Which is an example of a passive solar energy system

At the beginning of last year, tarind corporation budgeted $1,000,000 of fixed manufacturing overhead and chose a denominator le

maxonik [38]

The tari's total standard machine-hours allowed for last year's output 600,000 hours

Budgeted at start of year: $100,000 fixed manufacturing overhead for 500,000 machine hours

Standard = $100,000 / 500,000 hours = $0.2 fixed overhead / maching hour

At end of year, manufacturing overhead volume was $20,000 favorable which means

$20000 / $1=0.2 = 100000 additional hours.

Total Standard Machine Allowance Allowed for output = 500,000 + 100,000 = 600,000 hours.

the use of one machine running for an hour as a basis for cost estimation and operating effectiveness evaluation. In order to determine the contribution margin per machine hour for a specific product: a. Total cost per unit divided by the quantity of machine hours required to produce each unit of the target product. The manufacturing overhead cost divided by the activity driver yields the predetermined overhead rate. For instance, if machine hours were the activity driver, you would divide overhead costs by the anticipated number of machine hours.

Learn more about machine hours here:

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7 0

1 year ago

A 5000-lb truck is being used to lift a 1000-lb boulder B that is on a 200-lb pallet A. Knowing that the truck starts from rest

artcher [175]

Answer:

Explanation:

Total weight being moved = 5000+1000+200

= 6200 lb .

Force applied = 700 lb

= 700 x 32 = 22400 poundal .

acceleration (a) = 22400 / 6200

= 3.613 ft /s²

To know velocity after 6 ft we apply the formula

v² = u² + 2as

v² = 0 + 2 x 3.613 x 6

43.356

v = 6.58 ft/s

4 0

3 years ago

The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc

madreJ [45]

Answer:

Option B

$116 ft/s^{2}$

Explanation:

$\theta=2 rev=2(2\pi)=4\pi$

$\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-4^{2})$

$\omega_f=8.14 rads/s$

$v=r\omega=1.75*8.14=14.245 ft/s$

Centripetal acceleration $=\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}$

Tangential component=dr=2*1.75=3.5

$Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}$

5 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

A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the

tamaranim1 [39]

Answer:

Explanation:

Given data in question

mean stress = 50 MPa

amplitude stress = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first maximum stress and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

$\sigma _{m}$ = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2

50 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........1

and

225 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........2

from eqution 1 and 2 we get maximum and minimum stress

$\sigma _{maximum}$ = 275 MPa ............3

and $\sigma _{minimum}$ = -175 MPa ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio = $\sigma _{minimum}$ / $\sigma _{maximum}$

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = $\sigma _{maximum}$ - $\sigma _{minimum}$

magnitude = 275 - (-175) = 450 MPa

3 0

3 years ago