Answer:
a) 0.39795 kJ/K
b) 79.589.37 kJ
Explanation:
m = Mass of air = 2 kg
Temperature = 200 K
P₁ = Initial pressure = 300 kPa
P₂ = Final pressure = 600 kPa
R = mass-specific gas constant for air = 287.058 J/kgK
a) For isentropic process
∴ Entropy is generated in the process is 0.39795 kJ/K
b)
∴ Amount of lost work is 79.589.37 kJ
Answer: Not Enough Time
Often, the deadline date is decided before the project starts and is non-negotiable. This deadline results in a headlong rush to get started on the assumption, the sooner you begin coding, the sooner you'll finish.
A rush to start coding is almost always the wrong approach. It is important to spend the time to create a good design. Not having a good design leads to continuing changes throughout the development phase. When this happens, time and budget are consumed at a rapid rate.
Solution: Make time to create a good design. Don't be tempted to jump straight in and begin coding. Assign time to this task and the rest of the project will run much better. It will improve your reputation when you deliver something that fulfils the customers' expectations and works the first time correctly.
Explanation:
Answer:
probability P = 0.32
Explanation:
this is incomplete question
i found complete A manufactures makes integrated circuits that each have a resistance layer with a target thickness of 200 units. A circuit won't work well if this thickness varies too much from the target value. These thickness measurements are approximately normally distributed with a mean of 200 units and a standard deviation of 12 units. A random sample of 17 measurements is selected for a quality inspection. We can assume that the measurements in the sample are independent. What is the probability that the mean thickness in these 16 measurements x is farther than 3 units away from the target value?
solution
we know that Standard error is expess as
Standard error = 
Standard error = 
Standard error = 3
so here we get Z value for 3 units away are from mean are
mean = -1 and + 1
so here
probability P will be
probability P = P( z < -1 or z > 1)
probability P = 0.1587 + 0.1587
probability P = 0.3174
probability P = 0.32
Answer:
The transistor will be in amplifier mode and we will expect current will be higher than expected.
Answer:
The algorithm is as follows:
1. Declare Arr1 and Arr2
2. Get Input for Arr1 and Arr2
3. Initialize count to 0
4. For i in Arr2
4.1 For j in Arr1:
4.1.1 If i > j Then
4.1.1.1 count = count + 1
4.2 End j loop
4.3 Print count
4.4 count = 0
4.5 End i loop
5. End
Explanation:
This declares both arrays
1. Declare Arr1 and Arr2
This gets input for both arrays
2. Get Input for Arr1 and Arr2
This initializes count to 0
3. Initialize count to 0
This iterates through Arr2
4. For i in Arr2
This iterates through Arr1 (An inner loop)
4.1 For j in Arr1:
This checks if current element is greater than current element in Arr1
4.1.1 If i > j Then
If yes, count is incremented by 1
4.1.1.1 count = count + 1
This ends the inner loop
4.2 End j loop
Print count and set count to 0
<em>4.3 Print count</em>
<em>4.4 count = 0</em>
End the outer loop
4.5 End i loop
End the algorithm
5. End
