Answer:
Failure rate = 20%
MTBF = 880 hours
Explanation:
given data
batteries = 10
tested = 200 hours
one failed = 20 hours
another fail at = 140 hours
solution
we know that Mean Time between Failures is express as = (Total up time) ÷ (number of breakdowns) ....................1
so here Total up time will be
Total up time = 200 × 10
Total up time = 2000
and here
Number of breakdown = 1 at 20 hour and another at 140 hour = 2
so it will be = (Total up time) ÷ (number of breakdowns) .......2
=
= 1000
so here gap between occurrences is
gap between occurrences= 140 - 20
gap between occurrences = 120 hour
and
MTBF will be
MTBF = 1000 - 120
MTBF = 880 hours
and
Failure rate (FR) will be
Failure rate (FR) = 1 ÷ MTBF ................3
Failure rate (FR) = R÷T ......................4
as here R is the number of failures and T is total time
so Failure rate (FR) = 20%
Work package. Hope this helps!
Answer:
final displacement lf = 0.39 m
Explanation:
from change in momentum equation:
![\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]](https://tex.z-dn.net/?f=%5Cdelta%20p%20%3D%20m%20%5Csqrt%282g%20%2A%20y%2Fx%29%2A%20%5B%5Csqrt%20li%20%2B%20%5Csqrt%20lf%5D)
given: m = 0.4kg, y/x = 19/85, li = 1.9 m,
\delta p = 1.27 kg*m/s.
putting all value to get the final displacement value
![1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]](https://tex.z-dn.net/?f=1.27%20%3D%200.4%5Csqrt%282%2A9.81%20%2A%2819%2F85%29%29%2A%20%5B%5Csqrt%201.9%20%2B%20%5Csqrt%20lf%5D)
final displacement lf = 0.39 m
Answer:
1÷60 h
time equals distance upon speed