1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mezya [45]
3 years ago
11

Sarah and Raj take/takes me to a baseball game every year.

Engineering
1 answer:
timofeeve [1]3 years ago
4 0

Answer:

take

play

are

sits

crouches

direct

indirect

direct

direct

indirect

Explanation:

i had just taken the test

You might be interested in
When you are configuring data deduplication, you must choose a usage type for the volume you are configuring. Which of the follo
Finger [1]

Answer:

1. General purpose file server.

2. Virtual Desktop Infrastructure Server.

3. Virtualized Backup Server.

All except : Database server

Explanation:

As a simple definition, we can tell, data deduplication is an elimination of redundant data in data set and storing only one copy of the same data. It is done by identifying double byte patterns through data analysis, removing double data and replacing it with reference pointed to stored, single piece of data.

6 0
3 years ago
A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a
Xelga [282]

Answer:

Explanation:

Given the data in the question;

L = 46 in

Ga = 5 × 10³ ksi

Gs = 11 × 10³ ksi

Outside diameter da = 5 in

ds = 4 in

Tb = 3 kip.in

Now,

Ja = polar moment of Inertia of Aluminum;

Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 ) in^u

Js = polar moment of inertia of steel

Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )

Ta is torque transmitted by Aluminum  

Ts is torque transmitted by steel  

{composite member }

T = Ta + Ts ------ let this be equation m1

Now, we use the relation;

T/J = G∅/L

JG∅ = TL

∅ = TL/GJ

so, for aluminum rod ∅_{alu = TaLa/GaJa

for steel rod ∅_{steel = TsLs/GsJs

but we know that, ∅a = ∅s = ∅_B

so

[TaLa/GaJa]  =  [TsLs/GsJs]

also, we know that, La = Ls = L

∴ [Ta/GaJa]  =  [Ts/GsJs]

we solve for Ta

TaGsJs = TsGaJa  

Ta = TsGaJa / GsJs

we substitute

Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]

Ta = 0.66Ts

now, we substitute 0.66Ts for Ta and 3 for T in equation 1

T = Ta + Ts

3 = 0.66Ts + Ts

3 = 1.66Ts

Ts = 3 / 1.66

Ts = 1.8072 ≈ 1.81 kip-in

so

∅_{steel = TsLs / GsJs

we substitute

∅_{steel = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )

∅_{steel = 83.26 / 276460.1535

∅_{steel  = 0.000301

∅_{steel = 3.01 × 10⁻⁴ rad

so

∅_{steel = ∅_B = 3.01 × 10⁻⁴ rad

Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴  rad

5 0
2 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
2 years ago
A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to
Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 =  (16/3) * 10^2 V

c)  E = 64/15 J

d)  dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the final potential difference across each capacitor

(c) the final energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

                           Q_initial = C_1*V

                           Q_initial = 20*10^-6 * 800

                           Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

                          Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

                          V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

                          Q_2 = Q_1*C_2/C_1

                          Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

                          Q_initial = 1.5*Q_1

                          Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

                          V_2 = V_1 = Q_1 / C_1  

                                            = 10.67*10^-3 / 20*10^-6

                                            = (16/3) * 10^2 V

c)

- The energy in the system is:

                          E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

                           E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

                          E = 64/15 J

d)

- The decrease in energy of the capacitors is:

                           dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

                          dE = 0.5*10^-6*20*800^2 - (64/15)

                          dE = 32/5 - 64/15 = 32/15 J

5 0
3 years ago
What type of car engine is best for cold weather.
Komok [63]

Answer:Antifreeze/coolant

Explanation: keeps your engine cool in warm weather and keeps it from freezing up in the winter. A 50-50 mix of full strength coolant and water generally protects to around -30 degrees Fahrenheit. Make sure you check with the supplier or your owner's manual for the correct formulation

5 0
2 years ago
Other questions:
  • The moisture in hot, humid, stagnant air surrounding a cold-water pipeline continually diffuses to the cold surface where it con
    14·1 answer
  • The net potential energy EN between two adjacent ions, is sometimes represented by the expression
    13·1 answer
  • What is an ip<br> Number
    12·1 answer
  • What is the Principle of Entropy Increase?
    9·1 answer
  • Design an Armstrong indirect FM modulator to generate an FM signal with a carrier frequency 98.1 MHz and a frequency deviation △
    15·1 answer
  • Importance of civil engineering in nepal?​
    10·1 answer
  • Lets try to get to 100 sub before charismas day <br> Jordan Gracia 32 sub and 5 videos
    13·2 answers
  • 6
    10·2 answers
  • Deviations from the engineering drawing cannot be made without the approval of the
    15·2 answers
  • It is possible to design a reactor where the scy conductor and the nitrogen/ammonia electrode operate at different temperatures.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!