A particular electromagnetic wave travelling in vacuum is detected to have a frequency of 3 × 10 12 Hz. How much time will it ta
1 answer:
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Answer:
The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.
Explanation:
Here the first think you have to consider is the definition of the Reynolds number () for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:
where is the density of the fluid,
is the viscosity, D is the pipe diameter and v is the velocity of the fluid.
Now, we know that Re=2100. So the velocity is:
For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:
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.