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:
1.69 T
Explanation:
Applying,
F = BvqsinФ.................. Equation 1
Where F = Force, B = magnetic field, v = velocity, q = charge on an electron, Ф = angle between the electron and the field.
make B the subject of the equation,
B = F/(vqsinФ)............. Equation 2
From the question,
Given: F = 2.0×10⁻¹³ N, v = 7.4×10⁵ m/s, Ф = 90°
Constant: q = 1.60×10⁻¹⁹ C
Substitute into equation 2
B = 2.0×10⁻¹³/(7.4×10⁵×1.60×10⁻¹⁹×sin90°)
B = 0.169×10
B = 1.69 T