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aleksklad [387]
3 years ago
12

Where are revolved sections placed in a print? A) in between break lines B) cutting planes are used to identify their locations

C) in between section lines D) stand alone
Engineering
1 answer:
Svetach [21]3 years ago
6 0

Answer:

B. Cutting planes are used to identify their locations.

Explanation:

Revolved view is a cross section view of revolved 90 degrees around a cutting plane projections. The revolved view of print will differ from a cross sectional view. It includes a line nothing the axis of revolution for the view. The correct answer is B. The revolved section in the prints has cutting planes that will be used to identify their location.

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In details and step-by-step, show how you apply the Bubble Sort algorithm on the following list of values. Your answer should sh
astraxan [27]

( 12 17 18 19 25 )

<u>Explanation:</u>

<u>First Pass:</u>

( 19 18 25 17 12 ) –> ( 18 19 25 17 12 ), Here, algorithm compares the first two elements, and swaps since 19 > 18.

( 18 19 25 17 12 ) –> ( 18 19 25 17 12 ), Now, since these elements are already in order (25 > 19), algorithm does not swap them.

( 18 19 25 17 12 ) –> ( 18 19 17 25 12 ), Swap since 25 > 17

( 18 19 17 25 12 ) –> ( 18 19 17 12 25 ), Swap since 25 > 12

<u>Second Pass:</u>

( 18 19 17 12 25 ) –> ( 18 19 17 12 25 )

( 18 19 17 12 25 ) –> ( 18 17 19 12 25 ), Swap since 19 > 17

( 18 17 19 12 25 ) –> ( 18 17 12 19 25 ), Swap since 19 > 12

( 18 17 12 19 25 ) –> ( 18 17 12 19 25 )

<u>Third Pass:</u>

( 18 17 12 19 25 ) –> ( 17 18 12 19 25 ), Swap since 18 > 17

( 17 18 12 19 25 ) –> ( 17 12 18 19 25 ), Swap since 18 > 12

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

<u>Fourth Pass:</u>

( 17 12 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 17 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 18 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.

<u>Fifth Pass:</u>

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

6 0
3 years ago
The following C program asks the user for two input null-terminated strings, each stored in uninitialized 100-byte buffer, and c
marissa [1.9K]

Answer:

Code is given below:

Explanation:

.data  

str1: .space 20  

str2: .space 20  

msg1:.asciiz "Please enter string (max 20 characters): "  

msg2: .asciiz "\n Please enter string (max 20 chars): "  

msg3:.asciiz "\nSAME"  

msg4:.asciiz "\nNOT SAME"  

.text

.globl main

main:  

   li $v0,4        #loads msg1  

   la $a0,msg1  

   syscall

   li $v0,8

   la $a0,str1

   addi $a1,$zero,20

   syscall          #got string to manipulate

   li $v0,4        #loads msg2

   la $a0,msg2

   syscall

   li $v0,8

   la $a0,str2

   addi $a1,$zero,20

   syscall         #got string  

       la $a0,str1             #pass address of str1  

   la $a1,str2         #pass address of str2  

   jal methodComp      #call methodComp  

   beq $v0,$zero,ok    #check result  

   li $v0,4

   la $a0,msg4

   syscall

   j exit

ok:  

   li $v0,4  

   la $a0,msg3  

   syscall  

exit:  

   li $v0,10  

   syscall  

methodComp:  

   add $t0,$zero,$zero  

   add $t1,$zero,$a0  

   add $t2,$zero,$a1  

loop:  

   lb $t3($t1)         #load a byte from each string  

   lb $t4($t2)  

   beqz $t3,checkt2    #str1 end  

   beqz $t4,missmatch  

   slt $t5,$t3,$t4     #compare two bytes  

   bnez $t5,missmatch  

   addi $t1,$t1,1      #t1 points to the next byte of str1  

   addi $t2,$t2,1  

   j loop  

missmatch:    

   addi $v0,$zero,1  

   j endfunction  

checkt2:  

   bnez $t4,missmatch  

   add $v0,$zero,$zero  

endfunction:  

   jr $ra

3 0
3 years ago
Design a PI controller to improve the steady-state error. The system should operate with a damping ratio of 0.8. Compute the ove
blondinia [14]

Answer:

The MATLAB Code for this PI Controller will be:

Kp = 350;

Ki = 300;

Kd = 50;

C = pid(Kp,Ki,Kd)

T = feedback(C*P,1);

t = 0:0.01:2;

step(T,t)

Explanation:

When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.

Obtain an open-loop response and determine what needs to be improved

Add a proportional control to improve the rise time

Add a derivative control to reduce the overshoot

Add an integral control to reduce the steady-state error

Adjust each of the gains $K_p$, $K_i$, and $K_d$ until you obtain a desired overall response.

The further explanation is attached in the Word File.

Download docx
5 0
3 years ago
A 1-mm-diameter methanol droplet takes 1 min for complete evaporation at atmospheric condition. What will be the time taken for
Svet_ta [14]

Answer:

Time taken by the 1\mu m diameter droplet is 60 ns

Solution:

As per the question:

Diameter of the droplet, d = 1 mm = 0.001 m

Radius of the droplet, R = 0.0005 m

Time taken for complete evaporation, t = 1 min = 60 s

Diameter of the smaller droplet, d' = 1\times 10^{- 6} m

Diameter of the smaller droplet, R' = 0.5\times 10^{- 6} m

Now,

Volume of the droplet, V = \frac{4}{3}\pi R^{3}

Volume of the smaller droplet, V' = \frac{4}{3}\pi R'^{3}

Volume of the droplet ∝ Time taken for complete evaporation

Thus

\frac{V}{V'} = \frac{t}{t'}

where

t' = taken taken by smaller droplet

\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}

\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}

t' = 60\times 10^{- 9} s = 60 ns

5 0
3 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
3 years ago
Read 2 more answers
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