Answer:
hand tracing
Explanation:
as a programmer when we pretend computer in the debugging process by the step of each statement in recording
then there value of variable is hand tracing because as The hand tracking feature is the use of hands as an input method
so while recording value of each variable each step is hand tracing
Answer:
hello your question is incomplete attached below is the complete question
A) overall mean = 5.535, standard deviation ≈ 0.3239
B ) upper limit = 5.85, lower limit = 5.0
C) Not all the samples meet the contract specifications
D) fluctuation ( unstable Asphalt content )
Explanation:
B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday
The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85
The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0
attached below is the required plot
C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :
15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20
D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed
Did not engineer cables factoring wind shear
Answer:
The change in entropy is found to be 0.85244 KJ/k
Explanation:
In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.
P1/T1 = P2/T2
T2/T1 = P2/P1
T2/T1 = 180 KPa/120KPa
T2/T1 = 1.5
Now, the change in entropy is given as:
ΔS = m(s2 - s1)
where,
s2 = Cv ln(T2/T1)
s1 = R ln(V2/V1)
ΔS = change in entropy
m = mass of CO2 = 3.2 kg
Therefore,
ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]
Since, V1 = V2, therefore,
ΔS = mCv ln(T2/T1)
Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K
Therefore,
ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)
<u>ΔS = 0.85244 KJ/k</u>