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2 years ago

Explain three examples of workshop

Engineering

2 answers:

stealth61 [152]2 years ago

7 0

Answer:

Invitational Workshop

An invitational workshop is what many of us know. It’s what Lucy Calkins has made famous through the Reading and Writing Workshop. In the invitational workshop, the instructor usually hosts a minilesson. This minilesson is intended to meet the needs of the majority of children in the classroom. Afterward, the children are “invited” to employ the skills or strategy for the minilesson during workshop time, where students work independently or in small groups

Explanation:

Wewaii [24]2 years ago

6 0

Answer:

Here are 5 examples:

Invitational Workshop
Constructivist Workshop
Reflection Workshop
Conferencing Workshop
Choice Workshop

Explanation:

I hope this helps!

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3 years ago

THE MASS FOR OBJECT 1 is 107.01 grams what is the objects force in Newton’s answer please

Aleksandr [31]

Given that,

Mass of the object 1, m = 107.01 grams

To find,

Force on the object.

Solution,

The force acting on the object is gravitational force. The force is given by the formula as follow :

F = mg

g is acceleration due to gravity

F = 0.10701 kg × 9.8 m/s²

F = 1.048 N

So, the force acting on object 1 is 1.048 N.

5 0

3 years ago

Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's

pochemuha

Answer:

import java.util.Scanner;

public class PeopleWeights {

public static void main(String[] args) {

Scanner reader = new Scanner(System.in);

double weightOne = reader.nextDouble();

System.out.println("Enter 1st weight:");

double weightTwo = reader.nextDouble();

System.out.println("Enter 2nd weight :");

double weightThree = reader.nextDouble();

System.out.println("Enter 3rd weight :");

double weightFour = reader.nextDouble();

System.out.println("Enter 4th weight :");

double weightFive = reader.nextDouble();

System.out.println("Enter 5th weight :");

double sum = weightOne + weightTwo + weightThree + weightFour + weightFive;

double[] MyArr = new double[5];

MyArr[0] = weightOne;

MyArr[1] = weightTwo;

MyArr[2] = weightThree;

MyArr[3] = weightFour;

MyArr[4] = weightFive;

System.out.printf("You entered: " + "%.1f %.1f %.1f %.1f %.1f ", weightOne, weightTwo, weightThree, weightFour, weightFive);

double average = sum / 5;

System.out.println();

System.out.println("Total weight: " + sum);

System.out.println("Average weight: " + average);

double max = MyArr[0];

for (int counter = 1; counter < MyArr.length; counter++){

if (MyArr[counter] > max){

max = MyArr[counter];

}

System.out.println("Max weight: " + max);

}

import java.util.Scanner;

public class PeopleWeights {

public static void main(String[] args) {

Scanner reader = new Scanner(System.in);

double weightOne = reader.nextDouble();

System.out.println("Enter 1st weight:");

double weightTwo = reader.nextDouble();

System.out.println("Enter 2nd weight :");

double weightThree = reader.nextDouble();

System.out.println("Enter 3rd weight :");

double weightFour = reader.nextDouble();

System.out.println("Enter 4th weight :");

double weightFive = reader.nextDouble();

System.out.println("Enter 5th weight :");

double sum = weightOne + weightTwo + weightThree + weightFour + weightFive;

double[] MyArr = new double[5];

MyArr[0] = weightOne;

MyArr[1] = weightTwo;

MyArr[2] = weightThree;

MyArr[3] = weightFour;

MyArr[4] = weightFive;

System.out.printf("You entered: " + "%.1f %.1f %.1f %.1f %.1f ", weightOne, weightTwo, weightThree, weightFour, weightFive);

double average = sum / 5;

System.out.println();

System.out.println("Total weight: " + sum);

System.out.println("Average weight: " + average);

double max = MyArr[0];

for (int counter = 1; counter < MyArr.length; counter++){

if (MyArr[counter] > max){

max = MyArr[counter];

}

System.out.println("Max weight: " + max);

}

8 0

3 years ago

Read 2 more answers

A laboratory in the Y building keep a vacuum pressure of 0.1 kPa abs. What is the net force acting on the door considering the a

seropon [69]

Answer:

net force acting on the floor is 100 kN

Explanation:

Given data:

$P_{vaccum} = 0.1 kPa$

$P_{atm} = 101.325 kPa$

dimension of floor = 2 m \times 0.5 m

we know that

Net force can be calculated as follow

$f_{net} = P_{vaccum} \times area$

$f_{net} = 0.1\times 10^3 \times 2\times 0.5$

$f_{net} = 0.1\times 10^3 \times 1$

$f_{net} = 100 kN$

Therefore net force acting on the floor is 100 kN

7 0

3 years ago