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3 years ago

7

Write a single statement to print: user_word,user_number. Note that there is no space between the comma and user_number. Sample

output with inputs: 'Amy' 5

1 answer:

kykrilka [37]3 years ago

8 0

Answer:

cout<<"''<<user_word<<"' "<<user_number;

Explanation:

The above question was answered using C++ programming language.

The keyword cout represents print and it carries out print operation only.

It prints all variable in front of it.

Assume the values of user_word and user_number to be Charles and 20, respectively.

The output of the above instruction would be

'Charles' 20 just as it is in the sample output in the question.

In java programming language, it is

System.out.print("'"+user_word+"' "+user_number);

In Qbasic, it is

PRINT "'"+user_word+"' "+ user_number

You might be interested in

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati

Lynna [10]

Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

$\dot m = 5000 kg/s$

Assume Mechanical energy at exist is negligible

A) Take lake bottom as reference, and then kinetic and potential energy are taken as zero.

change in mechanical energy is givrn as

$e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}$

= 0.491 kJ/kg

$\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW$

$\eta_{OVERALL} = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76$

B) $\eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80$

c) $\dot W_{shaft} = \eta_{overall} \left | \Delta \dot E_{mec} \right | = 0.80(2455)$

$\dot W_{shaft} = 1964 kW$

7 0

4 years ago

What is 90 to the power of 46

Mnenie [13.5K]

Answer:Just multiply 90 by itself 46 times

Explanation:

do it

6 0

4 years ago

Consider a multiprocessor system and a multithreaded program written using the many-to-many threading model. Let the number of u

Montano1993 [528]

Answer:

At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.

4 0

3 years ago

This program will store roster and rating information for a soccer team. Coaches rate players during tryouts to ensure a balance

Flauer [41]

Answer:

#include <iostream>

#include <vector>

using namespace std;

int main() {

vector<int> jerseyNumber;

vector<int> rating;

int temp;

for (int i = 1; i <= 5; i++) {

cout << "Enter player " << i

<< "'s jersey number: ";

cin >> temp;

jerseyNumber.push_back(temp);

cout << "Enter player " << i

<< "'s rating: ";

cin >> temp;

rating.push_back(temp);

cout << endl;

}

cout << "ROSTER" << endl;

for (int i = 0; i < 5; i++)

cout << "Player " << i + 1 << " -- "

<< "Jersey number: " << jerseyNumber.at(i)

<< ", Rating: " << rating.at(i) << endl;

char option;

'

while (true) {

cout << "MENU" << endl;

cout << "a - Add player" << endl;

cout << "d - Remove player" << endl;

cout << "u - Update player rating" << endl;

cout << "r - Output players above a rating"

<< endl;

cout << "o - Output roster" << endl;

cout << "q - Quit" << endl << endl;

cout << "Choose an option: ";

cin >> option;

switch (option) {

case 'a':

case 'A':

cout << "Enter a new player's"

<< "jersey number: ";

cin >> temp;

jerseyNumber.push_back(temp);

cout << "Enter the player's rating: ";

cin >> temp;

rating.push_back(temp);

break;

case 'd':

case 'D':

cout << "Enter a jersey number: ";

cin >> temp;

int i;

for (i = 0; i < jerseyNumber.size();

i++) {

if (jerseyNumber.at(i) == temp) {

jerseyNumber.erase(

jerseyNumber.begin() + i);

rating.erase(rating.begin() + i);

break;

}

}

break;

case 'u':

case 'U':

cout << "Enter a jersey number: ";

cin >> temp;

for (int i = 0; i < jerseyNumber.size();

i++) {

if (jerseyNumber.at(i) == temp) {

cout << "Enter a new rating "

<< "for player: ";

cin >> temp;

rating.at(i) = temp;

break;

}

}

break;

case 'r':

case 'R':

cout << "Enter a rating: ";

cin >> temp;

cout << "\nABOVE " << temp << endl;

for (int i = 0; i < jerseyNumber.size();

i++)

if (rating.at(i) > temp)

cout << "Player " << i + 1

<< " -- "

<< "Jersey number: "

<< jerseyNumber.at(i)

<< ", Rating: "

<< rating.at(i) << endl;

break;

case 'o':

case 'O':

cout << "ROSTER" << endl;

for (int i = 0; i < jerseyNumber.size();

i++)

cout << "Player " << i + 1 << " -- "

<< "Jersey number: "

<< jerseyNumber.at(i) << ", Rating: "

<< rating.at(i) << endl;

break;

case 'q':

return 0;

default:

cout << "Invalid menu option."

<< " Try again." << endl;

}

}

}

Explanation:

4 0

3 years ago

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stre

velikii [3]

Complete Question

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.

Answer:

The stress is $\sigma = 10. 655 MPa$

Explanation:

From the question we are told that

The critical yield resolved shear stress is $\sigma = 2.9Mpa$

First we obtain the angle $\lambda$ between the slip direction [121] and [111]

$\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ]$

Where $u_1 ,u_2 ,v_1 , v_2 , w_1 , w_2$ are the directional indices

$\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ]$

$= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ]$

$= 61.87^0$

Next is to obtain the angle $\O$ between the direction [121] and [101]

$\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ]$

Substituting 1 for $u_1$ , 2 for $v_1$ , 1 for $w_1$ , 1 for $u_2$ , 0 for $v_2$ , and 1 for $w_2$

$\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ]$

$\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ]$

$= 54.74 ^o$

The stress is mathematically represented as

$\sigma = \frac{\tau_c}{cos \O cos \lambda }$

$= \frac{2.9}{cos 54.74^o cos 61.87^o}$

$= \frac{2.9}{0.2722}$

$\sigma = 10. 655 MPa$

5 0

3 years ago