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Liula [17]
3 years ago
11

Different Gauss quadrature formulae predict different values for the same integral a. True b. False

Engineering
1 answer:
scoray [572]3 years ago
5 0

Answer: False

Explanation:

The given statement are false as, the gauss quadrature predicted same value only when their is a minute error at one point and two point rule. It basically given the more precise answer and the answer are only changed by the decimal place. The gauss quadrature method are basically used for calculating the certain integral.

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) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p
andrey2020 [161]

Answer:

number of pulses produced =  162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

circumference = 314.16 mm

so now we get number of pulses produced

number of pulses produced = \frac{linear\ displacement}{circumference} × No of pulses per revolution .................2

number of pulses produced = \frac{200}{314.16} × 256

number of pulses produced =  162 pulses

5 0
3 years ago
Two balls are chosen randomly from an urn containing 8 white 4 black, and orange balls. Suppose that we win $ 2 for each black b
Scorpion4ik [409]
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8 0
3 years ago
What is electricity defined as
antoniya [11.8K]

Answer:

Exact answer: a form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current.

4 0
3 years ago
Which of the following explains why material properties present challenges for engineers?
Maurinko [17]

Answer:

Explanation:

They are altered by variables such as temperature hence making materials challenging when dealing with them.

4 0
4 years ago
import java.util.Scanner; public class FindSpecialValue { public static void main (String [] args) { Scanner scnr = new Scanner(
Hitman42 [59]

Answer:

Java program explained below

Explanation:

FindSpecialNumber.java

import java.util.Scanner;

public class FindSpecialNumber {

public static void main(String[] args) {

//Declaring variable

int number;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

//getting the input entered by the user

System.out.print("Enter a number :");

number = sc.nextInt();

/* Based on user entered number

* check whether it is special number or not

*/

if (number == -99 || number == 0 || number == 44) {

System.out.println("Special Number");

} else {

System.out.println("Not Special Number");

}

}

}

_______________

Output#1:

Enter a number :-99

Special Number

Output#2:

Enter a number :49

Not Special Number

7 0
3 years ago
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