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erastova [34]
2 years ago
7

What is the basic molecular mechanism for polymer plasticity, and how does it differ from that of ductile crystalline metals?

Engineering
1 answer:
sdas [7]2 years ago
4 0

Answer:

In Crystalline metals or materials, plasticity is examined from the perspective of the motion of linear defects or dislocations within the polymer chains.

Explanation:

When a temperature range below and near the glass transition temperature is reached, there is warping or contortion of structureless or malformed polymers. This warping happens as the polymer chains move over one another.

Unlike elasticity when requires or enables an object to resume its original dimensions, ductility is the quality of an element or material to change form albeit permanently.

Cheers

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When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note th
Amanda [17]

Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant

Differentiating the above equation:

\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\

\to V_A=  \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\

        = \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}

3 0
2 years ago
A pumping test was made in pervious gravels and sands extending to a depth of 50 ft. ,where a bed of clay was encountered. The n
Vikki [24]

Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.

Explanation:

6 0
2 years ago
Write a iterative function that finds the n-th integer of the Fibonacci sequence. Then build a minimal program (main function) t
Natasha2012 [34]

Answer:

Codes for each of the problems are explained below

Explanation:

PROBLEM 1 IN C++:

#include<iostream>

using namespace std;

//fib function that calculate nth integer of the fibonacci sequence.

void fib(int n){

  // l and r inital fibonacci values for n=1 and n=2;

  int l=1,r=1,c;

 

  //if n==1 or n==2 then print 1.

  if(n==1 || n==2){

      cout << 1;

      return;

  }

  //for loop runs n-2 times and calculates nth integer of fibonacci sequence.

  for(int i=0;i<n-2;i++){

      c=l+r;

      l=r;

      r=c;

      cout << "(" << i << "," << c << ") ";

  }

  //prints nth integer of the fibonacci sequence stored in c.

  cout << "\n" << c;

}

int main(){

  int n; //declared variable n

  cin >> n; //inputs n to find nth integer of the fibonacci sequence.

  fib(n);//calls function fib to calculate and print fibonacci number.

}

PROBLEM 2 IN PYTHON:

def fib(n):

   print("fib({})".format(n), end=' ')

   if n <= 1:

       return n

   else:

       return fib(n - 1) + fib(n - 2)

if __name__ == '__main__':

   n = int(input())

   result = fib(n)

   print()

   print(result)

7 0
2 years ago
Read 2 more answers
Problem 1: A catchment has the following Horton’s infiltration parameters: f0=280 mm/hr, fc=25 mm/hr and k = 2.5 hr-1 . For the
harkovskaia [24]

Answer:

Ponding will occur in 40mins

Explanation:

We say that the infiltration rate is the velocity or speed at which water enters into the soil. This often times is measured by the depth (in mm) of the water layer that can enter the soil in one hour. An infiltration rate of 15 mm/hour means that a water layer of 15 mm on the soil surface, will take one hour to infiltrate.

Consider checking attachment for the step by step solution.

6 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
max2010maxim [7]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

7 0
3 years ago
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