The centre of the circumstancribing circle of a triangle can be found by using the

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T

anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × $10^{4}$ ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates = 0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = $V_o \times e^{t\RC}$ ...........1

so R will be

R = $\frac{t}{C\times ln(\frac{V_o}{V})}$ ....................2

put here value

so for t min 11.5 μs

R = $\frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}$

R min = 28.173 ohm

and

for t max 6.33 ms

R max = $\frac{6.33}{11.5} \times 28.173$

R max = 1.55 × $10^{4}$ ohm

4 0

2 years ago

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

Are the wooden pillars shown in the image below, a dead load?

Brrunno [24]

Answer:

no

Explanation:

it's not a dead load because when load is put on the pillars it's not fully straining it's been slowly getting to be heavier in that period of time before it falls

6 0

3 years ago

What is the mass of a brass axle that has a volume of 0.318 cm?

NeX [460]

Answer:

2.7g

Explanation:

the mass of a brass axle that has a volume of 0.318 cm is 2.7g.

8 0

2 years ago

A vacuum gage connected to a tank reads 30 kPa at a location where the barometric reading is 755 mmHg. Determine the absolute pr

navik [9.2K]

Answer:

Absolute pressure=70.72 KPa

Explanation:

Given that Vacuum gauge pressure= 30 KPa

Barometer reading =755 mm Hg

We know that barometer always reads atmospheric pressure at given situation.So atmospheric pressure is equal to 755 mm Hg.

We know that P= ρ g h

Density of $Hg=13600 \frac{kg}{m^3}$

So P=13600 x 9.81 x 0.755

P=100.72 KPa

We know that

Absolute pressure=atmospheric pressure + gauge pressure

But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.

Absolute pressure=atmospheric pressure + gauge pressure

Absolute pressure=100.72 - 30 KPa

So

Absolute pressure=70.72 KPa

8 0

2 years ago