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miskamm [114]
2 years ago
13

. Write a program that computes and displays a 15 percent tip when the price of a meal is input by the user. (Hint: the tip is c

omputed by multiplying the price of the meal by 0.15.) You will need the following variables: MealPrice (a Float) Tip (a Float) reddit
Engineering
1 answer:
Eddi Din [679]2 years ago
3 0

Answer:

Explanation:

// The solution is solved using c++

#include <iostream>

using namespace std;

int main ()

{

 float meal;

 cout << "Please enter the price of the meal: ";

 cin >> meal;

 cout << "The Tip for the meal is " << 0.15*meal;

 cout << " The tip for the meal  " << 0.15* meal  << ".\n";

 return 0;

}

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Determine F12 and F21 for the following configurations: (a) A long semicircular duct with diameter of 0.1 meters: (b) A hemisphe
uysha [10]

Answer:

long duct: 1.0 and 0.424

Hemisphere 1.0 ; 0.125; 0.5

Explanation:

For a long duct:

By inspection, F_{12} = 1.0

Calculating by reciprocity, F_{21} = \frac{A_{1} }{A_{2}F_{12}  }  = \frac{2RL}{\frac{3}{4}*2\pi RL  }* 1.0\\                                                         = 0.424

Hemisphere:

By reciprocity gives = 0.125

using the summation rule: F_{21} + F_{22} + F_{23} = 1

However, because this is a hemisphere, the value will be= 0.5 * 1

                                                                                               = 0.5

6 0
2 years ago
Steam enters a turbine operating at steady state at 1 MPa, 200 °C and exits at 40 °C with a quality of 83%. Stray heat transfer
Andrei [34K]

Answer:

(a) Work out put=692.83\frac{KJ}{Kg}

(b) Change in specific entropy=0.0044\frac{KJ}{Kg-K}

Explanation:

Properties of steam at 1 MPa and 200°C

        h_1=2827.4\frac{KJ}{Kg},s_1=6.69\frac{KJ}{Kg-K}

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.

Properties of saturated steam at 40°C

      h_f= 167.5\frac{KJ}{Kg} ,h_g= 2537.4\frac{KJ}{Kg}

 s_f= 0.57\frac{KJ}{Kg-K} ,s_g= 8.25\frac{KJ}{Kg-K}

So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=167.5+0.83(2537.4-167.5)\frac{KJ}{Kg}

h_2=2134.57\frac{KJ}{Kg}

s_2=s_f+x(s_g-s_f)\frac{KJ}{Kg-K}

s_2=0.57+0.83(8.25-0.57)\frac{KJ}{Kg-K}

s_2=6.6944\frac{KJ}{Kg-K}

(a)

Work out put =h_1-h_2

                      =2827.4-2134.57 \frac{KJ}{Kg}

Work out put =692.83 \frac{KJ}{Kg}

(b) Change in specific entropy

     s_2-s_1=6.6944-6.69\frac{KJ}{Kg-K}

Change in specific entropy =0.0044\frac{KJ}{Kg-K}

3 0
2 years ago
Identify the four engineering economy symbols and their values from the following problem statement. Use a question mark with th
Vilka [71]

Answer:

The company will have $311,424 in its investment set-aside account.

Explanation:

To determine the amount of money that the company will have after 7 years with an interest rate of 11% per year, we must calculate the price to start with an increase of 11%, and then repeat the operation until reaching seven years:

Year 0: 150,000

Year 1: 150,000 x 1.11 = 166,500

Year 2: 166,000 x 1.11 = 184,815

Year 3: 184,815 x 1.11 = 205,144.65

Year 4: 205,144.65 x 1.11 = 227,710.56

Year 5: 227,710.56 x 1.11 = 252,758.72

Year 6: 252,758.72 x 1.11 = 280,562.18

Year 7: 280,562.18 x 1.11 = 311,424

3 0
2 years ago
A shipment of rebar that weighs 745 kg would weigh roughly how much in pounds​
Andre45 [30]

Answer:

Dont no but will check

Explanation:

6 0
2 years ago
At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribut
professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

The average time spent by the vehicle is given as

⇒ \frac{1}{\mu(1-\frac{\lambda}{\mu})}

thus,

on substituting the respective values, we get

Average time spent by the vehicle = \frac{1}{60(1-\frac{30}{60})}

or

Average time spent by the vehicle = \frac{1}{60(1-0.5)}

or

Average time spent by the vehicle = \frac{1}{60(0.5)}

or

Average time spent by the vehicle = \frac{1}{30} hr/veh

or

Average time spent by the vehicle = \frac{1}{30}\times60 min/veh

[ 1 hour = 60 minutes]

thus,

Average time spent by the vehicle = 2 min/veh

Hence,

Option (d) 2 min/veh

7 0
2 years ago
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