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yaroslaw [1]
3 years ago
8

he Animal Insurance Company:Writes a program that repeatedly prints the insurance fee to pay for a pet according to the followin

g rules: A dog that has been neutered costs $50. A dog that has not been neutered costs $80. A cat that has been neutered costs $40. A cat that has not been neutered costs $60. A bird or reptile costs $10. Any other animal generates an error message.
Engineering
1 answer:
nordsb [41]3 years ago
4 0

Answer:

Program code is given below.

Explanation:

#include <iostream>

using namespace std;03

int main()

{

char neutered, opselect, animal;

const int dogcost = 80, catcost = 60, birdOrReptilecost = 0;

cout << "What kind of animal do you have?\n";

cout << "d for dog, c for cat, b for bird or reptile and o for others.\n\n";

cin >> animal;

cout << "Answer the following questions\n";

cout << "With Y for yes or N for no.\n\n";

cout << "Is your pet neutered? ";

cin >> neutered;

cout << "Enter a select code: ";

cout << "\n1)A dog that has been neutered: ";

cout << "\n2)A dog that has not been neutered: ";

cout << "\n3)A cat that has been neutered costs: ";

cout << "\n4)A cat that has not been neutered costs: ";

cout << "\n5)A bird or reptile: ";

cin >> opselect;

switch (opselect)

{

  case 1:

    cout << "The price for dog that has been neutered is " << dogcost - 30;

    break;

  case 2:

    cout << "The price for dog that has not been neutered is " << dogcost << endl;

   break;

  case 3:

    cout << "The price for cat that has been neutered is " << catcost - 20 << endl;

   break

  case 4:

    cout << "The price for cat that has not been neutered is " << catcost << endl;

   break;

  case 5:

    cout << "The price for bird or reptile is " << birdOrReptilecost << endl;

break;

return 0;

}

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A tensile test was made on a tensile specimen, with a cylindrical gage section which had a diameter of 10 mm, and a length of 40
tamaranim1 [39]

Answer:

The answers are as follow:

a) 10 mm

b) 12.730 N/mm^{2}

c) 127.307 N/mm^{2}

d) 0.25

Explanation:

d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9

Force = F =1000 N

let us find initial area first, A1 = pi*r^{2} = 78.55 mm^{2}

using reduction in area formula : 0.9 = (A1 - A2 ) / A1

solving it will give,  A2 = 0.1 A1 = 7.855  mm^{2}

a) The specimen elongation is final length - initial length

50 - 40 = 10 mm

b) Engineering stress uses the original area for all stress calculations,

Engineering stress = force / original area  = F / A1 = 1000 / 78.55  

Engineering stress = 12.730 N / mm^{2}

c) True stress uses instantaneous area during stress calculations,

True fracture stress = force / final  area  = F / A2 = 1000 / 7.855

True Fracture stress = 127.30 N / mm^{2}

e) strain = change in length / original length

strain = 10 / 40  = 0.25

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lakkis [162]

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3 years ago
The electron concentration in silicon at T = 300 K is given by
puteri [66]

Answer:

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

Explanation:

From the question we are told that:

Temperature of silicon T=300k

Electron concentration n(x)=10^{16}\exp (\frac{-x}{18})

                                        \frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})

Electron diffusion coefficient is Dn = 25cm^2/s \approx 2.5*10^{-3}

Electron mobility is \mu n = 960 cm^2/V-s \approx0.096m/V

Electron current density Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2

Generally the equation for the semiconductor is mathematically given by

Jn=qb_n\frac{dn}{dx}+nq \mu E

Therefore

-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E

E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

7 0
3 years ago
1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are rem
BARSIC [14]

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁  = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2}   \right ]^{\dfrac{k}{k-1}}

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}

P_2= \dfrac{1060}{5^{1.33}}  = 124.65 \ kPa

\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2}   \right ]^{\dfrac{k}{k-1}}

\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2}   \right

5^{0.33} = \left \dfrac{T_1}{T_2}   \right

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3×c_v×(T₂ - T₁)

c_v = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86  kJ/(kg·K)

We use cv given by  (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3×c_v×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = U_g = h_g -pV_g

h_g = 2777.12 kJ/kg

V_g = 0.194349 m³/kg

p = 1000 kPa

U_{g1} = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, U_{g1} + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

3 0
3 years ago
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