Every time I take a photo, that photo has to be stored in a file somewhere within "My Files" correct?How would I be able to take that photo out of the file it was stored?
Answers
Cut that photo by
1. Left click your mouse on the photo
2. Click cut
Then enter the file where you want to transfer and press
1. ctrl+v
Answer:
you can go to your file and then select the phpto and hold on a little bit and choose the delete option
Related Questions
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3 , c 550 J/kg K, k 48 W/m K), which is initially at a uniform temperature of Ti 200 C and is to be heated to a minimum temperature of 550 C. Heating is effected in a gas-fired furnace, where products of combustion at T 800 C maintain a convection coefficient of h 250 W/m2 K on both surfaces of the plate. How long should the plate be left in the furnace
For steels, we can assume the endurance limit measured by rotating beam tests is the 50% of the ultimate tensile strength (UTS) as long as the UTS is equal to or less than 200 ksi (1400 MPa).True/False
_________ items are similar to the free issue items, but their access is limited. (CLO5) a)-Bin stock items free issue b)-Bin stock controlled issue c)-Critical or insurance spares d)-Rebuildable spares e)-consumables
How to find magnitude of net electric charge when current flows through a capacitor with initial charge?
Answers
Answer:
A. 450 btu/h
Explanation:
We solve this problem by using this formula:
Q = U x TD x area
U = U value of used material
TD = Temperature difference = 60°
Q = heat loss
Area = 3x5 = 15
We first find U
R = 1/u
2 = 1/U
U = 1/2 = 0.5
Then when we put these values into the formula above, we would have:
Q = 0.5 x 15 x 60
Q = 450Btu/h
Therefore 450btu/h is the answer
Answers
Answer:
Explanation:
mole fraction of propane after passing through the separator is
mole fractions of ethane and methane
after passing through separator are:
Change in partial pressures then can be written as:
where
and
are mole fractions after and before passing through the separator
Hence,
Answers
Answer:
Heat loss=85.9W/m^2
ΔT1(Steel)=0.04C
ΔT2(Brick1)=110.13C
ΔT3(Mwood)=343.6C
ΔT1(Brick2)=154.18C
Explanation:
raise the heat transfer equation from the air inside the wall to the outside air from the wall, because that is where you have the temperature data, to find the heat.
To find the temperatures you use the heat found in the previous step, and you use the conduction and convection equations in each wall layer.
I attached the procedure
Answer:
hi every body tommoro i have an phisics exam and i need help only for 20 min please if any one is free messege me instagram meeraalk99
Explanation:
please
Answers
The four relevant pressures in a Rankine cycle with one stage of reheat are P1, P2, P3, and P4.
For a Rankine cycle with one stage of reheat between turbines, there are typically four relevant pressures:
- Boiler pressure (P1): This is the pressure at which the water is heated in the boiler before entering the first turbine.
- High-pressure turbine outlet pressure (P2): This is the pressure at the outlet of the first turbine before the steam is sent to the reheater.
- Reheat pressure (P3): This is the pressure at which the steam is reheated before entering the second turbine.
- Low-pressure turbine outlet pressure (P4): This is the pressure at the outlet of the second turbine, which is also the condenser pressure.
To know more about Rankine cycle visit:
#SPJ11
Answers
Answer:
Output:-
Enter the five digit lottery number
Enter the digit 1 : 23
Enter the digit 2 : 44
Enter the digit 3 : 43
Enter the digit 4 : 66
Enter the digit 5 : 33
YOU LOSS!!
Computer Generated Lottery Number :
|12|38|47|48|49|
Lottery Number Of user:
|23|33|43|44|66|
Number Of digit matched: 0
Code:-
import java.util.Arrays;
import java.util.Random;
import java.util.Scanner;
public class Lottery {
int[] lotteryNumbers = new int[5];
public int[] getLotteryNumbers() {
return lotteryNumbers;
}
Lottery() {
Random randomVal = new Random();
for (int i = 0; i < lotteryNumbers.length; i++) {
lotteryNumbers[i] = randomVal.nextInt((50 - 1) + 1);
}
}
int compare(int[] personLottery) {
int count = 0;
Arrays.sort(lotteryNumbers);
Arrays.sort(personLottery);
for (int i = 0; i < lotteryNumbers.length; i++) {
if (lotteryNumbers[i] == personLottery[i]) {
count++;
}
}
return count;
}
public static void main(String[] args) {
int[] personLotteryNum = new int[5];
int matchNum;
Lottery lnum = new Lottery();
Scanner input = new Scanner(System.in);
System.out.println("Enten the five digit lottery number");
for (int i = 0; i < personLotteryNum.length; i++) {
System.out.println("Enter the digit " + (i + 1) + " :");
personLotteryNum[i] = input.nextInt();
}
matchNum = lnum.compare(personLotteryNum);
if (matchNum == 5)
System.out.println("YOU WIN!!");
else
System.out.println("YOU LOSS!!");
System.out.println("Computer Generated Lottery Number :");
System.out.print("|");
for (int i = 0; i < lnum.getLotteryNumbers().length; i++) {
System.out.print(lnum.getLotteryNumbers()[i] + "|");
}
System.out.println("\n\nLottery Number Of user:");
System.out.print("|");
for (int i = 0; i < personLotteryNum.length; i++) {
System.out.print(personLotteryNum[i] + "|");
}
System.out.println();
System.out.println("Number Of digit matched: " + matchNum);
}
}
Explanation:
Answers
Answer:
Less material waste and time.
Explanation:
Two advantages of forging vs machining would be that with forging there is much less waste of material. With machining you remove a large amount of material turning into not so valuable chips.
There is also a time factor, as machining can be very time intensive. This depends on the speed of the machining, newer machines tend to be very fast, and forging requires a lengthy heating, but for large parts the machining can be excessively long.