All categories

2 years ago

In a case where electrical current leakage from the circuit occurs, the GFCI would do the following:

Engineering

1 answer:

Jobisdone [24]2 years ago

5 0

Answer:

Sense the current leakage

Trip the circuit

Cut off the electricity

Explanation:

The correct order is -

Sense the current leakage

Trip the circuit

Cut off the electricity

Reason -

First step has to be taken to sense the electric current leakage without it being detected, it cant be dealt with.

The second step consists of tripping the circuit so at to remove the flow of electricity.

The third step is to Cut the electricity off so to stop the leakage and be safe.

You might be interested in

JAVA HADOOP MAPREDUCE

taurus [48]

Answer:

Explanation:

package PackageDemo;

import java.io.IOException;

import org.apache.hadoop.conf.Configuration;

import org.apache.hadoop.fs.Path;

import org.apache.hadoop.io.IntWritable;

import org.apache.hadoop.io.LongWritable;

import org.apache.hadoop.io.Text;

import org.apache.hadoop.mapreduce.Job;

import org.apache.hadoop.mapreduce.Mapper;

import org.apache.hadoop.mapreduce.Reducer;

import org.apache.hadoop.mapreduce.lib.input.FileInputFormat;

import org.apache.hadoop.mapreduce.lib.output.FileOutputFormat;

import org.apache.hadoop.util.GenericOptionsParser;

public class WordCount {

public static void main(String [] args) throws Exception

{

Configuration c=new Configuration();

String[] files=new GenericOptionsParser(c,args).getRemainingArgs();

Path input=new Path(files[0]);

Path output=new Path(files[1]);

Job j=new Job(c,"wordcount");

j.setJarByClass(WordCount.class);

j.setMapperClass(MapForWordCount.class);

j.setReducerClass(ReduceForWordCount.class);

j.setOutputKeyClass(Text.class);

j.setOutputValueClass(IntWritable.class);

FileInputFormat.addInputPath(j, input);

FileOutputFormat.setOutputPath(j, output);

System.exit(j.waitForCompletion(true)?0:1);

}

public static class MapForWordCount extends Mapper<LongWritable, Text, Text, IntWritable>{

public void map(LongWritable key, Text value, Context con) throws IOException, InterruptedException

{

String line = value.toString();

String[] words=line.split(",");

for(String word: words )

{

Text outputKey = new Text(word.toUpperCase().trim());

IntWritable outputValue = new IntWritable(1);

con.write(outputKey, outputValue);

}

public static class ReduceForWordCount extends Reducer<Text, IntWritable, Text, IntWritable>

{

public void reduce(Text word, Iterable<IntWritable> values, Context con) throws IOException, InterruptedException

{

int sum = 0;

for(IntWritable value : values)

{

sum += value.get();

}

con.write(word, new IntWritable(sum));

}

3 0

2 years ago

This is hard please help me you will give brainlist

Musya8 [376]

It spirals clockwise

5 0

2 years ago

Read 2 more answers

Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?

Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

x_average = ∑ $x_{i}$ / n

The tolerance or error is the current value over the mean value per 100

Δx₁ = x₁ / x_average

tolerance = | 100 -Δx₁ 100 |

bars indicate absolute value

let's look for these values for each case

x_average = (2.1700000+ 2.258571429) / 2

x_average = 2.2142857145

fluctuation for x₁

Δx₁ = 2.17000 / 2.2142857145

Tolerance = 100 - 97.999999991

Tolerance = 2.000000001%

fluctuation x₂

Δx₂ = 2.258571429 / 2.2142857145

Δx2 = 1.02

tolerance = 100 - 102.000000009

tolerance 2.000000001%

x_average = (2.2 + 2.29) / 2

x_average = 2,245

fluctuation x₁

Δx₁ = 2.2 / 2.245

Δx₁ = 0.9799554

tolerance = 100 - 97,999

Tolerance = 2.00446%

fluctuation x₂

Δx₂ = 2.29 / 2.245

Δx₂ = 1.0200445

Tolerance = 2.00445%

x_average = (2.211445 +2.3) / 2

x_average = 2.2557225

Δx₁ = 2.211445 / 2.2557225 = 0.9803710

tolerance = 100 - 98.0371

tolerance = 1.96%

Δx₂ = 2.3 / 2.2557225 = 1.024624

tolerance = 100 -101.962896

tolerance = 1.96%

x_average = (2.20144927 + 2.29130435) / 2

x_average = 2.24637681

Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

tolerance = 100 - 98.000043

tolerance = 2.000002%

Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

tolerance = 2.0000002%

x_average = (2 +2,3) / 2

x_average = 2.15

Δx₁ = 2 / 2.15 = 0.93023

tolerance = 100 -93.023

tolerance = 6.98%

Δx₂ = 2.3 / 2.15 = 1.0698

tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0

2 years ago

den301095 [7]

Jae pain seems the most off

4 0

2 years ago

Shortly after the introduction of a new coin, newspapers published articles claiming the coin is biased. The stories were based

garik1379 [7]

Answer:

(a) 0.12924

(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

Explanation:

(a)

n=200 for fair coin getting head, p= 0.5

Expectation = np =200*0.5=100

Variance = np(1 - p) = 100(1-0.5)=100*0.5=50

Standard deviation, $s = \sqrt {variance}=\sqrt {50}= 7.071068$

Z value for 108, $z =\frac {108-100}{7.071068}= 1.131371$

P( x ≥108) = P( z >1.13)= 0.12924

(b)

Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

3 0

2 years ago