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amm1812
3 years ago
8

In many problems where the potential energy is considered between two objects, the potential energy is defined as zero when the

particles are in infinite distance apart. Using the definition of potential energy, explain why the potential energy would be positive if the force between the particles is repulsive and negative if the force between the particles is attractive for noninfinite distances.
Engineering
1 answer:
algol [13]3 years ago
4 0

Answer:

potential energy=ε Qq/r. Putting values of charge along with sign and calculating the potential energy will tell whether the resultant force is attractive or repulsive.

Explanation:

potential energy=ε Qq/r

q and Q are charges on each particle

So in case the particles have attractive force between them, one particle will be negative and the other will be positiive. The result will be negative.

In case the particles have repulsive forces, the charge on both particles will be either positive or negative. The result will be positive

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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);

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}

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));

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3 0
3 years ago
Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t
raketka [301]

Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

Average shear = -7.5 MPa

Explanation:

Given

A = 45

B = -60

C = 30

a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

P1 = 53 MPa

Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

P1 = -68 MPa

Tan 2a = C/{(A-B)/2}

Tan 2a = 30/(45+60)/2

a = 14.87 degree

Principal stress

p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

b) Shear stress in plane

Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

Average = (45-(-60))/2 = -7.5 MPa

5 0
3 years ago
I am standing on the upper deck of the football stadium. I have an egg in my hand. I am going to drop it and you are going to tr
Alina [70]

Answer:

Δx = 25 ft.

Explanation:

Assuming that the person on the ground starts running at the same time as the egg is dropped, we have two simultaneous trajectories:

1 ) Egg falling:

If the egg is dropped, and we neglect the air resistance, we can use the kinematic equation that relates the distance and fall time, as follows:

yf-y₀ = 1/2* g* t²

If we take the up direction as positive, we can solve for t as follows:

0-100 ft = 1/2* (-32.15 ft/s²)* t²

⇒ t = \sqrt{(100*2)/32.15} = 2.5 sec.

2) Person on the ground running away:

In order to be able to run away, and then return to catch the egg, running at constant speed, he must run during exactly the half of the time that the egg is falling, i.e., 1.25 sec.

We can get the distance at which he can reach, applying the definition of velocity:

v = (xf-x₀) / (tfi-t₀)

If we choose t₀=0 and x₀ = 0 , we can solve for xf, as follows:

xf = v*t = 20 ft/sec*1.25 sec = 25 ft.

8 0
3 years ago
Which of the following is not an electronic device ?
PolarNik [594]

Answer:

B

Explanation:

it's does not transmit any energy

6 0
2 years ago
Earth completes one full ____ on its axis every 24 hours
mars1129 [50]

Answer:

rotation

Explanation:

I just answered this!

7 0
3 years ago
Read 2 more answers
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