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Gekata [30.6K]
2 years ago
13

Q5

Engineering
1 answer:
Klio2033 [76]2 years ago
5 0

The C++ code that would draw all the iterations in the selection sort process on the array is given below:

<h3>C++ Code</h3>

#include <stdio.h>

#include <stdlib.h>

int main() {

   int i, temp1, temp2;

   int string2[16] = { 0, 4, 2, 5, 1, 5, 6, 2, 6, 89, 21, 32, 31, 5, 32, 12 };

   _Bool check = 1;

   while (check) {

       temp1 = string2[i];

       temp2 = string2[i + 1];

       if (temp1 < temp2) {

           string2[i + 1] = temp1;

           string2[i] = temp2;

           i = 0;

       } else {

           i++;

           if (i = 15) {

               check = !check;

           }

       }

   }

   

   return 0;

}

Read more about C++ programming here:

brainly.com/question/20339175

#SPJ1

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The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is
nexus9112 [7]

Answer:

The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25 mm; it is made of a brass for which the allowable shearing stress is 50 MPa.

4 0
3 years ago
A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?
blagie [28]

Answer:

3A

Explanation:

Using Ohms law U=I×R solve for I by I=U/R

4 0
3 years ago
A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t
Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa

The design factor for metals is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}

The design factor for plastic and ceramics is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5

This material will break.

3 0
3 years ago
a) A total charge Q = 23.6 μC is deposited uniformly on the surface of a hollow sphere with radius R = 26.1 cm. Use ε0 = 8.85419
dusya [7]

Answer:

(a) E = 0 N/C

(b) E = 0 N/C

(c) E = 7.78 x10^5 N/C

Explanation:

We are given a hollow sphere with following parameters:

Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C

R = radius of sphere = 26.1 cm = 0.261 m

Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²

The formula for the electric field intensity is:

E = (1/4πεo)(Q/r²)

where, r = the distance from center of sphere where the intensity is to be found.

(a)

At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.

<u>E = 0 N/C</u>

(b)

Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).

<u>E = 0 N/C</u>

(c)

Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:

E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]

<u>E = 7.78 x10^5 N/C</u>

4 0
3 years ago
Injection melding is a process that
Bumek [7]

Answer:

mark brainliest :)

Explanation:

Injection Molding. Injection molding is the most commonly used manufacturing process for the fabrication of plastic parts. ... The plastic is melted in the injection molding machine and then injected into the mold, where it cools and solidifies into the final part.

Materials: Thermoplastics

5 0
3 years ago
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