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bixtya [17]
4 years ago
10

List all information for each order item. Include an item total, which can be calculated by multiplying the Quantity and Paid ea

ch columns. Use a column alias for the calculated value to show the heading "Item Total" in the output
Engineering
1 answer:
Zigmanuir [339]4 years ago
3 0

Answer:

#include<iostream>

Using namespace std;

int main()

{

int n, qty;

double price, amount;

cout<<"Number of items ";

cin>>n;

cout<<"ITEM<<"\t"<<"QUANTITY"<<"\t"<<"PRICE"<<"\t"<<"ITEM TOTAL";

for(int i= 1; I<= n; i++)

{

cin>>qty;

cin>>price;

amount = qty * price;

cout<<i<<"\t "<<qty<<"\t"<<price<<"\t"<<amount;

}

}

Explanation

The above program is written in C++ programming language

5 variables are declared and used in the program

n is declared as an integer to represent the total number of items

qty is declared as integer to represent the total quantity of each item

price is declared as double to represent the amount of each individual item

amount is declared as double to represent the total amount of an item; it is gotten by qty * price

i is declared as integer to iterated between each items

The amount of each item is calculated within the iteration and also printed immediately

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4 years ago
The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and
garik1379 [7]

Solution :

The spring is expanded by 2 times of the block when it moves down an inclined by x times.

Here, $x_1$ = 39 mm

        x_2 = 225 mm

a). From the work energy principal,

   Work forces = kinetic energy

$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$

$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$

$9.75= 7.5(V^2_2-0.08^2)$

$1.3= V^2_2-0.08^2$

$V_2=1.14\ m/s$

b). calculating the distance travelled by the block before it comes to rest.

Substitute the value of V_2 in (1),

$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$

$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$

$98.43x - 100(4x^2+0.156x)+0.048=0$

$98.43x - 400x^2-15.6x+0.048=0$

$82.83x - 400x^2+0.048=0$

$  400x^2- 82.83x-0.048=0$

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3 years ago
Using the tables for water determine the specified property data at the indicated state. For H2O at T = 140 °C and v = 0.2 m3/kg
Dennis_Churaev [7]

Answer:

h = 1429.74\,\frac{kJ}{kg}

Explanation:

The determination of any further properties requires the knowledge of two independent properties. (Temperature and specific volume in this case). The specific volumes for saturated liquid and vapor at 140 °C are, respectively:

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Since \nu_{f} < \nu < \nu_{g}, it is a liquid-vapor mixture. The quality of the mixture is:

x = \frac{\nu-\nu_{f}}{\nu_{g}-\nu_{f}}

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h_{f} = 589.16\,\frac{kJ}{kg}

h_{g} = 2733.5\,\frac{kJ}{kg}

The specific enthalpy is:

h = h_{f}+x\cdot (h_{g}-h_{f})

h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)

h = 1429.74\,\frac{kJ}{kg}

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