All categories

3 years ago

Yeah this question might be difficult as most of the brainly community is math. Hope I can find at last one robotics person. ;-;

Engineering

2 answers:

tatyana61 [14]3 years ago

7 0

Answer:

soryr

Explanation:

Vesnalui [34]3 years ago

6 0

Servos are used in radio-controlled airplanes to position control surfaces like elevators, rudders, walking a robot, or operating grippers.

You might be interested in

Which of the following team members would not be involved in the design of

dimulka [17.4K]

Answer:

Writer

Explanation:

5 0

3 years ago

Your program will be a line editor. A line editor is an editor where all operations are performed by entering commands at the co

soldier1979 [14.2K]

Answer:

Java program given below

Explanation:

import java.util.*;

import java.io.*;

public class Lineeditor

{

private static Node head;

class Node

{

int data;

Node next;

public Node()

{data = 0; next = null;}

public Node(int x, Node n)

{data = x; next =n;}

}

public void Displaylist(Node q)

{if (q != null)

{

System.out.println(q.data);

Displaylist(q.next);

}

public void Buildlist()

{Node q = new Node(0,null);

head = q;

String oneLine;

try{BufferedReader indata = new

BufferedReader(new InputStreamReader(System.in)); // read data from terminals

System.out.println("Please enter a command or a line of text: ");

oneLine = indata.readLine(); // always need the following two lines to read data

head.data = Integer.parseInt(oneLine);

for (int i=1; i<=head.data; i++)

{System.out.println("Please enter another command or a new line of text:");

oneLine = indata.readLine();

int num = Integer.parseInt(oneLine);

Node p = new Node(num,null);

q.next = p;

q = p;}

}catch(Exception e)

{ System.out.println("Error --" + e.toString());}

}

public static void main(String[] args)

{Lineeditor mylist = new Lineeditor();

mylist.Buildlist();

mylist.Displaylist(head);

}

7 0

3 years ago

What type of slab and beam used in construction of space neddle

Sav [38]

jrjrkeekkekekkwkkakkllalllalallalllalalaallalalaalalalalalallallallllallalalallaaallalallllllallllllllalaalalalaaaaalalaaaaaaalgjgiejxpwunfifjruritiririirieoeowowowowowowowowooeowowowoeeoeowowowowowowowoowowwowowowoozoeisiaokseekxidjdkdjfidjfjdjfjfjrifjrifjdirjdjrjfjrjfjrjfjrfuejwwuxmaneanfjkaosndjxieneamalhaqzeeshanvhorahfuensiwjakaksjdhfhfnfhfndjxnxmakaalalalwlwlwwow

5 0

3 years ago

Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at

sergeinik [125]

Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

$h_{f} = 173.358 \\h_{fg} = 2402.522$

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

$h_{2a} = 489.752\\h_{2b} = 313.2$

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

$h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876$

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

$x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119$

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

$\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a} - h_{4a}) - (h_{2a} - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26 - 2241.448938 ) - (489.752 - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}$

Heat transfer rate through boiler

$Q_{in} = mass flow * (h_{3a} - h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W$

Heat transfer rate through condenser

$Q_{out} = mass flow * (h_{4a} - h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W$

Thermal Efficiency

$n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.06485$

Part b) @ 4 MPa

mass flow

$\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b} - h_{4b}) - (h_{2b} - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14 - 2405.54119 ) - (313.12 - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}$