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
spayn [35]
3 years ago
12

The typical Canadian worker is able to produce 100 board feet (a unit of measure) of lumber or 1000 light bulbs per year. The wo

rld price of a board foot of lumber is $20 and that of a light bulb is $1. Draw the production possibilities frontier for the average Canadian worker by manipulating the line segment labeled PPF.
Engineering
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

Explanation:

In the Canadian country, it is possible that the country is able to produce either 100 units of lumber or 1000 lightbulbs. Then, 100 lumber is equal to 1000 bulbs in the home, and hence, one lumber is equivalent to 10 bulbs. Therefore, the opportunity cost of lumber in the home is ten bulbs.

Thus, the country's production possibilities frontier PPF is a linear combination of 100 lumber and 1000 bulbs. The graphical diagram of the production possibilities frontier PPF can be seen in the image below.

You might be interested in
CS3733: Homework/Practice 05 Suppose we would like to write a program called monitor which allows two other programs to communic
valina [46]

Answer:

#include<stdio.h>

#include<stdlib.h>

#include<unistd.h>

#include<sys/types.h>

#include<string.h>

#include<pthread.h>

//#include<sys/wait.h>

int main(int argc, char** argv)

{

int fd1[2];

int fd2[2];

int fd3[2];

int fd4[2];

char message[] = "abcd";

char input_str[100];

pid_t p,q;

if (pipe(fd1)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd2)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd3)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd4)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

p = fork();

if (p < 0)

{

 fprintf(stderr, "fork Failed" );

return 1;

}

// child process-1

else if (p == 0)

{

 close(fd1[0]);// Close reading end of first pipe

 char concat_str[100];

 printf("\n\tEnter meaaage:"):

 scanf("%s",concat_str);

 write(fd1[1], concat_str, strlen(concat_str)+1);

 // Concatenate a fixed string with it

 int k = strlen(concat_str);

 int i;

 for (i=0; i<strlen(fixed_str); i++)

 {

  concat_str[k++] = fixed_str[i];

 }

 concat_str[k] = '\0';//string ends with '\0'

 // Close both writting ends

 close(fd1[1]);

 wait(NULL);

//.......................................................................

 close(fd2[1]);

 read(fd2[0], concat_str, 100);

 if(strcmp(concat_str,"invalid")==0)

 {

 printf("\n\tmessage not send");

 }

 else

 {

  printf("\n\tmessage send to prog_2(child_2).");

 }

 close(fd2[0]);//close reading end of pipe 2

 exit(0);

}

else

{

 close(fd1[1]);//Close writting end of first pipe

 char concat_str[100];

 read(fd1[0], concal_str, strlen(concat_str)+1);

 close(fd1[0]);

 close(fd2[0]);//Close writing end of second pipe

 if(/*check if msg is valid or not*/)

 {

  //if not then

  write(fd2[1], "invalid",sizeof(concat_str));

  return 0;

 }

 else

 {

  //if yes then

  write(fd2[1], "valid",sizeof(concat_str));

  close(fd2[1]);

  q=fork();//create chile process 2

  if(q>0)

  {

   close(fd3[0]);/*close read head offd3[] */

   write(fd3[1],concat_str,sizeof(concat_str);//write message by monitor(main process) using fd3[1]

   close(fd3[1]);

   wait(NULL);//wait till child_process_2 send ACK

   //...........................................................

   close(fd4[1]);

   read(fd4[0],concat_str,100);

   close(fd4[0]);

   if(sctcmp(concat_str,"ack")==0)

   {

    printf("Messageof child process_1 is received by child process_2");

   }

   else

   {

    printf("Messageof child process_1 is not received by child process_2");

   }

  }

  else

  {

   if(p<0)

   {

    printf("Chiile_Procrss_2 not cheated");

   }

   else

   {

     

    close(fd3[1]);//Close writing end of first pipe

    char concat_str[100];

    read(fd3[0], concal_str, strlen(concat_str)+1);

    close(fd3[0]);

    close(fd4[0]);//Close writing end of second pipe

    write(fd4[1], "ack",sizeof(concat_str));

     

   }

  }

 }

 close(fd2[1]);

}

}

8 0
4 years ago
A 1020 CD steel shaft is to transmit 15 kW while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a
vladimir2022 [97]

Answer:

diameter is 14 mm

Explanation:

given data

power = 15 kW

rotation N = 1750 rpm

factor of safety = 3

to find out

minimum diameter

solution

we will apply here power formula to find T that is

power = 2π×N×T / 60    .................1

put here value

15 ×10^{3} = 2π×1750×T / 60

so

T = 81.84 Nm

and

torsion = T / Z                        ..........2

here Z is section modulus i.e = πd³/ 16

so from equation 2

torsion = 81.84 / πd³/ 16

so torsion = 416.75 / / d³     .................3

so from shear stress theory

torsion = σy / factor of safety

so here σy = 530 for 1020 steel

so

torsion = σy / factor of safety

416.75 / d³ = 530 × 10^{6} / 3

so d = 0.0133 m

so diameter is 14 mm

3 0
3 years ago
Refrigerant 134a enters an air conditioner compressor at 4 bar, 20 C, and is compressed at steady state to 12 bar, 80 C. The vol
sleet_krkn [62]

Answer:

Q=15.7Kw

Explanation:

From the question we are told that:

Initial Pressure P_1=4bar

Initial Temperature T_1=20 C

Final Pressure  P_2=12 bar

Final Temperature T_2=80C

Work Output W= 60 kJ/kg

Generally Specific Energy from table is

At initial state

 P_1=4bar \& T_1=20 C

 E_1=262.96KJ/Kg

With

Specific Volume V'=0.05397m^3/kg

At Final state

 P_2=12 bar \& P_2=80C

 E_1=310.24KJ/Kg

Generally the equation for The Process is mathematically given by

 m_1E_1+w=m_2E_2+Q

Assuming Mass to be Equal

 m_1=m_1

Where

 m=\frac{V}{V'}

 m=frac{0.06666}{V'=0.05397m^3/kg}

 m=1.24

Therefore

 1.24*262.96+60)=1.24*310.24+Q

 Q=15.7Kw

4 0
3 years ago
A 4-pole, 60-Hz, 690-V, delta-connected, three-phase induction motor develops 20 HP at full-load slip of 4%. 1) Determine the to
gladu [14]

Answer:

1. i. 20 Nm ii. 4.85 HP

2. 16.5 %

Explanation:

1) Determine the torque and the power developed at 4% slip when a reduced voltage of 340V is applied.

i. Torque

Since slip is constant at 4 %,torque, T ∝ V² where V = voltage

Now, T₂/T₁ = V₂²/V₁² where T₁ = torque at 690 V = P/2πN where P = power = 20 HP = 20 × 746 W = 14920 W, N = rotor speed = N'(1 - s) where s = slip = 4% = 0.04 and N' = synchronous speed = 120f/p where f = frequency = 60 Hz and p = number of poles = 4.

So, N' = 120 × 60/4 = 30 × 60 = 1800 rpm

So, N = N'(1 - s) = 1800 rpm(1 - 0.04) = 1800 rpm(0.96) = 1728 rpm = 1728/60 = 28.8 rps

So, T = P/2πN = 14920 W/(2π × 28.8rps) = 14920 W/180.96 = 82.45 Nm

T₂ = torque at 340 V, V₁ = 690 V and V₂ = 340 V

So, T₂/T₁ = V₂²/V₁²

T₂ = (V₂²/V₁²)T₁

T₂ = (V₂/V₁)²T₁

T₂ = (340 V/690 V)²82.45 Nm

T₂ = (0.4928)²82.45 Nm

T₂ = (0.2428)82.45 Nm

T₂ = 20.02 Nm

T₂ ≅ 20 Nm

ii. Power

P = 2πT₂N'

= 2π × 20 Nm × 28.8 rps

= 1152π W

= 3619.11 W

converting to HP

= 3619.11 W/746 W

= 4.85 HP

2) What must be the new slip for the motor to develop the same torque when the reduced voltage is applied

Since torque T ∝ sV² where s = slip and V = voltage,

T₂/T₁ = s₂V₂²/s₁V₁²

where T₁ = torque at slip, s₁ = 4% and voltage V₁ = 690 V and T₂ = torque at slip, s₂ = unknown and voltage V₂ = 340 V

If the torque is the same, T₁ = T₂ ⇒ T₂T₁ = 1

So,

T₂/T₁ = s₂V₂²/s₁V₁²

1 = s₂V₂²/s₁V₁²

s₂V₂² = s₁V₁²

s₂ = s₁V₁²/V₂²

s₂ = s₁(V₁/V₂)²

substituting the values of the variables into the equation, we have

s₂ = s₁(V₁/V₂)²

s₂ = 4%(690/340)²

s₂ = 4%(2.0294)²

s₂ = 4%(4.119)

s₂ = 16.47 %

s₂ ≅ 16.5 %

3 0
3 years ago
How can you organize a PowerPoint presentation so it engages the audience?
Yakvenalex [24]

Answer:

Use a story structure,

Explanation: Usually story engage the audience.

7 0
3 years ago
Other questions:
  • What’s another name for a service overcurrent device?
    11·1 answer
  • As Becky was driving "Old Betsy," the family station wagon, the engine finally quit, being worn out after 171,000 miles. It can
    12·1 answer
  • Which phrase best describes a safety-critical system? A. a system that faces a very high risk of failure B. a system isolated fr
    13·1 answer
  • How an AK 47 gun was works​
    14·1 answer
  • Where the velocity is highest in the radial direction? Why?
    9·1 answer
  • In this type of projection, the angles between the three axes are different:- A) Isometric B) Axonometric C) Trimetric D) Dimetn
    11·1 answer
  • An Otto cycle with air as the working fluid has a compression ratio of 8.2. Under cold air standard conditions, what is the ther
    14·1 answer
  • You rent an apartment that costs $1800 per month during the first year, but the rent is set to go up 11,5% per year. What would
    12·1 answer
  • 8. What is the purpose of the 300 Log?
    12·1 answer
  • There are some sections of the SDS that are not mandatory.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!