1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Lyrx [107]
3 years ago
15

NASA’s Ames research center in Mountain View, California has a centrifuge built to simulate high-g environments. It spins horizo

ntally in uniform circular motion, with the payload (possibly a human!) in a box at the end of a 29-foot arm. Its maximum rotation rate is 50 revolutions per minute (rpm). (a) What is the centripetal acceleration experienced by an object in the payload box when it is spinning at its maximum rate? (b) How many gs is this acceleration, and could a human survive it (take the maximum survivable g-force to be 8g). (c) What would the rotation rate be to produce this maximum g-force of 8g?
Engineering
1 answer:
AVprozaik [17]3 years ago
6 0

Answer:

a) a_{c}=795.06 ft/s, b) approximately 25g's, c) 28.46 rpm

Explanation:

In order to solve this problem, we must first do a drawing of the situation to better visualize it (look at attached picture).

a)

Now, the very first thing we must do is convert the frequency to angular speed, we can do this like this:

\omega = \frac{50rev}{s}*\frac{2\pi rad}{1 rev}*\frac{1min}{60s}

so we get that:

\omega=5.236 rad/s

once we have the angular speed, we can use it to find the centripetal acceleration by using the following formula:

a_{c}=\omega^{2}r

which yields:

a_{c}=(5.236rad/s)^{2}(29ft)=795.06 ft/s^{2}

b)

we can use it to figure out how many g's this represents, we know that:

1g=32.2ft/s^{2}

so, we can do the conversion

795.06ft/s^{2}*\frac{1g}{32.2ft/s^{2}}=24.69g's

so we have approximately 25 g's on the machine.

c)

In order to produce a maximum acceleration of 8g's we can do the following.

First, turn the 8g's to ft/s^{2}

8g*\frac{32.2ft/s^{2}}{1g}=257.6ft/s^{2}

next, we use the same formula, but this time we solve for the angular speed:

a_{c}=\omega^{2}r

which solves to:

\omega=\sqrt{\frac{a_{c}}{r}}

and substitute the provided data:

\omega=\sqrt{\frac{257.6ft/s^{2}}{29ft}}

which yields:

\omega=2.98rad/s

we can use this to find our frequency, like this:

f=2.98rad/s*\frac{1rev}{2\pi rad}*\frac{60s}{1min}=28.46rev/min

so the frequency the machine should have to reach an acceleration of 8g's is:

f=28.46rpm

You might be interested in
Why is it important to stop climate change?
vampirchik [111]

Answer:

avoiding cutting down tree carelessy

Explanation:

people cutting down tree due to high population in order to find land for building this house so government should encourage people to have less children in the families and train them that when they are cutting trees should plants 10 tree inorder to recovery tree that is take off.

3 0
2 years ago
Read 2 more answers
A ________ can be installed in a cast-iron block to repair a worn or cracked cylinder. Question 24 options:
VLD [36.1K]

Answer:

Cylinder sleeve

Explanation:

7 0
2 years ago
Consider a simply supported rectangular beam with a span length of 10 ft, a width of 8 in, and an effective depth of 20 in. Norm
yulyashka [42]

Answer:

beam with a span length of 10 ft, a width of 8 in, and an effective depth of 20 in. Normal weight concrete is used for the beam. This beam carries a total factored load of 9.4 kips. The beam is reinforced with tensile steel, which continues uninterrupted into the support. The concrete has a strength of 4000 psi, and the yield strength of the steel is 60,000 psi. Using No. 3 bars and 60,000 psi steel for stirrups, do the followings:

8 0
2 years ago
Consider a multiprocessor system and a multithreaded program written using the many-to-many threading model. Let the number of u
Montano1993 [528]

Answer:

At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.

4 0
2 years ago
g For this project you are required to perform Matrix operations (Addition, Subtraction and Multiplication). For each of the ope
Kruka [31]

Answer:

C++ code is explained below

Explanation:

#include<iostream>

using namespace std;

//Function Declarations

void add();

void sub();

void mul();

//Main Code Displays Menu And Take User Input

int main()

{

  int choice;

  cout << "\nMenu";

  cout << "\nChoice 1:addition";

  cout << "\nChoice 2:subtraction";

  cout << "\nChoice 3:multiplication";

  cout << "\nChoice 0:exit";

 

  cout << "\n\nEnter your choice: ";

 

  cin >> choice;

 

  cout << "\n";

 

  switch(choice)

  {

      case 1: add();

              break;

             

      case 2: sub();

              break;

             

      case 3: mul();

              break;

     

      case 0: cout << "Exited";

              exit(1);

     

      default: cout << "Invalid";      

  }

  main();  

}

//Addition Of Matrix

void add()

{

  int rows1,cols1,i,j,rows2,cols2;

 

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

 

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

 

   int m1[rows1][cols1];

 

  //Taking First Matrix

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

  //Printing 1st Matrix

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

     

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

  int m2[rows2][cols2];

  //Taking Second Matrix

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

  //Displaying second Matrix

  cout << "\n";

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m2[i][j] << " ";

      cout << "\n";

  }

  //Displaying Sum of m1 & m2

  if(rows1 == rows2 && cols1 == cols2)

  {

      cout << "\n";

      for(i=0;i<rows1;i++)

      {

          for(j=0;j<cols1;j++)

              cout << m1[i][j]+m2[i][j] << " ";

          cout << "\n";  

      }

  }

  else

      cout << "operation is not supported";

     

  main();

 

}

void sub()

{

  int rows1,cols1,i,j,k,rows2,cols2;

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

 

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

 

   int m1[rows1][cols1];

 

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

 

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

     

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

  int m2[rows2][cols2];

 

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

 

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

  cout << "\n";

  //Displaying Subtraction of m1 & m2

  if(rows1 == rows2 && cols1 == cols2)

  {

      for(i=0;i<rows1;i++)

      {

          for(j=0;j<cols1;j++)

              cout << m1[i][j]-m2[i][j] << " ";

          cout << "\n";  

      }

  }

  else

      cout << "operation is not supported";

     

  main();

 

}

void mul()

{

  int rows1,cols1,i,j,k,rows2,cols2,mul[10][10];

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

 

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

 

   int m1[rows1][cols1];

 

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

  cout << "\n";

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

     

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

  int m2[rows2][cols2];

 

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

  cout << "\n";

  //Displaying Matrix 2

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m2[i][j] << " ";

      cout << "\n";

  }

     

  if(cols1!=rows2)

      cout << "operation is not supported";

  else

  {

      //Initializing results as 0

      for(i = 0; i < rows1; ++i)

  for(j = 0; j < cols2; ++j)

  mul[i][j]=0;

// Multiplying matrix m1 and m2 and storing in array mul.

  for(i = 0; i < rows1; i++)

  for(j = 0; j < cols2; j++)

  for(k = 0; k < cols1; k++)

  mul[i][j] += m1[i][k] * m2[k][j];

// Displaying the result.

  cout << "\n";

  for(i = 0; i < rows1; ++i)

      for(j = 0; j < cols2; ++j)

      {

      cout << " " << mul[i][j];

      if(j == cols2-1)

      cout << endl;

      }

      }  

  main();

 }

5 0
3 years ago
Other questions:
  • When you are configuring data deduplication, you must choose a usage type for the volume you are configuring. Which of the follo
    8·1 answer
  • A steam pipe passes through a chemical plant, where wind passes in cross-flow over the outside of the pipe. The steam is saturat
    13·1 answer
  • suppose we number the bytes in a w-bit word from 0 (less significant) to w/8-1 (most significant). write code for the followign
    11·1 answer
  • Its an opinion!!!!
    8·1 answer
  • What is matrix ? answer plzzz 27 point you got answer​
    15·2 answers
  • A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu
    6·1 answer
  • Given the latent heat of fusion (melting) and the latent heat of vaporisation for water are Δhs = 333.2 kJ/kg and Δhv = 2257 kJ/
    15·1 answer
  • Consider a sphere made of stainless steel with diameter of 25 cm. It is heated to temperature of 300°C for some chemical tests.
    12·1 answer
  • Identify renewable energy sources you will propose. Explain the key elements to your solution and the basic technical principles
    5·1 answer
  • To understand the concept of moment of a force and how to calculate it using a scalar formulation.
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!