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Nimfa-mama [501]
3 years ago
6

4. Which of the following is a mechanical property of a material? Conductivity o Strength Heat resistivity Weight​

Engineering
2 answers:
stiv31 [10]3 years ago
4 0

Answer:

The Mechanical Properties include Elasticity, Plasticity, Ductility, Malleability, Hardness, Toughness, Brittleness, Tenacity, Fatigue, Fatigue resistance, Impact Resistance property, Machineability, Strength, Strain Energy, Resilience, Proof Resilience, Modulus of Resilience, Creep, Rupture, and Modulus of Toughness.

Darina [25.2K]3 years ago
4 0
Answer:

It’s conductivity.
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3 years ago
9.21 LAB: Sorting TV Shows (dictionaries and lists) Write a program that first reads in the name of an input file and then reads
natka813 [3]

The following code or the program will be used

<u>Explanation:</u>

def readFile(filename):

   dict = {}

   with open(filename, 'r') as infile:

       lines = infile.readlines()

       for index in range(0, len(lines) - 1, 2):

           if lines[index].strip()=='':continue

           count = int(lines[index].strip())

           name = lines[index + 1].strip()

           if count in dict.keys():

               name_list = dict.get(count)

               name_list.append(name)

               name_list.sort()

           else:

               dict[count] = [name]

           print(count,name)

   return dict

def output_keys(dict, filename):

   with open(filename,'w+') as outfile:

       for key in sorted(dict.keys()):

           outfile.write('{}: {}\n'.format(key,';'.join(dict.get(key))))

           print('{}: {}\n'.format(key,';'.join(dict.get(key))))  

def output_titles(dict, filename):

   titles = []

   for title in dict.values():

       titles.extend(title)

   with open(filename,'w+') as outfile:

       for title in sorted(titles):

           outfile.write('{}\n'.format(title))

           print(title)

def main():

   filename = input('Enter input file name: ')

   dict = readFile(filename)

   if dict is None:

       print('Error: Invalid file name provided: {}'.format(filename))

       return

   print(dict)

   output_filename_1 ='output_keys.txt'

   output_filename_2 ='output_titles.txt'

   output_keys(dict,output_filename_1)

   output_titles(dict,output_filename_2)  

main()

8 0
3 years ago
A 60-m-long steel wire is subjected to a 6-kN tensile load. Knowing that E = 200 GPa and that the length of the rod increases by
Talja [164]

Answer:

(a) 6.91 mm (b) 160 MPa

Explanation:

Solution

Given that:

E = 200 GPa

The rod length = 48 mm

P =P¹ = 6 kN

Recall that,

1 kN = 10^3 N

1 m =10^3 mm

I GPa = 10^9 N/m²

Thus

The rod deformation is stated as follows:

δ = PL/AE-------(1)

σ = P/A----------(2)

Now,

(a) We substitute the values in equation and obtain the following:

48 * 10 ^⁻3 m =  (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]

Thus, we simplify

A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²

A =0.0375 * 10 ^⁻3 m²

A =37.5 mm²

A = π/4 d²

Thus,

d² = 4A /π

After inserting the values we have,

d = √37.5 * 4/3.14 mm

= 6.9116 mm

or d = 6.91 mm

Therefore, the smallest that should be used is 6.91 mm

(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)

Thus,

σ = P/A

σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²

σ= 160 MPa

Note: I MPa = 10^6 N/m²

Hence the the corresponding normal stress is σ= 160 MPa

5 0
3 years ago
Technician A says that Safety Data Sheets contain detailed health information about hazardous materials. Technician B says that
omeli [17]

Answer:

C

Explanation:

The answer is C - both technician A & B

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I have to invent something for people 65pluse so I need to know what are something old people would bye.
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Phones would be a great thing.
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