Answer:
Step 1: State your null and alternate hypothesis. ...
Step 2: Collect data. ...
Step 3: Perform a statistical test. ...
Step 4: Decide whether the null hypothesis is supported or refuted. ...
Step 5: Present your findings.
Answer:
B A and C
Explanation:
Given:
Specimen σ
σ
A +450 -150
B +300 -300
C +500 -200
Solution:
Compute the mean stress
σ
= (σ
+ σ
)/2
σ
= (450 + (-150)) / 2
= (450 - 150) / 2
= 300/2
σ
= 150 MPa
σ
= (300 + (-300))/2
= (300 - 300) / 2
= 0/2
σ
= 0 MPa
σ
= (500 + (-200))/2
= (500 - 200) / 2
= 300/2
σ
= 150 MPa
Compute stress amplitude:
σ
= (σ
- σ
)/2
σ
= (450 - (-150)) / 2
= (450 + 150) / 2
= 600/2
σ
= 300 MPa
σ
= (300- (-300)) / 2
= (300 + 300) / 2
= 600/2
σ
= 300 MPa
σ
= (500 - (-200))/2
= (500 + 200) / 2
= 700 / 2
σ
= 350 MPa
From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.
Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.
In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.
Answer:
improper imput validation
Explanation:
Answer:
domestic, public, commercial, and industrial uses.
Answer:
the pressure at the end of the combustion is 2.68 MPa
Solution:
As per the question:
Initial Pressure, P = 0.95\ MPa
Temperature before combustion,
= 273 + 425 = 698 K
Temperature after combustion,
= 1973 K
Now,
To calculate the pressure at the end of combustion, P':
By using the Pressure-Temperature relation from Gay- Lussac's law:


