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3 years ago

Task Three :Write a C++ program to read temperature

Engineering

1 answer:

leva [86]3 years ago

3 0

Answer:

Write a program in C++ to convert temperature in Fahrenheit to Celsius.

Sample Solution:

C++ Code :

#include <iostream>

using namespace std;

int main()

{

float frh, cel;

cout << "\n\n Convert temperature in Fahrenheit to Celsius :\n";

cout << "---------------------------------------------------\n";

cout << " Input the temperature in Fahrenheit : ";

cin >> frh;

cel = ((frh * 5.0)-(5.0 * 32))/9;

cout << " The temperature in Fahrenheit : " << frh << endl;

cout << " The temperature in Celsius : " << cel << endl;

cout << endl;

return 0;

Explanation:

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Write an ALP to separate odd and even numbers from an array of N numbers; arrange odd

Marta_Voda [28]

Below is the program to separate odd and even numbers

<u>Explanation</u>:

mov ah,00

mov al,[BX]

mov dl,al

div dh

cmp ah,00

je EVEN1

mov [DI],dl

add OddAdd,dl

INC DI

INC BX

Loop L1

jmp CAL

mov [SI],dl

add Even Add,dl

INC SI

INC BX

Loop L1

mov ax,0000

mov bx,0000

mov al,OddAdd

mov bl,EvenAdd

MOV ax,4C00h

int 21h

end

The above program separates odd and even numbers from the array using 8086 microprocessor. It has odd numbers in 2000h and even numbers in 3000h.

6 0

4 years ago

The slotted link is pinned at O, and as a result of rotation it drives the peg P along the horizontal guide. Compute the magnitu

MariettaO [177]

Answer:

Find attached of the diagram that complete the question.

Velocity = 1.5 sec²θ m/s

Acceleration = 9 sec²θtanθ m/s²

Explanation:

Calculating the velocity:

V = dx/dt

x = 0.5tanθ

θ = 3t

ö = 3

V = d(0.5tanθ)/dt

= 0.5*sec²θ*ö

= 3*0.5sec²θ

= 1.5 sec²θ m/s

Calculating the acceleration, we have;

a = dv/dt

= d(1.5sec²θ)/dt

= 1.5*2sec²θtanθ*ö

= 1.5*2sec²θtanθ*3

= 9sec²θtanθ m/s²

3 0

4 years ago

The specific gravity of a substance that has mass of 10 kg and occupies a volume of 0.02 m^3 is a) 0.5 b) 1.5 c) 2.5 d) 3.5 e) n

zhuklara [117]

Answer:

Specific gravity is 0.5 so option (a) is correct option

Explanation:

We have given mass of substance m =10 kg

Volume $V=0.02m^3$

Density of substance is given by $Density\ d=\frac{mass}{volume}=\frac{10}{0.02}=500kg/m^3$

Specific gravity is given by $specific\ gravity=\frac{density\ of\ substance}{density\ f\ water}=\frac{500}{1000}=0.5$

So option (a) is correct option

5 0

4 years ago

Technician A says that a part is considered bent if it has a sharp bend with a small radius over a short distance. Technician B

zhuklara [117]

In the above case with Technician A and B, non of them are correct in what they have said.

This is known to be a sharp bend that is known to have a little radius over a short distance.

A Bend is said to be the slow alteration or change in shape that exist between the damaged and undamaged area.

Therefore, looking at the definition, you can see that In the above case with Technician A and B, non of them are correct in what they have said.

Learn more about Technician from

brainly.com/question/18428188

#SPJ1

8 0

2 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