electric current passes through a person. Both hazards have already been discussed. Here we will concentrate on systems and devices that prevent electrical hazards. Figure 1 shows the schematic for a simple AC circuit with no safety features. This is not how power is distributed in practice. Modern household and industrial wiring require the three-wire system, shown schematically in Figure 2, which has several safety features. First is the familiar circuit breaker (or fuse) to prevent thermal overload. Second, there is a protective case around the appliance, such as a toaster or refrigerator. The case’s safety feature is that it prevents a person from touching exposed wires and coming into electrical contact with the circuit, helping prevent shocks.

Hope this helps! : )

You might be interested in

valkas [14]

Answer:

Check the 2nd, 3rd and 4th statements.

Explanation:

4 0

3 years ago

Which of the followong parts does not rotate during starter operation? A. Commutator segments B. Armature windings c. Field wind

photoshop1234 [79]

Answer: B

Explanation: unless newer models added wingding to code inside fused computer...wingdings on a window ...not a motor

8 0

3 years ago

Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user

Thepotemich [5.8K]

Answer:

import java.util.Scanner;

public class FindMatchValue {

public static void main (String [] args) {

Scanner scnr = new Scanner(System.in);

final int NUM_VALS = 4;

int[] userValues = new int[NUM_VALS];

int i;

int matchValue;

int numMatches = -99; // Assign numMatches with 0 before your for loop

matchValue = scnr.nextInt();

for (i = 0; i < userValues.length; ++i) {

userValues[i] = scnr.nextInt();

}

/* Your solution goes here */

numMatches = 0;

for (i = 0; i < userValues.length; ++i) {

if(userValues[i] == matchValue) {

numMatches++;

}

}

System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

}

}

8 0

3 years ago

If a 9V battery produces a current of 3 A through a load, what is the resistance of the load

Elden [556K]

3 ohms hope this helps :D ❤

7 0

3 years ago

Read 2 more answers

A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

3 years ago