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2 years ago

The shear force diagram is always the slope of the bending moment diagram. a)True b)- False

Engineering

1 answer:

lara [203]2 years ago

8 0

Answer:

True

Explanation:

Shear force diagram is a diagram which is drawn by calculating the shear force either to the right of the section or to the left of the section .

shear force is also be define as the change of moment w.r.t to distance

V= $\frac{\mathrm{d}M }{\mathrm{d} x}$

where V is shear force and M is bending moment.

from the equation we can clearly say that shear force diagram is slope of bending moment diagram.

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Typical noise associated with failed cv joint

igor_vitrenko [27]

Answer:

A worn inner CV joint often makes a clunking noise during starts and stops.

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2 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

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3 years ago

C++ - Green Crud Fibonacci programThe following program is to be written with a loop. You are to write this program three times

Fynjy0 [20]

Answer:

Below is the required code:

Explanation:

Using for loop

#include <iostream>

using namespace std;

int main()

{

//Initial crud size

int init = 0;

int newCrud;

int next=0;

//Number of days to simulate

int no_days;

int day;

cout << "Enter initial amount of green crud: ";

cin >> newCrud;

cout << "Enter number of days to simulate: ";

cin >> no_days;

for (day = 10; day<=no_days; day++)

{

if (day % 10 == 0)

{

next = newCrud + init;

}

newCrud = init;

init = next;

}

if (no_days < 5)

cout << "\nCrud reproduce only after 5 days minimum.Hence the current amount is "

<< newCrud << " pounds.";

else

cout << "On day " << no_days << " you have " << init

<< " pounds of green crud." << endl;

cout << "\nWould you like to continue? (y or n): ";

cin >> ans;

return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Using while loop

Program:

#include <iostream>

using namespace std;

int main()

{

char ans='y';

while (ans == 'Y' || ans == 'y')

{

//Initial crud size

int init = 0;

int newCrud;

int next=0;

//Number of days to simulate

int no_days;

int day;

cout << "Enter initial amount of green crud:

cin >> newCrud;

cout << "Enter number of days to simulate:

cin >> no_days;

for (day = 10; day<=no_days; day++)

{

if (day % 10 == 0)

{

next = newCrud + init;

}

newCrud = init;

init = next;

}

if (no_days < 5)

cout << "\nCrud reproduce only after 5 days

minimum.Hence the current amount is "

<< newCrud << " pounds.";

else

cout << "On day " << no_days << " you have "

<< init

<< " pounds of green crud." << endl;

cout << "\nWould you like to continue? (y or

n): ";

cin >> ans;

}

return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

Using do while loop

Program:

#include <iostream>

using namespace std;

int main()

{

char ans;

{

//Initial crud size

int init = 0;

int newCrud;

int next=0;

//Number of days to simulate

int no_days;

int day;

cout << "Enter initial amount of green crud: ";

cin >> newCrud;

cout << "Enter number of days to simulate: ";

cin >> no_days;

for (day = 10; day<=no_days; day++)

{

if (day % 10 == 0)

{

next = newCrud + init;

}

newCrud = init;

init = next;

}

if (no_days < 5)

cout << "\nCrud reproduce only after 5 days

minimum.Hence the current amount is "

<< newCrud << " pounds.";

else

cout << "On day " << no_days << " you have " <<

init << " pounds of green crud." << endl;

cout << "\nWould you like to continue? (y or n):

cin >> ans;

} while (ans == 'Y' || ans == 'y');

return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

7 0

2 years ago

What’s the answer please help

irina1246 [14]

Answer:

It B

Explanation:

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2 years ago

A bathtub faucet has a maximum flow rate of 3 gal/min. The tub is rectangular (3 ft long x 2 ft wide x 1.5 ft deep). Although th

Anvisha [2.4K]

Answer:

Time taken = 136.32 minutes

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

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3 years ago