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3 years ago

In order to lift a lighter object on the other side, a boy placed 155 N of

Engineering

1 answer:

const2013 [10]3 years ago

4 0

Answer:

0.58 m

Explanation:

In order to find the vertical distance the load lifts, we first need to find the magnitude of the resistance force. So, Rd = Fd' where R = resistance force = ?, d = resistance distance = 3.5 m, F = effort force = 155 N and d' = effort distance = 1.5 m.

So R = Fd'/d = 155 N × 1.5 m/3.5 m = 232.5 Nm/3.5 m = 66.43 N

Now, the vertical work done by the effort = vertical work done by load

Fy = Ry' where y = vertical distance moved by effort = 0.25 m and y' = vertical distance moved by resistance force

So, Fy = Ry'

y' = Fy/R

= 155 N 0.25 m/66.43 N

= 38.75 Nm/66.43 N

= 0.58 m

So, the vertical distance the lighter object moves is y' = 0.58 m

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For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger sh

Paul [167]

Answer:

the pressure drop is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere $dp$ = 0.003 m

particle density = 1300 kg/m³

the bed void fraction $\in =$ 0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas $\rho$ = 0.15 mol/dm ⁻³

viscosity of methane gas $\mu$ = 1.429 x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³ to kg/m³

SO; we have :

Density = 0.15 mol/dm ⁻³

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density = $0.1 5 *\dfrac{16}{0.1^3}$

Density = 2400

Density $\rho_f$ = 2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas = 4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

$Re = \dfrac{dV \rho}{\mu}$

$Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}$

$Re=2276.317705$

For Re > 1000

$\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}$

$\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}$

$\Delta P=8575.755212*2.5$

$\Delta = 21439.38803 \ Pa$

To atm ; we have

$\Delta P = \dfrac{21439.38803 }{101325}$

$\Delta P =0.2115903087 \ atm$

ΔP ≅ 0.21159 atm

Thus; the pressure drop is 0.21159 atm

4 0

3 years ago

Listed below are snippets from a prgram to perform input validation for

leva [86]

Answer:

#include <iostream>

#include <string>

#include "user.h"

#include "password.h"

using namespace Authenticate;

using namespace std;

int main()

{

inputUserName();

inputPassword();

cout << "Your username is " << getUserName() <<

" and your password is: " <<

getPassword() << endl;

return 0;

}

user.h:

#ifndef USER_H

#define USER_H

#include <string>

using namespace std;

namespace Authenticate

{

namespace

{

bool isvalid();

}

void inputUserName();

string getUserName();

}

#endif

user.cpp:

#include <iostream>

#include "user.h"

namespace Authenticate

{

string username="";

namespace

{

bool isvalid()

{

if(username.length() == 8)

return true;

else

return false;

}

void inputUserName(){

{

cout << "Enter your username (8 letters only)" << endl;

cin >> username;

}

while(!isvalid());

}

string getUserName()

{

return username;

}

password.h:

#ifndef PASSWORD_h

#define PASSWORD_h

#include <string>

using namespace std;

namespace Authenticate

{

namespace

{

bool isValid();

}

void inputPassword();

string getPassword();

}

#endif

password.cpp:

#include <iostream>

#include <string>

using namespace std;

namespace Authenticate

{

string password="";

namespace

{

bool isValid()

{

if(password.length() >= 8)

{

for(int i=0; i<password.length(); i++)

if(password[i] >= '0' && password[i] <= '9')

return true;

return false;

}

else

return false;

}

void inputPassword(){

{

cout << "Enter your password (at least 8 characters " <<

"and at leat one non-letter)" << endl;

cin >> password;

}

while(!isValid());

}

string getPassword()

{

return password;

}

3 0

3 years ago

Implement the function lastChars() that takes a list of strings as a parameter and prints to the screen the last character of ea

Liono4ka [1.6K]

Answer:

The following program is in C++.

#include <bits/stdc++.h>

using namespace std;

void lastChars(string s)

{

int l=s.length();

if(l!=0)

{

cout<<"The last character of the string is: "<<s[l-1];

}

int main() {

string s;//declaring a string..

getline(cin,s);//taking input of the string..

lastChars(s);//calling the function..

return 0;

}

Input:-

Alex is going home

Output:-

The last character of the string is: e

Explanation:

In the function lastChars() there is one argument that is a string.I have declared a integer variable l that stores the length of the string.If the length of the string is not 0.Then printing the last character of the string.In the main function I have called the function lastChars() with the string s that is prompted from the user.

8 0

3 years ago

Provide an argument justifying the following claim: The average (as defined here) of two Java ints i and j is representable as a

ahrayia [7]

Answer:

public static int average(int j, int k) {

return (int)(( (long)(i) + (long)(j) ) /2 );

}

Explanation:

The above code returns the average of two integer variables

Line 1 of the code declares a method along with 2 variables

Method declared: average of integer data type

Variables: j and k of type integer, respectively

Line 2 calculates the average of the two variables and returns the value of the average.

The first of two integers to average is j

The second of two integers to average is k

The last parameter ensures average using (j+k)/2

3 0

3 years ago

Question 5

likoan [24]

Answer:

worth of being noticed

Explanation:

i had this in iready it’s correct trust me

8 0

3 years ago

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