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2 years ago

Am I the only one who cannot see the answers?? Does anybody know how to solve this?

Engineering

1 answer:

Debora [2.8K]2 years ago

4 0

Answer:

no, it seems that everyone is having the same issue. If you use the app you can still find the answers and see them.

Explanation:

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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

What is the name of the first federation vessel in star trek hint 17?

Nitella [24]

Answer:

The Fesarius

Explanation:

It's the first vessel

7 0

2 years ago

A Venturi meter (see below) uses the principles of the Bernoulli equation to measure the velocity of a flow in a reactor system.

LenaWriter [7]

Answer:

Q=0.000604 m³/s

Explanation:

Given that

d₁=5 cm

d₂=1 cm

P= 30 KPa

Density of water ,ρ=1000 kg/m³

As we know that volume flow rate Q given as

$Q=A_1A_2\sqrt{\dfrac{\dfrac{2\Delta P}{\rho}}{A_1^2-A_2^2}}$

$A_1=\dfrac{\pi}{4}\times 0.05^2\ m^2$

A₁=0.0019 m²

$A_2\dfrac{\pi}{4}\times 0.01^2\ m^2$

A₂=0.000078 m²

$Q=0.0019 \times 0.000078 \sqrt{\dfrac{\dfrac{2\times 30\times 1000}{1000}}{0.0019^2-0.000078^2}}\ m^3/s$

Q=0.000604 m³/s

7 0

3 years ago

What kind of job does Malcolm have?

Effectus [21]

Huh? what do you mean?

8 0

3 years ago

Read 2 more answers

(a) If 5 x 10^17 phosphorus atoms per cm3 are add to silicon as a substitutional impurity, determine the percentage of silicon a

Y_Kistochka [10]

Answer:

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice = 0.001 %

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice with boron atoms = 0.4 × $10^{-5}$ %

Explanation:

No. of phosphorus atoms = 5 × $10^{17} \ cm^{-3}$

The volume occupied by a single Si atom

$V_{si} = \frac{a^{3} }{8}$

$V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}$

$V_{si} =$ 2 × $10^{-23}$ $\frac{cm^{3} }{atom}$

$n_{si} = \frac{1}{V_{si} }$

$n_{si}$ = 5 × $10^{22}$ $\frac{atoms}{cm^{3} }$

$PCT = \frac{N_p}{N_{si}} 100$

Put the values in above equation we get

$PCT = \frac{5 (10^{17} )}{5 (10^{22}) } 100$

PCT = $10^{-3} = 0.001$ %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

(b).

No. of boron atoms = 2 × $10^{15} \ cm^{-3}$

The volume occupied by a single Si atom

$V_{si} = \frac{a^{3} }{8}$

$V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}$

$V_{si} =$ 2 × $10^{-23}$ $\frac{cm^{3} }{atom}$

$n_{si} = \frac{1}{V_{si} }$

$n_{si}$ = 5 × $10^{22}$ $\frac{atoms}{cm^{3} }$

$PCT = \frac{N_p}{N_{si}} 100$

Put the values in above equation we get

$PCT = \frac{2 (10^{15} )}{5 (10^{22}) } 100$

PCT = 0.4 × $10^{-5}$ %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

7 0

2 years ago