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Inessa [10]
3 years ago
7

Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To

o small". If greater than 10, print "Too large". Otherwise, compute and print 6 * bagOunces followed by " seconds". End with a newline. Example output for ounces = 7:
Engineering
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

#include <iostream>

using namespace std;

void PrintPopcornTime(int bagOunces) {

if(bagOunces < 3){

 cout << "Too small";

 cout << endl;

}

else if(bagOunces > 10){

 cout << "Too large";

 cout << endl;

}

else{

 cout << (6 * bagOunces) << " seconds" << endl;

}

}

int main() {

  PrintPopcornTime(7);

  return 0;

}

Explanation:

Using C++ to write the program. In line 1 we define the header "#include <iostream>"  that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.

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The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo
Tresset [83]

Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)

consider the equation of bernoulli at the top of the pool

P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)

where P_1=P_0 atm pressure

At the top of the pool v_1=0m/s, substitute in V_1 in equation (2)

P_0+\rho gh_1 =constant ...(3)

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)

where P_2=P_0 atm pressure and h_2=0m

P_0+\rho v_1^2 =constant ...(6)

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

        P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)    

Hence velocity is v_2=(\sqrt{2gh_1})m/s

d.) consider (7)

v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)

This is the norminal value of velocity  

e.) consider the equation of flow rate interval of v and t

flow(t)=\frac{dv}{dt}(m^3/s) hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2

6 0
2 years ago
A thermoelectric refrigerator is powered by a 16-V power supply that draws 2.9 A of current. If the refrigerator cools down 3.1
Viktor [21]

Answer:

COP = 0.090

Explanation:

The general formula for COP is:

COP = Desired Output/Required Input

Here,

Desired Output = Heat removed from water while cooling

Desired Output = (Specific Heat of Water)(Mass of Water)(Change in Temperature)/Time

Desired Output = [(4180 J/kg.k)(3.1 kg)(25 - 11)k]/[(12 hr)(3600 sec/hr)]

Desired Output = 4.199 W

And the required input can be given as electrical power:

Required Input = Electrical Power = (Current)(Voltage)

Required Input = (2.9 A)(16 V) = 46.4 W

Therefore:

COP = 4.199 W/46.4 W

<u>COP = 0.090</u>

8 0
3 years ago
Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/
kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k

Obtain the following properties at 10kPa from the table "saturated water"

h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K

Calculate the enthalpy at exit of the turbine using the energy balance equation.

\frac{dE}{dt}=Q-W+m(h_1-h_2)

Since, the process is isentropic process Q=0

0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg

Use the isentropic relations:

s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87

Calculate the enthalpy at isentropic state 2s.

h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg

a.)

Calculate the isentropic turbine efficiency.

\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%

b.)

Find the quality of the water at state 2

since h_f at 10KPa <h_2<h_g at 10KPa

Therefore, state 2 is in two-phase region.

h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9

Calculate the entropy at state 2.

s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K

Calculate the rate of entropy production.

S=\frac{Q}{T}+m(s_2-s_1)

since, Q = 0

S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k

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1 year ago
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A thrust angle can be defined as an imaginary line which is drawn perpendicularly from the centerline of the rear axle of a vehicle, down the centerline.

This ultimately implies that, the thrust angle is a reference to the centerline (wheelbase) of a vehicle, and it confirms that the two wheels on both sides are properly angled within specification.

Read more on thrust angle here: brainly.com/question/13000914

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