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%
Answer:
W = 0.060 J
v_2 = 0.18 m/s
Explanation:
solution:
for the spring:
W = 1/2*k*x_1^2 - 1/2*k*x_2^2
x_1 = -0.025 m and x_2 = 0
W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2
W = 0.060 J
the work-energy theorem,
W_tot = K_2 - K_1 = ΔK
with K = 1/2*m*v^2
v_2 = √2*W/m
v_2 = 0.18 m/s
Answer:
The appropriate response is "
". A further explanation is described below.
Explanation:
The torque (
) produced by the force on the dam will be:
⇒ 
On applying integration both sides, we get
⇒ 
⇒ 
⇒ ![=pgL[\frac{h^3}{2} -\frac{h^3}{3} ]](https://tex.z-dn.net/?f=%3DpgL%5B%5Cfrac%7Bh%5E3%7D%7B2%7D%20-%5Cfrac%7Bh%5E3%7D%7B3%7D%20%5D)
⇒ 
The correct answer would be left
The percentage of energy available to each organism is always 10 percent .