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Volgvan
2 years ago
6

Transcription machinery assembles at _______________.

Engineering
1 answer:
vivado [14]2 years ago
3 0

Answer:

The base

Explanation:

You might be interested in
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
Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,
Lyrx [107]

Answer:

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0

Explanation:

v_0 = Initial velocity of ball A

v_A=v_0\cos45^{\circ}

v_B = Initial velocity of ball B = 0

(v_A)_n' = Final velocity of ball A

v_B' = Final velocity of ball B

e = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

Coefficient of restitution is given by

e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

Adding the above two equations we get

2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0

\boldsymbol{\therefore v_B'=0.6364v_0}

(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0

From the conservation of momentum along the plane of contact we have

(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}

v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0.

5 0
3 years ago
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core tem
Llana [10]

Answer:

The average thickness of the blubber is<u> 0.077 m</u>

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

7 0
3 years ago
Spring-loaded rack guide yokes are made of ______ and have a spring that pushes on the back side of the rack to help reduce the
Kobotan [32]

Answer:

metal

Explanation:

A yoke assembly are use in an assembly of a rack and pinion steering gear for a vehicle. The spring loaded yokes guided with a rack are made up of metals. It consists of a spring made of steel which pushes the back side of the rack to reduce the play that occurs between the pinion and the rack and still allow the relative motion.

6 0
3 years ago
. Two rods, with masses MA and MB having a coefficient of restitution, e, move
GarryVolchara [31]

Answer:

a) V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}

V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}

b) U_A = 3.66 m/s

V_B = 4.32 m/s

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let U_A be the initial velocity of rod A

Let U_B be the initial velocity of rod B

Let V_A be the final velocity of rod A

Let V_B be the final velocity of rod B

Using the principle of conservation of momentum:

M_AU_A + M_BU_B = M_AV_A + M_BV_B............(1)

Coefficient of restitution, e = \frac{V_B - V_A}{U_A - U_B}

V_A = V_B - e(U_A - U_B)........................(2)

Substitute equation (2) into equation (1)

M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B..............(3)

Solving for V_B in equation (3) above:

V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}....................(4)

From equation (2):

V_B = V_A + e(U_A -U_B)......(5)

Substitute equation (5) into (1)

M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))..........(6)

Solving for V_A in equation (6) above:

V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}.........(7)

b)

M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?

Rod A is said to be at rest after the impact, V_A = 0 m/s

Substitute these parameters into equation (7)

0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s

To calculate the final velocity, V_B, substitute the given parameters into (4):

V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s

c) Impulse, I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)

I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))

I = 0 kg m/s^2

d) %\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%

%\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%

% \triangle KE = -47.85 \%

7 0
3 years ago
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