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sesenic [268]
2 years ago
14

What is the magnitude of the speed of a

Engineering
1 answer:
Mars2501 [29]2 years ago
6 0

The equatorial diameter of the earth is 12,756 km.

Therefore the equatorial radius is

r = 12756/2 = 6378 km.

The angular velocity of the earth is  

ω = (2π radians)/(24 hours) = 0.2618 rad/h

The tangential velocity of a person at the equator is

v = rω

  = (6378 km)*( 0.2618  rad/h)

  =  1670 km/h

Answer: 1670 km/h

I hope this helps you.

Brainly, thanks you, rating are all appreciated greatly!

Smile and have an amazing day ; )

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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
3 years ago
A thin metal disk of mass m=2.00 x 10^-3 kg and radius R=2.20cm is attached at its center to a long fiber. When the disk is turn
topjm [15]

Answer:

k = 1.91 × 10^-5 N m rad^-1

Workings and Solution to this question can be viewed in the screenshot below:

6 0
3 years ago
AC motor characteristics require the applied voltage to be proportionally adjusted by an AC drive whenever the frequency is chan
Margarita [4]
The answer is false
6 0
3 years ago
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A) For Well A, provide a cross-section sketch that shows (i) ground elevation, (ii) casing height, (iii) depth to
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Don’t go on that file will give a virus! Sorry just looking out and I don’t know how to comment!
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3 years ago
Your family has asked you to estimate the operating costs of your clothes dryer for the year. The clothes dryer in your home has
trasher [3.6K]

Answer:

The costs to run the dryer for one year are $ 9.03.

Explanation:

Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:

1 watt = 0.001 kilowatt

2250/45 = 50 watts per minute

45 x 365 = 16,425 / 60 = 273.75 hours of consumption

50 x 60 = 300 watt = 0.3 kw / h

0.3 x 273.75 = 82.125

82.125 x 0.11 = 9.03

Therefore, the costs to run the dryer for one year are $ 9.03.

8 0
2 years ago
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