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3 years ago

Another way to explain the definition of "virtual condition" is to say?

Engineering

1 answer:

zvonat [6]3 years ago

6 0

Answer:

theoretical extreme boundary condition of a feature of size generated by the collective effects of MMC and any applicable geometric tolerances.

Explanation:

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Does anyone know what this is

sammy [17]

Answer:

Looks like mold that got frosted over

Explanation:

4 0

2 years ago

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This question is 100 points<br> I NEED HELP!!!

Mamont248 [21]

Answer:

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3 years ago

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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

Four of the minterms of the completely specified function f(a, b, c, d) are m0, m1, m4, and m5.

Sveta_85 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The required additional minterms for f so that f has eight primary implicants with two literals and no other prime implicant are $m_{2},m_{3},m_{7},m_{8},m_{11},m_{12},m_{13},m_{14}$ and $m_{15}$

b) The essential prime implicant are $c' d',a'b',ab$ and $cd$

c) The minimum sum-of-product expression for f are

$a'b' +ab +c'd'+cd+a'c',\\ a'b'+ab+c'd'+cd+a'd,\\a'b'+ab+c'd'+cd+bc' and \\ a'b'+ab+c'd' +cd+bd$

Explanation:

The explanation is shown on the second third and fourth image

8 0

3 years ago

A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such t

bekas [8.4K]

Answer:

a)Δs = 834 mm

b)V=1122 mm/s

$a=450\ mm/s^2$

Explanation:

Given that

$s = 15t^3 - 3t\ mm$

When t= 2 s

$s = 15t^3 - 3t\ mm$

$s = 15\times 2^3 - 3\times 2\ mm$

s= 114 mm

At t= 4 s

$s = 15t^3 - 3t\ mm$

$s = 15\times 4^3- 3\times 4\ mm$

s= 948 mm

So the displacement between 2 s to 4 s

Δs = 948 - 114 mm

Δs = 834 mm

We know that velocity V

$V=\dfrac{ds}{dt}$

$\dfrac{ds}{dt}=45t^2-3$

At t= 5 s

$V=45t^2-3$

$V=45\times 5^2-3$

V=1122 mm/s

We know that acceleration a

$a=\dfrac{d^2s}{dt^2}$

$\dfrac{d^2s}{dt^2}=90t$

a= 90 t

a = 90 x 5

$a=450\ mm/s^2$

4 0

3 years ago