Answer:
Check the 2nd, 3rd and 4th statements.
Explanation:
Answer:
a. is correct.
Explanation:
During servicing you should examine each bearing carefully for chips, pits, and scratches, so Technician A is correct.
Discoloration of a bearingbis not acceptable wear, so Technician B is false.
Combine the answers: Only Technician A is correct therefore answer a. is the right answer.
Answer:
e = 3.97*10^-4
Explanation:
1.8 mm = 0.0018 m
2.6*10^4 mm = 26 m
Elongation is The ratio between the stretched length and the original length.
e = L/L0
This is calculated with Hooke's law:
e = σ/E
Where
σ: normal stress
E: elastic constant
σ = P/A
Where
P: normal load
A: cross section
A = π/4 * d^2
Therefore:
e = P / (A * E)
e = 4 * P / (π * d^2 * E)
e = 4 * 290 / (π * 0.0018^2 * 207*10^9) = 3.97*10^-4
Answer:
h=1.122652m
Explanation:
Assuming density of air = 1.2kg/m³
the differential pressure is given by:
Answer:
I=0.3636
Explanation:
See the attached picture for explanation.
