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OleMash [197]
4 years ago
14

Output all combinations of character variables a, b, and c. If a = 'x', b = 'y', and c = 'z', then the output is: xyz xzy yxz yz

x zxy zyx Your code will be tested in three different programs, with a, b, c assigned with 'x', 'y', 'z', then with '#', '$', '%', then with '1', '2', '3'.
Engineering
1 answer:
gladu [14]4 years ago
7 0

Answer & Explanation:

//written in java

public class Main {

   public static void main(String[] args) {

       //declare a char variable for a, b, c

       char a;

       char b;

       char c;

       //assign a b and c

       //a b and c can be replaced for with  

       // '#', '$', '%', then with '1', '2', '3'

       // for further testing

       a = 'x';

       b = 'y';

       c = 'z';

       //output for all possible combination for a, b, c.

       System.out.println("" + a + b + c + " " + a + c + b + " " + b + a + c +

               " " + b + c + a + " " + c + a + b + " " + c + b + a);

   }

}

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Explain why the following scenario fails to meet the criteria for proper reverse engineering.
avanturin [10]

Answer:

he must document or remember the order he took it apart so he put it back together

Explanation:

5 0
2 years ago
Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adia
harina [27]

Answer:

Power = 371.28 kW

Explanation:

Initial pressure, P1 = 5 bar

Final pressure, P2 = 1 bar

Initial temperature, T1 = 320°C

Final temperature, T2 = 160°C

Volume flow rate, V = 0.65m³/s

From steam tables at state 1,

h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK

v1 = 0.5416 m³/kg

Mass flow rate, m = V/v1

m = 1.2 kg/s

From steam tables, at state 2

h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK

Power developed, P = m(h1 - h2)

P = 1.2(3105.6-2796.2)

P = 371.28 kW

8 0
4 years ago
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
-BARSIC- [3]

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

7 0
3 years ago
What color is an orange? its for my 8th grade
Ksenya-84 [330]

Answer:

hmmmm i think orange I may be wrong....

Explanation:

5 0
3 years ago
The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio
Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

Answer:

v=23.6m/s

Explanation:

Velocity v_c=18.0km/h

Radius r=21m

initial velocity uu=19=>5.27778

Generally the equation for Angle is mathematically given by

\theta=\frac{v_c}{2r}

\theta=\frac{18}{2*21}

\theta=0.45

\theta=25.7831 \textdegree

Generally

Height of mass

h=\frac{rsin\theta}{\theta}

h=\frac{21sin25.78}{0.45}

h=20.3m

Generally the equation for Work Energy is mathematically given by

0.5mv_0^2+mgh=0.5mv^2

Therefore

v=\sqrt{u^2+2gh}

v=\sqrt{=5.27778^2+2*9.81*20.3}

v=23.6m/s

3 0
3 years ago
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