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erica [24]
4 years ago
6

Oliver is designing a new children's slide to increase speed at which a child can descend.His first design involved steel becaus

e it is a very strong material that can also be smooth. H e found this was not a good option after all. Which explanation most logically explains why the steel was not optimal for this project.
Engineering
1 answer:
goldenfox [79]4 years ago
7 0

Answer:

Steel can rusts easily in the presence of moisture and oxygen, and tarnishes as the rust progresses.

Explanation:

Steel is an alloy of carbon and iron. it is a very useful alloy that is found in almost any engineered piece. The problem with steel is that it is very susceptible to rust, and the cost of maintaining it in order to prevent rust is very high. Steel rusts in the presence of moisture and oxygen (rust is an oxidation-reduction process). Using it as a water slide exposes it constantly to moisture, and to prevent it from rusting in this case will involve a lot of maintenance cost, which is why steel is not advisable to be used in this case.

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Question 1 : Replacement [42 Pts] Consider the following page reference string:
MakcuM [25]

Answer:

Optimal

Time 123456789101112

RS. ecbeagdcegda

F0. eeeeeeeee e e a

F1 c c c c c c c c c c c

F2 b b b g g g g g g g

F3 a a d d d d d d

Page fault? * * * * * * *

Total page fault:7

2. LRU

Time 1 2 3 4 5 6 7 8 9 10 11 12

RS e c b e a g d c e g d a

F0 e e e e e e e c c c c a

F1 c c c c g g g g g g g

F2 b b b b d d d d d d

F3 a a a a e e e e

Page fault? Y Y Y N Y Y Y Y Y N N Y Total page fault:9

3. LRU approximation algorithm: Second chance

Time 1 2 3 4 5 6 7 8 9 10 11 12

RS e c b e a g d c e g d a

F0 0,e 0,e 0,e 1,e 1,e 0,e 0,e 0,e 1,e 1,e 1,e 0,e

F1 0,c 0,c 0,c 0,c 0,g 0,g 0,g 0,g 1,g 1,g 0,g

F20,b0,b0,b0,b0,d0,d0,d0,d1,d0,dF30,a0,a0,a0,c0,c0,c0,c0,a

Page fault? YYYNYYYYNNNY

Total page fault: 8

4 0
3 years ago
Read 2 more answers
1 // Lab 2 tryIt2A 2 #include 3 using namespace std; 4 5 int main() 6 { int x = 1, y = 3; 7 int X = 2, Y = 4; 8 9 cout <<
padilas [110]

Answer:

Here is the complete program:

#include <iostream>

 using namespace std;    

 int main()

 {  int x = 1, y = 3;  

 int X = 2, Y = 4;  

 cout << "tryIt 2A" <<endl;

   cout << x << y << endl;  

   cout << "x" << "y" << endl;  

   cout << X << " " << Y << endl;

   cout << 2 * x + y << endl;  

   cout << 2 * X + Y << endl;  

   //cout << x + 2*y << endl;  

   cout << "x = ";  

   cout << x;  

   cout << " y = ";  

   cout << y;        

   return 0;

   }

Explanation:

I will explain the code line by line in the comment with each line of code and the output of each cout statement.

  • int x = 1, y = 3;  

This statement assigns value 1 to integer variable x and 3 to int variable y

  • int X = 2, Y = 4;  

This statement assigns value 2 to integer variable X and 4 to int variable Y As C++ is a case sensitive language so variable x and y are different from variables X and Y.

  • cout << "tryIt 2A" <<endl;

This statement has cout which is used to display output on the screen. So the output displayed by this cout statement is:

tryIt2A

  • cout << x << y << endl;  

This statement will print the values stored in x and y variables. So output displayed by cout statement here is 1 and 3. As there is not space or next line specified in the statement so output displayed will look like this:

13

  • cout << "x" << "y" << endl;  

This statement will display x and y but these are not the variable x and y. They are enclosed in double quotation marks so they are treated as strings not variables so the output displayed is:

xy

  • cout << X << " " << Y << endl;

This statement will print the values stored in X and Y variables. So output displayed by cout statement here is 2 and 4. As there is  space " " specified in the statement so 2 and 4 are displayed with a space between them so the output displayed will look like this:

2 4

  • cout << 2 * x + y << endl;  

This statement has an arithmetic operation in which 2 is multiplied by the values stored in variable x and then the result is added by value of y. So  2*1 = 2 and 2 + 3 = 5. So the result produced by this cout statement is:

5

  • cout << 2 * X + Y << endl;  

This will work same as above cout statement but the only difference is that the values of capital X and Y variables are calculated here. So 2 * 2 = 4 and then 4 + 4 = 8. The result produced by this cout statement is:

8

  • //cout << x + 2*y << endl;  

This is a comment because before this statement // is written which is used for single line comment. So compiler ignores comments and will not compile this statement.

  •    cout << "x = ";  

This will display "x = " as it is not variable but it is treated as a line to be displayed on the screen. So cout statement displays:

x =

  • cout << x;

This will print the value stored in x variable as there are no double quotes around x so it is a variable which contains value 1. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:

x = 1

  • cout << " y = ";

This will display "y = " as it is not variable but it is treated as a line to be displayed on the screen. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen

x = 1 y =

  • cout << y;    

This will print the value stored in y variable as there are no double quotes around y so it is a variable which contains value 3. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:

x = 1 y = 3

So the output of the entire program along with the program is attached as screenshot.

6 0
4 years ago
Write a Pig script and run it in local mode on this data to find out the top 10 states according to the land area. Since you wan
ELEN [110]

Explanation:

Apache Pig script execution modes

Local mode: In 'local mode', you can run the pig script on the local file system. In this case, you don't need to store the data in the Hadoop HDFS file system, instead you can work with the data stored in the local file system.

MapReduce mode: In 'MapReduce mode', the data must be stored in the HDFS file system and you can process the data with the help of pig script.

Apache Pig Script in MapReduce mode

Let's say our task is to read data from a data file and display the required contents in the terminal as output.

The sample data file contains the following data:

Txt information file - Apache Pig Script - Edureka

Save the text file with the name 'information.txt'

The sample data file contains five First Name, Last Name, Mobile Number, City, and Profession columns separated by the tab key. Our task is to read the contents of this HDFS file and display all the columns of these records.

To process this data using Pig, this file must be present in Apache Hadoop HDFS.

Command: hadoop fs –copyFromLocal /home/edureka/information.txt / edureka

3 0
3 years ago
Suppose that the weights for newborn kittens are normally distributed with a mean of 125 grams and a standard deviation of 15 gr
kherson [118]

(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

(b) For a kitten to be at 90th percentile, the minimum weight is 146.45 g.

<h3>Weight distribution of the kitten</h3>

In a normal distribution curve;

  • 2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%
  • 1 standard deviation (d) below the mean (M), (M - d) is at 16 %
  • 1 standard deviation (d) above the mean (M), (M + d) is at 84%
  • 2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%

M - 2d = 125 g - 2(15g) = 95 g

M - d = 125 g - 15 g = 110 g

95 g is at 2% and 110 g is at 16%

(16% - 2%) = 14%

(110 - 95) = 15 g

14% / 15g = 0.93%/g

From 95 g to 99 g:

99 g - 95 g  = 4 g

4g x 0.93%/g = 3.72%

99 g will be at:

(2% + 3.72%) = 5.72%

Thus, if a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

<h3>Weight of the kitten in the 90th percentile</h3>

M + d = 125 + 15 = 140 g      (at 84%)

M + 2d = 125 + 2(15) = 155 g   ( at 98%)

155 g - 140 g = 15 g

14% / 15g = 0.93%/g

84% + x(0.93%/g) = 90%

84 + 0.93x = 90

0.93x = 6

x = 6.45 g

weight of a kitten in 90th percentile = 140 g + 6.45 g  = 146.45 g

Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g

Learn more about standard deviation here: brainly.com/question/475676

#SPJ1

7 0
2 years ago
A 750-turn solenoid, 24 cm long, has a diameter of 2.3 cm . A 19-turn coil is wound tightly around the center of the solenoid. P
oksian1 [2.3K]

Answer: 2.26x10^-4 v

Explanation:

Lenght of the selonoid = 24x10^-2m

Diameter of the selonoid = 2.3cm

The radius will then be = 1.15cm = 1.15x10^-2m

The area of the selonoid = ¶r^2 = 3.142 x (1.15x10^-2)^2 = 0.000415m^2.

Number of turns on selonoid N1 is 750

For the small center coil, number of turns N2 is 19.

There is a change in current dI/dt from 0 to 5.1 in 0.7s, dI/dt = (5.1-0)/0.7

dI/dt = 7.29A/s.

Induced EMF on selonoid due to magnetic Flux due to changing current in small coil is given as;

E = -M(dI/dt), where M is the mutual inductance of the coils.

but M = (u°AN1N2)/L, where u°= 4¶x10^-7,

A = area of selonoid,

L = Lenght of selonoid.

M = (4¶X10^-7X0.000415X750X19)/(24X10^-2)

M = 3.096X10^-5H

Induced EMF E = 3.096X10^-5 x 7.29

E = 2.26x10^-4V

6 0
4 years ago
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