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Dmitriy789 [7]
2 years ago
5

Write a function "funthree" that will print a box of characters. The function will always receive as the first input argument th

e character to print. If there are only two input arguments, they will be the character followed by a number, which is the number of rows and columns of the box to print. If, however, there are three input arguments, they will be the character, the number of rows, and the number of columns for the box of characters. Here are examples of calling the function:
Engineering
1 answer:
stiv31 [10]2 years ago
4 0

Answer: i8g7ieusgr7eytuu7esieso87 sugr7pi8y8hiuh9yehroe998 rydyh9 t9ry9 7 fdgerje 78re87es 7yehr87ehtu9hu7b puuhihugogi;ghi uhugyug fhglhfu fufbiup hughghuihu fhuihihiuhg uhfuhuig8fguh hguihfhjliigihfuhf;gjihgh hfuyt8uyiiohiodohi

Explanation: so first you are going to hyigfgegidii9thdf5dh8yy7 gdiuigp 87hgiuf 8yhijh gihoighhgd

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You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
Ymorist [56]

Answer:

t'_{1\2} = 6.6 sec

Explanation:

the half life of the given circuit is given by

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

t'_{1\2} =R'Cln2

Divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

putting all value we get new half life

t'_{1\2} = 3 * \frac{88}{40}  = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
2 years ago
Consider the following class definitions: class smart class superSmart: public smart { { public: public: void print() const; voi
arsen [322]

Answer:

a) There is no any private member of smart which are public members of superSmart.

Explanation:

5 0
2 years ago
A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. If the
Bad White [126]

Answer:

=>> 167.3 kpa.

=>> 60° from horizontal face.

Explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this kind of question;

=>> "A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. "

=>>"If the deviator stress is 120 kPa and the shear strength parameters are a friction angle of 30° and a cohesion of 5 kPa."

The orientation of this plane with respect to the major principle stress plane = 50 tan^2 (45 + 30/2) + 10 tan ( 45 + 30/2).

magnitude of the stresses on the failure plane = 167.3 kpa.

The orientation of this plane with respect to the major principle stress plane => x = 60 cos 60° = 30kpa.

y = 60 sin 60° = 30√3 = sheer stress.

the orientation of this plane with respect to the major principle stress plane.

Theta = 45 + 15 = 60°.

5 0
2 years ago
A thin-walled tube with a diameter of 12 mm and length of 25 m is used to carry exhaust gas from a smoke stack to the laboratory
nlexa [21]

Answer:

(a)  h₁   = 204.45 W/m²k

(b) h₀ = 46.80 W/m².k

(c) T = T = 15.50°C

Explanation:

Given Data;

Diameter = 12mm

Length = 25 m

Entry temperature = 200°C

Flow rate = 0.006 kg/s

velocity = 2.5 m/s.

Step 1: Calculating the mean temperature;

(200 + 15)/2

Mean temperature = 107.5°C = 380.5 K

The properties of air at mean temperature 380.5 K are given as:

v = 24.2689*10⁻⁶m²/s

a = 35.024*10⁻⁶m²/s

μ    = 221.6 *10⁻⁷Ns/m²

k = 0.0323 W/m.k

Cp = 1012 J/kg.k

Step 2: Calculating the prantl number using the formula;

Pr = v/a

   = 24.2689*10⁻⁶/ 35.024*10⁻⁶

   = 0.693

Step3: Calculating the reynolds number using the formula;

Re = 4m/πDμ

    = 4 *0.006/π*12*10⁻³ * 221.6 *10⁻⁷

    = 0.024/8.355*10⁻⁷

    = 28725

Since Re is greater than 2000, the flow is turbulent. Nu becomes;

Nu = 0.023Re^0.8 *Pr^0.3

Nu = 0.023 * 28725^0.8 * 0.693^0.3

     = 75.955

(a) calculating the heat transfer coefficient:

Nu = hD/k

h = Nu *k/D

  = (75.955 * 0.0323)/12*10^-3

h   = 204.45 W/m²k

(b)

Properties of air at 15°C

v = 14.82 *10⁻⁶m²/s

k = 0.0253 W/m.k

a = 20.873 *10⁻⁶m²/s

Pr(outside) = v/a

                  = 14.82 *10⁻⁶/20.873 *10⁻⁶

                 = 0.71

Re(outside) = VD/v

                   = 2.5 * 12*10⁻³/14.82*10⁻⁶

                    =2024.29

Using Zakauskus correlation,

Nu = 0.26Re^0.6 * Pr^0.37 * (Pr(outside)/Pr)^1/4

    = 0.26 * 2024.29^0.6 *  0.71^0.37 * (0.71/0.693)^1/4

    = 22.199

Nu = h₀D/k

h₀ = Nu*k/D

     = 22.199* 0.0253/12*10⁻³

h₀ = 46.80 W/m².k

 (c)

Calculating the overall heat transfer coefficient using the formula;

1/U =1/h₁ +1/h₀

1/U = 1/204.45 + 1/46.80

1/U = 0.026259

U = 1/0.026259

U = 38.08

Calculating the temperature of the exhaust using the formula;

T -T₀/T₁-T₀ = e^-[uπDL/Cpm]

T - 15/200-15 = e^-[38.08*π*12*10⁻³*25/1012*0.006]

T - 15/185 = e^-5.911

T -15 = 185 * 0.002709

T = 15+0.50

T = 15.50°C

6 0
2 years ago
Water at 70 kPa and 1008C is compressed isentropically in a closed system to 4 MPa. Determine the final temperature of the water
Nikitich [7]

Answer:

The answer is "909.3928  KJ".

Explanation:

70 \ kPa  \ \ and \ \ 100^{\circ}C \\\\s_i= 7.56162\ \frac{kJ}{kgK}\\\\u_i= 2509.39 \ \frac{kJ}{kg}\\\\

The method is isentropic since the cylinders are shielded.

Calculating the work:

w= u_2-u_i \\\\

   = 3418.7728-2509.38 \\\\=909.3928 \ KJ

3 0
2 years ago
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