A 5000-lb truck is being used to lift a 1000-lb boulder B that is on a 200-lb pallet A. Knowing that the truck starts from rest
1 answer:
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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:
The answer is "".
Explanation:
The amount of kilograms, which travel in a thick sheet of hydrogen:
calculating the value of
calculating the value of
Answer:
4.26
Explanation:
The wavelength λ is given by:
Phase constant (β) = 2π/λ
βl = 2π/λ × l
l = 2 cm = 0.02 m
βl = 2π/0.0625 × 0.02=2.01 rad = 115.3°
1 rad = 180/π degrees