What is the minimum clamp time for gluing a panel?

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 0.50 m3 drum was filled with 0.49 m3 of liquid water at 25oC and the remaining volume was water vapor without any air. The dru

timurjin [86]

Answer:

There is not going to be pressure build up in the system,that is isobaric process.

Explanation:

Assumptions to be made

1. No mass is gained or lost during the heating process.

2. There are no friction losses,so work is transmitted efficiently.

3. It was started the water in the drum and its surrounding have same temperature.

4. This system is closed,so there is no mass transfer across its boundaries.

5 0

3 years ago

An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit

kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²) ......................1

here ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99 .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k × ( $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$ )

put here value

0.005 = 120/k × ( $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$ )

k ×∈ × √(1-2∈²) = 1200 ......................3

so by equation 3 and 2

$\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}$

$\varepsilon^{2} - \varepsilon^{4} = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}$

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

7 0

3 years ago

12.28 LAB: Output values in a list below a user defined amount - functions Write a program that first gets a list of integers fr

Anastaziya [24]

Answer:

def output_ints_less_than_or_equal_to_threshold(user_values, upper_threshold):

for value in user_values:

if value < upper_threshold:

print(value)

def get_user_values():

n = int(input())

lst = []

for i in range(n):

lst.append(int(input()))

return lst

if __name__ == '__main__':

userValues = get_user_values()

upperThreshold = int(input())

output_ints_less_than_or_equal_to_threshold(userValues, upperThreshold)

8 0

3 years ago

Water is stored in a tank which has vent open to the atmosphere. The water level is 1.0 m below the top the tank and the water i

sp2606 [1]

Answer:

6.99 x 10⁻³ m³ / s

Explanation:

Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.

h = difference in the height of water column at two ends of delivery pipe

6 - 1 = 5 m

Velocity of flow of water

v = √2gh

= √ (2 x 9.8 x 5)

= 9.9 m /s

Volume of water flowing per unit time

velocity x cross sectional area

= 9.9 x 3.14 x .015²

= 6.99 x 10⁻³ m³ / s

7 0

3 years ago