What is a dynamic load? *

Which atom bond in atomic interaction combines electrons, filling its valence zone a) Van der Vaals bond; b) a covalent bond; c)

Setler79 [48]

Answer: Covalent bond

Explanation: Covalent bond is the bond that gets created when there is a sharing of electrons among atoms and hence creating atomic interaction. The bond formed is from the shared pair because they allow the atoms or ions to achieve stability by completely filling the outer shell of the electron and thus form the covalent bond .Therefore, the correct option is the option(b) .

7 0

3 years ago

Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment

aleksandrvk [35]

Answer:

//This Program is written in C++

// Comments are used for explanatory purpose

#include <iostream>

using namespace std;

enum mailbox{open, close};

int box[149];

void closeAllBoxes();

void OpenClose();

void printAll();

int main()

{

closeAllBoxes();

OpenClose();

printAll();

return 0;

}

void closeAllBoxes()

{

for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150

{

box[i] = close; //Close all boxes

}

void OpenClose()

{

for(int i = 2; i < 150; i++) {

for(int j = i; j < 150; j += i) {

if (box[j] == close) //Open box if box is closed

box[j] = open;

else

box[j] = close; // Close box if box is opened

}

// At the end of this test, all boxes would be closed

}

void printAll()

{

for (int x = 0; x < 150; x++) //use this to test

{

if (box[x] = 1)

{

cout << "Mailbox #" << x+1 << " is closed" << endl;

// Print all close boxes

}

Explanation:

7 0

3 years ago

List irreversibilities

natali 33 [55]

Answer:

Some of the irreversibilities are listed below:

Plastic deformation of solids
Transfer of heat over finite difference of temperature
When two fluids are mixed together the process is irreversible
Combustion of a gas
Current flowing through a finite resistor
Diffusion and free compression or expansion of gas
Relative motion of body with force of friction
Processes involving chemical reactions(spontaneous)

6 0

3 years ago

This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Cons

telo118 [61]

Explanation:

(a)

Here, distance between hosts A and B is m meters and, propagation speed along the link is s meter/sec

Hence, propagation delay, $d_{prop} = m/sec (s)$

(b)

Here, size of the packet is L bits

And the transmission rate of the link is R bps

Hence, the transmission time of the packet, $d_{trans} = L/R$

(c)

As we know, end-to-end delay or total no delay,

$\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}$

$Here, $\mathrm{d}_{\text {rroc }}$ and $\mathrm{d}_{\text {quat }}$ \\Hence, $\mathrm{d}_{\text {rodal }}=\mathrm{d}_{\text {trass }}+\mathrm{d}_{\text {prop }}$ \\We know, $\mathrm{d}_{\text {trax }}=\mathrm{L} / \mathrm{R}$ sec and $\mathrm{d}_{\text {vapp }}=\mathrm{m} / \mathrm{s}$ sec$ $\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }$

(d)

The expression, time $time $t=d_{\text {trans }}$$ means the\at time since transmission started is equal to transmission delay.

As we know, transmission delay is the time taken by host to push out the packet.

Hence, at $time $t=d_{\text {trans }}$$ the last bit of the packet has been pushed out or transmitted.

(e)

If $\ d_{prop} >d_{trans}$

Then, at $time $t=d_{\text {trans }}$$ the bit has been transmitted from host A, but to condition (1), the first bit has not reached B.

(f)

If $\ d_{prop}$

Then, at $time $t=d_{\text {trans }}$$ , the first bit has reached destination on B

Here, $s=2.5 \times 10^{8} \mathrm{sec}$

$\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}$

It's given that $\ d_{prop} =d_{trans}$

Hence,

$\begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}$

5 0

3 years ago

A Simply supported wood beam with overhang is subjected to uniformly distributed load q. The beam has a rectangular cross sectio

irinina [24]

Answer:

q = 61.71 KN/m

Explanation:

We know that shear force at one end of the beam is;

F = wl/2

Where;

w is the uniformly distributed load and l is the span.

Thus, in this question, q is the distributed load, so;

F = ql/2

Area of beam section = breadth x depth

In this case,

Area = 200 × 250 = 50000 mm²

We are given allowable shear stress of τa=1.8MPa. This can also be written as τa = 1.8 N/mm²

We know that formula for average shear stress is;

τ_avg = Force/Area

Thus, Force = τ_avg x Area

However, we are given maximum allowable shear stress as 1.8and we know that; τ_max = 1.5 × τ_avg

Thus, τ_avg = 1.8/1.5 = 1.2

Hence;

Force = 1.2 × 50,000 = 60000 N

We need

So from the earlier equation F = ql/2,we can get; 60000 = ql/2

ql = 120000 - - - - - (1)

Now, to the bending stress, we know that section modulus of a rectangular section is;

Z = bd²/6

So,for this question, we have;

Z = (200 × 250²)/6

Z = 2083333.33 mm²

Maximum bending moment of a simply supported beam is wl²/8

So,in this case, M = ql²/8

So,formula for maximum bending stress = M/Z

So, plugging in the values, we have ;

σ_max = (ql²/8) / 2083333.33

We are given σ= 14 MPa or 14 N/mm²

Thus;

14 = (ql²/8) / 2083333.33

ql² = 14 × 2083333.33 × 8

ql² = 233333332.96 - - - eq(2)

From equation 1,we saw that;ql = 120000.

Putting this for ql in equation 2,we will get;

120000l = 233333332.96

l = 233333332.96/120000

l = 1944.44 mm

So from eq 1,q = 120000/l

q = 120000/1944.44

q = 61.71 KN/m

6 0

3 years ago