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Sav [38]
3 years ago
7

WHICH TASK BEST FITS THE ROLE OF A DESIGN ENGINEER ?

Engineering
1 answer:
pshichka [43]3 years ago
5 0

Answer: drafting blueprints

Explanation:

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A seamless pipe 800mm diameter contains a fluid under a pressure of 2N/mm2. If the permissible tensile stress is 100N/mm2, find
Bad White [126]

Answer:

8 mm

Explanation:

Given:

Diameter, D = 800 mm

Pressure, P = 2 N/mm²

Permissible tensile stress, σ = 100 N/mm²

Now,

for the pipes, we have the relation as:

\sigma=\frac{\textup{PD}}{\textup{2t}}

where, t is the thickness

on substituting the respective values, we get

100=\frac{\textup{2\times800}}{\textup{2t}}

or

t = 8 mm

Hence, the minimum thickness of pipe is 8 mm

3 0
3 years ago
C#: Arrays - Ask the user how many students names they want to store. You will create two parallel arrays (e.g. 2 arrays with th
zhenek [66]

Answer:

  1. using System;      
  2. public class Program
  3. {
  4. public static void Main()
  5. {
  6.  Console.WriteLine("Enter number of students: ");
  7.  int num = Convert.ToInt32(Console.ReadLine());
  8.  string [] firstName = new string[num];
  9.  string [] lastName = new string[num];
  10.  
  11.  for(int i=0 ; i < num; i++){
  12.   Console.WriteLine("Enter first name: ");
  13.   firstName[i] = Console.ReadLine();
  14.    
  15.   Console.WriteLine("Enter last name: ");
  16.   lastName[i] = Console.ReadLine();
  17.  }
  18.  
  19.  for(int j=0; j < num; j++){
  20.   Console.WriteLine(lastName[j] + "," + firstName[j]);
  21.  }
  22. }
  23. }

Explanation:

Firstly, prompt user to enter number of student to be stored (Line 6- 7). Next, create two array, firstName and lastName with num size (Line 8-9).

Create a for-loop to repeat for num times and prompt user to enter first name and last name and then store them in the firstName and lastName array, respectively (Line 11 - 17).

Create another for loop to traverse through the lastName and firstName array and display the last name and first name by following the format given in the question (Line 19 - 21).

4 0
3 years ago
For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 1.7. If, afte
luda_lava [24]

Answer:

It would take approximately 305 s to go to 99% completion

Explanation:

Given that:

y = 50% = 0.5

n = 1.7

t = 100 s

We need to first find the parameter k from the equation below.

exp(-kt^n)=1-y

taking the natural logarithm of both sides:

-kt^n=ln(1-y)\\kt^n=-ln(1-y)\\k=-\frac{ln(1-y)}{t^n}

Substituting values:

k=-\frac{ln(1-y)}{t^n}= -\frac{ln(1-0.5)}{100^1.7} = 2.76*10^{-4}

Also

t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}

Substituting values and y = 99% = 0.99

t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s

∴ t ≅ 305 s

It would take approximately 305 s to go to 99% completion

8 0
3 years ago
Read 2 more answers
Describe at least one way you take advantage or could take advantage of each of the different forms of energy
Scorpion4ik [409]

Answer:

Kinetic energy can be used to develop electric energy which can be used as electricity.

Explanation:

The kinetic energy can be harnessed; much like some hydro power technologies harness water movement. A way to convert this kinetic energy into electric energy is through piezoelectric. By applying a mechanical stress to a piezoelectric crystal or material an electric current will be created and can be harvested.

Kinetic energy is also generated by the human body when it is in motion. Studies have also been done using kinetic energy and then converting it to other types of energy, which is then used to power everything from flashlights to radios and more.

4 0
3 years ago
A hypothetical metal has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.413 nm, 0.665 nm, and 0.87
Inessa [10]

Answer:

atomic radius  R = 0.157 nm

metal atomic weight = 72.27 g/mol

Explanation:

given data

parameters a =  0.413 nm

parameters b = 0.665 nm

parameters c =  0.876 nm

atomic packing factor = 0.536

density = 3.99 g/cm³

to find out

atomic radius and  atomic weight

solution

we apply  here atomic packing factor (x) that is

atomic packing factor (x) = \frac{volume(sphere)}{volume(unit\ cell)}  ..................1

put here value we get

atomic packing factor = \frac{8*(4/3)*\pi R^3}{3*a*b*c}

R = (\frac{3(x)(abc)}{32\pi })^{1/3}

R =  (\frac{3(0.536)(0.413*0.665*0.876)}{32\pi })^{1/3}

atomic radius  R = 0.157 nm

and

now we get here metal atomic weight that is

metal atomic weight = \frac{\rho (abc)(N_A)}{no\ of\ atom}   ....................2

metal atomic weight = \frac{3.99 (0.413*0.665*0.876)(6.023*10^{23})}{8}  

metal atomic weight = 72.27 g/mol

6 0
3 years ago
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