Answer:
Estimated number of indigenous faults remaining undetected is 6
Explanation:
The maximum likelihood estimate of indigenous faults is given by,
here,
= the number of unseeded faults = 6
= number of seeded faults = 30
= number of seeded faults found = 15
So NF will be calculated as,

And the estimate of faults remaining is
= 12 - 6 = 6
Answer should be C hopefully
Answer: 383.22K
Explanation:
L = 3m, w = 1.5m
Area A = 3 x 1.5 = 4.5m2
Q' = 750W/m2 (heat from sun) ,
& = 0.87
Q = &Q' = 0. 87x750 = 652.5W/m2
E = QA = 652.5 x 4.5 = 2936.25W
T(sur) = 300K, T(panel) = ?
Using E = §€A(T^4(panel) - T^4(sur))
§ = Stefan constant = 5.7x10^-8
€ = emmisivity = 0.85
2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)
T(panel) = 383.22K
See image for further details.
Answer:
σ = 391.2 MPa
Explanation:
The relation between true stress and true strain is given as:
σ = k εⁿ
where,
σ = true stress = 365 MPa
k = constant
ε = true strain = Change in Length/Original Length
ε = (61.8 - 54.8)/54.8 = 0.128
n = strain hardening exponent = 0.2
Therefore,
365 MPa = K (0.128)^0.2
K = 365 MPa/(0.128)^0.2
k = 550.62 MPa
Now, we have the following data:
σ = true stress = ?
k = constant = 550.62 MPa
ε = true strain = Change in Length/Original Length
ε = (64.7 - 54.8)/54.8 = 0.181
n = strain hardening exponent = 0.2
Therefore,
σ = (550.62 MPa)(0.181)^0.2
<u>σ = 391.2 MPa</u>