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mylen [45]
3 years ago
9

A renewable item is something that is capable of being replaced naturally.

Engineering
2 answers:
rodikova [14]3 years ago
7 0

Answer:

true corect me

Explanation:

djverab [1.8K]3 years ago
6 0
The answer is False.
You might be interested in
The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both
DedPeter [7]

Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

                               L^2 = x_a^2 + y_b^2   ....... 1

                               y_b = sqrt( 20^2 - 16^2 )

                               y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

                               0 = 2*x_a*v_a + 2*y_b*v_b

                               0 = x_a*v_a + y_b*v_b   ..... 2

                               0 = 16*36 + 12*v_b

                               v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

                               0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

                               0 = 9 + 4*a_a/3 + 16 + a_b

                               4*a_a/3 + a_b = -25

                               4*a_a + 3*a_b = -75  .... 3

- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

                               F_x = m_a*a_a

                               P - R_r*16/20 = m_a*a_a

                               

- For Slider B, Apply Newton's second law of motion in y direction:

                               F_y = m_b*a_b

                               - R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

                               P - 4*m_b*a_b / 3 = m_a*a_a

                               3P = 3*m_a*a_a + 4*m_b*a_b  ... 4

- Where,                  P = Is the force acting on slider A

                               P , m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.                    

5 0
3 years ago
State three active materials of a lead acid cell​
igomit [66]

Answer:

lead dioxide,sulfate and lead acid

6 0
3 years ago
Read 2 more answers
Carl why is there a dead man in the living room?
scoray [572]
You always need some company
3 0
3 years ago
You will create an array manipulation program that allows the user to do pretty much whatever they want to an array. When launch
enyata [817]

Answer:

Check the explanation

Explanation:

#include <iostream>

using namespace std;

void insert(int* arr, int* size, int value, int position){

if(position<0 || position>=*size){

cout<<"position is greater than size of the array"<<endl;

return ;

}

*size = *size + 1 ;

for(int i=*size;i>position;i--){

arr[i] = arr[i-1];

}

arr[position] = value ;

}

void print(int arr[], int size){

for(int i=0;i<size;i++){

cout<< arr[i] <<" ";

}

cout<<" "<<endl;

}

void remove(int* arr, int* size, int position){

* size = * size - 1 ;

for(int i=position;i<*size;i++){

arr[i] = arr[i+1];

}

}

int count(int arr[], int size, int target){

int total = 0 ;

for(int i=0;i<size;i++){

if(arr[i] == target)

total += 1 ;

}

return total ;

}

int main()

{

int size;

cout<<"Enter the initial size of the array:";

cin>>size;

int arr[size],val;

cout<<"Enter the values to fill the array:"<<endl;

for(int i=0;i<size;i++){

cin>>val;

arr[i] = val ;

}

int choice = 5,value,position,target ;

do{

cout<<"Make a selection:"<<endl;

cout<<"1) Insert"<<endl;

cout<<"2) Remove"<<endl;

cout<<"3) Count"<<endl;

cout<<"4) Print"<<endl;

cout<<"5) Exit"<<endl;

cout<<"Choice:";

cin>>choice;

switch(choice){

case 1:

cout << "Enter the value:";

cin>>value;

cout << "Enter the position:";

cin>>position;

insert(arr,&size,value,position);

break;

case 2:

cout << "Enter the position:";

cin>>position;

remove(arr,&size,position);

break;

case 3:

cout<<"Enter the target value:";

cin>>target;

cout <<"The number of times "<<target<<" occured in your array is:" <<count(arr,size,target)<<endl;

break;

case 4:

print(arr,size);

break;

case 5:

cout <<"Thank you..."<<endl;

break;

default:

cout << "Invalid choice..."<<endl;

}

}while(choice!=5);

return 0;

}

Kindly check the attached images below for the code output.

3 0
3 years ago
How many trips would one rubber-tired Herrywampus have to make to backfill a space with a geometrical volume of 5400 cubic yard?
nikklg [1K]

Answer:

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

Explanation:

a) By volume.

The shrinkage factor is:

\frac{5400cu-yd}{1-0.25} =7200cu-yd

The volume at loose is:

V_{loose} =V_{bank} (1+swell-factor)=7200(1+0.2)=8640cu-yd

If the Herrywampus has a capacity of 30 cubic yard:

\frac{8640cu-yd}{30cu-yd/trip} =288trip

b) By weight

The swell factor in terms of percent swell is equal to:

pounds-per-cubic-yard-loose=\frac{pounds-per-cubic-yard-bank}{\frac{percent-swell}{100}+1 }

pounds-per-cubic-yard-loose=\frac{3000}{\frac{20}{100} +1} =2500lb/cu-yd

The weight of backfill is:

8640cu-yd*2500\frac{lb}{cu-yd} *\frac{1ton}{2000lb} =10800ton

The Herrywampus has a capacity of 40 ton:

\frac{10800}{40ton/trip} =270trip

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

8 0
3 years ago
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