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tangare [24]
4 years ago
6

What should you following distance be at night

Engineering
1 answer:
Darina [25.2K]4 years ago
6 0
6 feet because of quarantine? I really didn’t understand the question.
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What is the aim of reviewing a research paper?
qaws [65]

Answer:

Purpose of review papers

They carefully identify and synthesize relevant literature to evaluate a specific research question, substantive domain, theoretical approach, or methodology and thereby provide readers with a state-of-the-art understanding of the research topic.

3 0
3 years ago
Read 2 more answers
9. What is the Tribal Assistance Coordination Group (TAC-G)?
Alchen [17]
C is the best answer
5 0
3 years ago
The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that deliv
Nataliya [291]

The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is $\frac{3}{8}$  in.

A) Determine the maximum shear stress developed in the shaft.

$\tau_{max}$ = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

$ \phi $ = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

                              $ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$

                              $= 56.67 \pi \text{ rad/s}$

Power $= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$

          = 1424500 ft. lb/s

Torque, $T = \frac{P}{\omega}$

                 $=\frac{1424500}{56.67 \pi}$

                 = 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

$\tau_{max} = \frac{T_c}{J}$

        $=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$

      = 2.93 ksi

B). Angle of twist :

     $\phi = \frac{TL}{JG}$

         $=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$

         = 0.08002 rad

         = 4.58°

6 0
3 years ago
Multiply each element in origList with the corresponding value in offsetAmount. Print each product followed by a space.Ex: If or
n200080 [17]

Answer:

Here is the JAVA program:  

import java.util.Scanner; // to take input from user

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4; // size is fixed that is 4 and assigned to NUM_VALS

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

//two lists origList[] and offsetAmount[] are assigned values

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

String product=""; // product variable to store the product of 2 lists

for(i = 0; i <= origList.length - 1; i++){

/* loop starts with i at 0th position or index and ends when the end of the origList is reached */

/* multiples each element of origList to corresponding element of offsetAmount and stores result in the form of character string in product*/

   product+= Integer.toString(origList[i] *= offsetAmount[i]) + " ";  }

 System.out.println(product); }}   //displays the product of both lists

Output:

80 180 80 400

Explanation:

If you want to print the product of origList with corresponding value in offsetAmount in vertical form you can do this in the following way:

import java.util.Scanner;

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4;

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

for(i = 0; i <= origList.length - 1; i++){

 origList[i] *= offsetAmount[i];

System.out.println(origList[i]); } }}

Output:

80                                                                                                                            

180                                                                                                                          

80                                                                                                                            

400

The program along with the output is attached as screenshot with the input given in the example.

8 0
3 years ago
The magic square is an arrangement of numbers in a square grid in such a way that the sum of the numbers in each row, and in eac
Yuki888 [10]

Answer:

See the explanation for the answer;

Explanation:

Matlab code is as given below

-------------------------------------------------------------------------------------Start of code

% Program to create a magic square and verify it

clc

M= magic(5)

% To find sum of elements of each row

r1= sum(M(1,:))      % Sum of row 1

r2= sum(M(2,:))      % Sum of row 2

r3= sum(M(3,:))      % Sum of row 3

r4= sum(M(4,:))      % Sum of row 4

r5= sum(M(5,:))      % Sum of row 5

% To find sum of each coloumn

c1= sum(M(:,1))      % Sum of coloumn 1

c2= sum(M(:,2))      % Sum of coloumn 2

c2= sum(M(:,3))      % Sum of coloumn 3

c2= sum(M(:,4))      % Sum of coloumn 4

c2= sum(M(:,5))      % Sum of coloumn 5

% To find sum of diagonal

d1= sum(diag(M))               % Sum of principal diagonal elements

d2= sum(diag(fliplr(M)))      

-------------------------------------------------------------------------------End of code

Following results are obtained when executed.

M =

   17    24     1     8    15

   23     5     7    14    16

    4     6    13    20    22

   10    12    19    21     3

   11    18    25     2     9

r1 =

   65

r2 =

   65

r3 =

   65

r4 =

   65

r5 =

   65

c1 =

   65

c2 =

   65

c2 =

   65

c2 =

   65

c2 =

   65

d1 =

   65

d2 =

   65

3 0
3 years ago
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