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miv72 [106K]
3 years ago
15

State three active materials of a lead acid cell​

Engineering
2 answers:
igomit [66]3 years ago
6 0

Answer:

lead dioxide,sulfate and lead acid

Karolina [17]3 years ago
3 0
Lead, dioxide, solfate and lead acid are correct
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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();

     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();

     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
What is ductile to brittle transition temperature (DBTT)?
vova2212 [387]

Answer: The answer is given below

Explanation:

Ductile to brittle transition temperature (DBTT) is the temperature when there is a pronounced reduction in the ability of a material to absorb force without being fractured.

It shuld be noted that when this happens, the material transitions from ductile to brittle which is known as ductile to brittle transition temperature(DBTT).

DBTT

7 0
3 years ago
Which of the following are true about algorithms? (You may select more than one)
r-ruslan [8.4K]
Have a orderf. Fr

R
F
F
D
Djeiiekee
7 0
4 years ago
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In poor weather, you should __ your following distance
Ratling [72]

Answer:

I think reduce your following distance

5 0
4 years ago
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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0
3 years ago
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