Answer:
74 Ω
Explanation:
Data provided in the question:
Maximum value of the current that can be provided = 500 mA
= 500 × 10⁻³ A
Applied voltage set for the system = 37 V
Now,
The smallest resistance that can be measured
= [ Applied voltage ] ÷ [ Maximum current that can be applied ]
= 37 ÷ ( 500 × 10⁻³ )
= 74 Ω
Answer:
The best first sentence of your e-mail will be-
<u>Weather forecasters are predicting a blizzard this afternoon, so, as a result of this news, our supervisor has decided to close the office at noon so people can travel home safely.</u>
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Answer:
The solution for the given problem is done below.
Explanation:
M1 = 2.0
= 0.3636
= 0.5289
= 0.7934
Isentropic Flow Chart: M1 = 2.0 , 
= 1.8
T1 = 
(1.8)(288K) = 653.4 K.
In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.
At the inlet,
T02= 
= (1.8)(288K) = 518.4 K.
Q= Cp(T02-T01) = 
= 135.7*
J/Kg.
Answer:
//This Program is written in C++
// Comments are used for explanatory purpose
#include <iostream>
using namespace std;
enum mailbox{open, close};
int box[149];
void closeAllBoxes();
void OpenClose();
void printAll();
int main()
{
closeAllBoxes();
OpenClose();
printAll();
return 0;
}
void closeAllBoxes()
{
for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150
{
box[i] = close; //Close all boxes
}
}
void OpenClose()
{
for(int i = 2; i < 150; i++) {
for(int j = i; j < 150; j += i) {
if (box[j] == close) //Open box if box is closed
box[j] = open;
else
box[j] = close; // Close box if box is opened
}
}
// At the end of this test, all boxes would be closed
}
void printAll()
{
for (int x = 0; x < 150; x++) //use this to test
{
if (box[x] = 1)
{
cout << "Mailbox #" << x+1 << " is closed" << endl;
// Print all close boxes
}
}
}
Explanation:
Answer:
See explanation below
Explanation:
Hypo-eutectoid steel has less than 0,8% of C in its composition.
It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.
Ferrite has a higher tensile strength than cementite but cementite is harder.
Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:
Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel
Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel
Hyper-eutectoid steel is harder than Hyper-eutectoid steel
Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.
When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because
1. It is harder
2. It has low cost
3. It is lighter
When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because
1. It is ductile
2. It has high tensile strength
3. It is durable
