Answer:
The magnitude of the load can be computed because it is mandatory in order to produce the change in length ( elongation )
Explanation:
Yield strength = 275 Mpa
Tensile strength = 380 Mpa
elastic modulus = 103 GPa
The magnitude of the load can be computed because it is mandatory in order to produce the change in length ( elongation ) .
Given that the yield strength, elastic modulus and strain that is experienced by the test spectrum are given
strain = yield strength / elastic modulus
= 0.0027
Answer:
a. 164 °F b. 91.11 °C c. 1439.54 kJ
Explanation:
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.
So, Δ°F = 212 °F - 48 °F = 164 °F
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
To find the degree change in Celsius, we convert the initial and final temperature to Celsius.
°C = 5(°F - 32)/9
So, 48 °F in Celsius is
°C₁ = 5(48 - 32)/9
°C₁ = 5(16)/9
°C₁ = 80/9
°C₁ = 8.89 °C
Also, 212 °F in Celsius is
°C₂ = 5(212 - 32)/9
°C₂ = 5(180)/9
°C₂ = 5(20)
°C₂ = 100 °C
So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.
So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?
Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise = 15.8 kJ/°C × 91.11 °C = 1439.54 kJ
Answer:
r=0.228m
Explanation:
The equation that defines the states of a gas according to its thermodynamic properties is given by the general equation of ideal gases
PV=nRT
where
P=pressure =5bar=500.000Pa
V=volume
n=moles=10
R = universal constant for ideal gases = 8.31J / (K.mol)
T=temperature=80F=299.8K
solvig For V
V=(nRT)/P

we know that the volume of a sphere is

solving for r
![r=\sqrt[3]{ \frac{3 V}{4\pi } }](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3%20V%7D%7B4%5Cpi%20%7D%20%7D)
solving
![r=\sqrt[3]{ \frac{3 (0.049)}{4\pi } }\\r=0.228m](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3%20%280.049%29%7D%7B4%5Cpi%20%7D%20%7D%5C%5Cr%3D0.228m)
Answer:
Explanation:
import java.util.Scanner;
public class NestedLoops {
public static void main (String [] args) {
Scanner scnr = new Scanner(System.in);
int numRows;
int numColumns;
int currentRow;
int currentColumn;
char currentColumnLetter;
numRows = scnr.nextInt();
numColumns = scnr.nextInt();
for (currentRow = 0; currentRow < numRows; currentRow++) {
currentColumnLetter = 'A';
for (currentColumn = 0; currentColumn < numColumns; currentColumn++) {
System.out.print(currentRow + 1);
System.out.print(currentColumnLetter + " ");
currentColumnLetter++;
}
}
System.out.println("");
}
}
2
3
1A 1B 1C 2A 2B 2C
thanks
Answer:
Therefore, the horizontal displacement of end F of rod EF is 0.4797 mm
Explanation:
solution is mentioned in steps.