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V125BC [204]
2 years ago
11

Traveling a average speed at 55mph how many miles will you travel in9 hours

Engineering
1 answer:
Katarina [22]2 years ago
6 0
You will go 495 miles in 9 hours
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What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C
Anvisha [2.4K]

Answer:

331809.5gallon/hr or 92.16gallon/s

Explanation:

What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C

convert 9.4 acre to inches we have=5.896*10^7

How to calculate Peak runoff discharge

1. take the dimension of the roof

2. multiply the dimension by the n umber of inches of rainfall

3. Divide by 231 to get gallon equivalence (because 1 gallon = 231 cubic inches)

5.896*10^7*1.3

7.66*10^7 cubic inches/hr

1 gallon=231 cubic inches

7.66*10^7 cubic inches=331809.5gallon/hr or 92.16gallon/s

this is gotten by converting 1 hr to seconds

331809.5gallon/hr /3600s=92.16gallon/s

8 0
3 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
3 years ago
Which of the following leaf modifications is beneficial in a low-light environment, e.g., the understory of a rainforest? (More
Mnenie [13.5K]

Answer: A, B and D.

Explanation:

A) If a leaf is iridescent and reflects light or colors from other objects, it helps to stand out from the rest in a low-light environment.

B) The anthocyanin pigments give the leaf a different coloration, so it has a different impact in a darker o flat colored space.

D) Having a large surface area will help to stand out from plants with small leaves.

5 0
3 years ago
A forklift raises a 90.5 kg crate 1.80 m. (a) Showing all your work and using unity conversion ratios, calculate the work done b
klasskru [66]

Answer:

(a) Work done is 1.59642 kJ

(b) Useful power supplied = 0.1298 kW

Explanation:

(a) Work done = mass of crate × acceleration due to gravity × distance = 90.5 kg × 9.8 m/s^2 × 1.8 m = 1596.42 kgm^2/s^2 × 1 J/1 kgm^2/s^2 × 1 kJ/1000 J = 1.59642 kJ

(b) Useful power supplied = work done ÷ time = 1.59642 kJ ÷ 12.3 s = 0.1298 kJ/s × 1 kW/1 kJ/s = 0.1298 kW

5 0
3 years ago
Read 2 more answers
According to the department of labor. Installing electrical systems in a new construction is less complicated than maintaining e
kobusy [5.1K]

<u>Explanation:</u>

This is the case because installing electrical systems in a new construction does not involve spending time identifying problems in the electrical wiring, but an electrician maintaining electrical equipment in an existing structure first needs to;

  • identify the problems, next
  • spend time-solving the identified problems.

8 0
3 years ago
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