Why do giant stars become planetary nebulas while supergiant stars become supernovas when their nuclear fusion slows and is over
1 answer:
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Answer:
theoretical fracture strength = 16919.98 MPa
Explanation:
given data
Length (L) = 0.28 mm = 0.28 × 10⁻³ m
radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m
Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa
solution
we get here theoretical fracture strength s that is express as
theoretical fracture strength = .............................1
put here value and we get
theoretical fracture strength =
theoretical fracture strength =
theoretical fracture strength = 16919.98 MPa
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 rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.
Explanation:
Given the data in the question;
From the first law of thermodynamics;
dQ = dU + dW ------ let this be equation 1
where dQ is the heat transfer, dU is internal energy and dW is the work done.
from the question, the process is isothermal ( internally reversible process )
Thus, the change in internal energy is 0
dU = 0
given that; Air is compressed by a 40-kW compressor from P1 to P2
since it is compressed, dW = -40 kW
we substitute into equation 1
dQ = 0 + ( -40 kW )
dQ = -40 kW
Now, change in entropy of air is;
ΔS = dQ / T
given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K
so we substitute
ΔS = -40 kW / 298.15 K
ΔS = -0.13416 ≈ -0.1342 kW/K
Therefore, the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.