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agasfer [191]
3 years ago
10

For metallurgical reasons, it is desirable to melt the weld metal with minimum energy input. Which one of the following heat sou

rces is most consistent with this objective: (a) high power, (b) high power density, (c) low power, or (d) low power density?
Engineering
1 answer:
stellarik [79]3 years ago
8 0

Answer:

The correct option is;

(a) High power density

Explanation:

The power density of a material is the amount of power per unit volume of the material. Power density, in the context of transformers, fuel cells, batteries, motors, and power supply units, is measured with respect to the volume, and the units is given as W/m³

A system such as a capacitor with an high power density has the capacity to put out large energy amount from a small volume. A capacitor with a high power density, can produce the same power output as a car battery and is said to have a high power density.

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Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user
Thepotemich [5.8K]

Answer:

import java.util.Scanner;

public class FindMatchValue {

  public static void main (String [] args) {

     Scanner scnr = new Scanner(System.in);

     final int NUM_VALS = 4;

     int[] userValues = new int[NUM_VALS];

     int i;

     int matchValue;

     int numMatches = -99; // Assign numMatches with 0 before your for loop

     matchValue = scnr.nextInt();

     for (i = 0; i < userValues.length; ++i) {

        userValues[i] = scnr.nextInt();

     }

     /* Your solution goes here */

         numMatches = 0;

     for (i = 0; i < userValues.length; ++i) {

        if(userValues[i] == matchValue) {

                       numMatches++;

                }

     }

     System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

  }

}

8 0
3 years ago
Let S = { p q |p, q are prime numbers greater than 0} and E = {0, −2, 2, −4, 4, −6, 6, · · · } be the set of even integers. . Pr
Finger [1]

Answer:

prove that | S | = | E | ; every element of S there is an Image on E , while not every element on E has an image on S

Explanation:

Given that S = { p q |p, q are prime numbers greater than 0}

                    E = {0, −2, 2, −4, 4, −6, 6, · · · }

To prove  by constructing a bijection from S to E

detailed  solution attached below

After the bijection :

<em>prove that | S | = | E |</em> :  every element of S there is an Image on E , while not every element on E has an image on S

∴ we can say sets E and S are infinite sets

7 0
3 years ago
A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the densit
alexdok [17]

Answer:

5.31\frac{kg}{m^3}

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to \frac{m}{M}, where m is the mass and M the molar mass of the gas, and the density is \frac{m}{V}.

For air M=28.66*10^{-3}\frac{kg}{mol} and \frac{5}{9}R=K

So, 598.59 R*\frac{5}{9}=332.55K

pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}

7 0
3 years ago
Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at
makkiz [27]

Answer:

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

Explanation:

The Young's module is:

E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }

E = \frac{\sigma\cdot L_{o}}{\dot L \cdot \Delta t}

Let assume that both specimens have the same geometry and load rate. Then:

E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}

The displacement rate for steel is:

\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}

\dot L_{steel} = \left(\frac{10000\,ksi}{29000\,ksi}\right)\cdot (0.001\,\frac{in}{min} )

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

7 0
3 years ago
Read 2 more answers
rt this question on a NEW page Give ONE word/tem for each of the following descriptions by choosing a word/term from the list be
Shalnov [3]

Answer:

oh I am sorry I can't understand your question.

6 0
3 years ago
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