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IRINA_888 [86]
2 years ago
5

Hi, everyone I'm a high school student in Texas. My engineering teacher is asking us to find an active engineer to complete a li

st of questions, but I don't know any engineers. so If anyone can help I would be appreciated. Thanks

Engineering
1 answer:
jenyasd209 [6]2 years ago
6 0

Answer:ill look it up

Explanation:

You might be interested in
A circular section of material is tested. The original specimen is 200 mm long and has a diameter of 13 mm. When loaded to its p
n200080 [17]

Answer:

modulus of elasticity = 100.45 Gpa,

proportional limit = 150.68 N/mm^2.

Explanation:

We are given the following parameters or data in the question as;

=> "The original specimen = 200 mm long and has a diameter of 13 mm."

=> "When loaded to its proportional limit, the specimen elongates by 0.3 mm."

=> " The total axial load is 20 kN"

Step one: Calculate the area

Area = π/ 4 × c^2.

Area = π/ 4 × 13^2 = 132.73 mm^2.

Step two: determine the stress induced.

stress induced = load/ area= 20 × 1000/132.73 = 150.68 N/mm^2.

Step three: determine the strain rate:

The strain rate = change in length/original length = 0.3/ 200 = 0.0015.

Step four: determine the modulus of elasticity.

modulus of elasticity = stress/strain = 150.68/0.0015 = 100453.33 N/mm^2 = 100.45 Gpa.

Step five: determine the proportional limit.

proportional limit = 20 × 1000/132.73 = 150.68 N/mm^2.

7 0
3 years ago
Read 2 more answers
Developed an automated program in any language which take 12 dependent variable and corresponding independent variables and show
otez555 [7]

Answer:

Explanation:

var generator = new Random(1);

// Now the nextGaussian() function returns a normal distribution of random numbers with the following parameters: a mean of zero and a standard deviation of one

var draw = function() {

var num = generator.nextGaussian();

var standardDeviation = 60;

var mean = 2003;

// Multiply by the standard deviation and add the mean.

var x = standardDeviation * num + mean;

noStroke();

fill(214, 159, 214, 10);

ellipse(x, 200, 16, 16); };

Hope this will be helpful

3 0
2 years ago
Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f
natita [175]

Answer:

Complete question is:

write the following decorators and apply them to a single function (applying multiple decorators to a single function):

1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <strong> and </strong> to the argument of the decorator. The return value of the wrapper should look like: return “<strong>” + func() + “</strong>”

2. The decorator will return the wrapper per usual.

3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <em> and </em> to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “<em>” + func() + “</em>.

4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.

5. Apply both decorators (by @ operator to greetings()).

6. Invoke the greetings() function and capture the result.

Code :

def strong_decorator(func):

def func_wrapper(name):

return "<strong>{0}</strong>".format(func(name))

return func_wrapper

def em_decorator(func):

def func_wrapper(name):

return "<em>{0}</em>".format(func(name))

return func_wrapper

@strong_decorator

@em_decorator

def Greetings(name):

return "{0}".format(name)

print(Greetings("Hello"))

Explanation:

5 0
3 years ago
What is the shape of the output signal on a rigexpert analyzer?
Gekata [30.6K]

Answer:

Output signal shape: square, from 0.1 to 230 MHz. Output power: -10 dBm (at a load of 50 Ohms).

Explanation:

8 0
2 years ago
Read 2 more answers
Identify factors that can cause a process to become out of control. Give several examples of such factors.
Oliga [24]

Answer:

Explained

Explanation:

This situation can occur because of various factors such as:

  • Gradual deterioration of lubrication and coolant.
  • change of environmental condition such as temperature, humidity, moisture, etc.
  • Change in the properties of incoming raw material
  • An increase or decrease in the temperature of the heat treating operation
  • Debris interfering with the manufacturing process.
4 0
2 years ago
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