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NARA [144]
3 years ago
6

Consider the following class definitions: class smart class superSmart: public smart { { public: public: void print() const; voi

d print() const; void set(int, int); void set(int, int, int); int sum(); int manipulate(); smart(); superSmart(); smart(int, int); superSmart(int, int, int); private: private: int x; int z; int y; int secret(); }; }; Which private members, if any, of smart are public members of superSmart
Engineering
1 answer:
arsen [322]3 years ago
5 0

Answer:

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

Explanation:

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Cutting and abrasive machining are the two major material processes. List the differences between Cutting tool and Abrasive mach
STatiana [176]

Answer:

Explained

Explanation:

Cutting tools:

 1. Cutting tools can either be single point or multi point.

2. Cutting tools can have variety of material depending on use like ceramics, diamonds, metals, CBN, etc.

3.Cutting tools have definite shapes and geometry.

Abrasive machining tools

1. Abrasive tools are always multi point tools.

2. Abrasive tools composed of abrasives bounded in medium of resin or metal.

3. They do not have definite geometry of shape

7 0
4 years ago
For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in.
makkiz [27]

Answer:

  • nP  ≈ 4.9
  • nL =  1.50

Explanation:

GIVEN DATA

external load applied (p) = 85 kips

bolt stiffness ( Kb ) = 3(10^6) Ibf / in

Member stiffness (Km) = 12(10^6) Ibf / in

Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8

Number of bolts = 6

assumptions

for unified screw threads UNC and UNF

tensile stress area ( A ) = 0.1419 in^2

SAE specifications for steel bolts for grade 8

we have

Minimum proff strength ( Sp) = 120 kpsi

Minimum tensile strength (St) = 150 Kpsi

Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips

Given the following values

Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip

Preload stress

αi = 0.75Sp = 0.75 * 120 = 90 kpsi

stiffness constant

C = \frac{Kb}{Kb + Km}  = \frac{3}{3+2} = 0.2

A) yielding factor of safety

nP = \frac{sPAt}{Cp + Fi} = \frac{120* 0.1419}{0.2*14.17 + 12.771}

nP = 77.028 / 15.605 = 4.94 ≈ 4.9

B) Determine the overload factor safety

nL = \frac{SpAt - Fi}{CP} = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17

= 17.028 - 12.771 / 2.834

= 1.50

3 0
3 years ago
code a class named sllqueue that uses a double-ended singly linked list to implement a queue as described in this chapter. provi
julia-pushkina [17]

Answer:

See attached file please.

Explanation:

See attached file for detailed explanation and code.

import java.util.*;

class LinklistImplementQueue {  

public static void main(String[] args)

{

Scanner scan = new Scanner(System.in);

/* Creating object of class SLLQueue */    

SLLQueue lq = new SLLQueue();

char ch;

do

{

System.out.println("\nQueue Operations");

System.out.println("1. ENQUEUE");

System.out.println("2. DEQUEUE");

int choice = scan.nextInt();

.

.

.

.

See attached file for complete code.

Download txt
6 0
3 years ago
8. Removing damage by tapping it out with a hammer
creativ13 [48]

Answer:

tapping out I think sorry if I'm wrong

7 0
3 years ago
One of our wifi network standards is IEEE 802.11ac. It can run at 6.77 Gbit/s data rate. Calculate the symbol rate for 801.11ac
Lena [83]

Answer: Symbol rate, Fs = 0.846

Explanation:

The attachment below shows the calculations

7 0
3 years ago
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