Technician a s ays both an ohmmeter and a self-powered test light may be used to test for continuity. technician b says both may
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Answer:
vacancy formation energy of Ni is 1.400 eV
Explanation:
given data
number of vacancies in Ni = 4.7 x
atomic weight = 58.69 g/mol
density = 8.8 g/cm³
solution
we get here N that is
N = ...........1
N =
N =
and here no of vacancy will be
Nv = .................2
put here value
take ln both side
Qv = 1.400 eV
so vacancy formation energy of Ni is 1.400 eV
Answer:
(a) The mean time to fail is 9491.22 hours
The standard deviation time to fail is 9491.22 hours
(b) 0.5905
(c) 3.915 × 10⁻¹²
(d) 2.63 × 10⁻⁵
Explanation:
(a) We put time to fail = t
∴ For an exponential distribution, we have f(t) =
Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);
e^(1000·λ) - 0.1·e^(1000·λ) = 1
0.9·e^(1000·λ) = 1
1000·λ = ㏑(1/0.9)
λ = 1.054 × 10⁻⁴
Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours
The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours
b) Here we have to integrate from 5000 to ∞ as follows;
(c) The Poisson distribution is presented as follows;
p(x = 3) = 3.915 × 10⁻¹²
d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours
The Cumulative Distribution Function is given as follows;
p( t ≤ 1/4) .