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saw5 [17]
3 years ago
8

1. A wood board is one of a dozen different parts in a homemade robot kit. The width, depth, and height dimensions of the board

are 7.5 x 14 x 1.75 inches, respectively. The board is made from southern yellow pine, which has an air dry weight density of .025 lb/in.3. a. What is the volume of the wood board? Precision = 0.00
Engineering
1 answer:
otez555 [7]3 years ago
3 0

Answer:

183.75 cubic inches.

Explanation:

The volume of the wood board is determine by means of this expression:

V = w \cdot h \cdot l

By replacing variables:

V = (7.5 in) \cdot (14 in) \cdot (1.75 in)\\V = 183.75 in^{3}

You might be interested in
Do plastic materials have high or low ductility? Explain why.​
Flura [38]
The impact behavior of plastic materials is strongly dependent upon the temperature. At high temperatures, materials are more ductile and have high impact toughness. At low temperatures, some plastics that would be ductile at room temperature become brittle.
3 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
Propane is to be compressed from 0.4 MPa and 360 K to 4 MPa using a two-stage compressor. An interstage cooler returns the tempe
kow [346]

Answer:

a. 81 kj/kg

b. 420.625K

c.  101.24kj/kg

Explanation:

\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }

t1 = 360

p1 = 0.4mpa

p2 = 1.20

y = 1.13

substitute these values into the equation

\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }

\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347

when we cross multiply

t2 = 360 * 1.1347

= 408.5

a. the work required in the firs compressor

w=c(t2-t1)

c=1.67x10³

t1 = 360

t2 = 408.5

w = 1670(408.5-360)

= 1670*48.5

= 80995 J

= 81KJ/kg

b. n=\frac{t2-t1}{t'2-t1}

n = 80%

t2 = 408.5

t1 = 360

0.80 = 408.5-360 ÷ t'2-360

0.80 =\frac{48.5}{t'2-360}

cross multiply to get the value of t'2

0.80(t'2-360) = 48.5

0.80t'2 - 288 = 48.5

0.8t'2 = 48.5+288

0.8t'2 = 336.5

t'2 = 336.5/0.8

= 420.625

this is the temperature at the exit of the first compressor

c. cooling requirement

w = c(t2-t1)

= 1.67x10³(420.625-360)

= 1670*60.625

= 101243.75

= 101.24kj/kg

8 0
3 years ago
A soil weighs 2,520 lbs/CY in its in situ condition, 1,970 lb/CY in its loose condition after excavation, and 3,025 lbs/CY in an
azamat

Answer:

load factor = 0.782

Shrink Factor = 0.833

no of truck is  62500

Explanation:

given data

soil weighs in situ condition  =  2,520 lbs/CY

soil weighs in loose condition  =  1,970 lb/CY

soil weighs in embanked state = 3,025 lbs/CY

average volume = 16 LCY

soil from a borrow pit = 1 million CCY

solution

first we get here Load Factor that is express as

load factor = \frac{1,970}{2550}

load factor = 0.782

and Shrink Factor will be as

Shrink Factor = \frac{2520}{3025}

Shrink Factor = 0.833

and

no of truck will be

no of truck = \frac{1000000}{16}

no of truck is  62500

6 0
4 years ago
Consider a single, porous, spherical, inert mineral particle. The pores inside the particle are filled with liquid water (specie
cluponka [151]

Answer:

Explanation:

kindly check the attachment below for detailed explanations. Thanks

7 0
4 years ago
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