Answer: 1.04N
Explanation:
Given
q1 = 2*10^-6C
q2 = 3.6*10^-6C
r = 0.25m
k = 9*10^9
Magnitude of electrostatic force can be calculated by using coulomb's law. Coulomb's law states that, "the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them."
F =(kq1q2) / r²
F = (9*10^9 * 2*10^-6 * 3.6*10^-6) / 0.25²
F = 0.0648/0.0625
F = 1.04N
The type of electrostatic force between the charges is the repulsive force
Answer:
Electric Field = E = 36.848 N/C
Explanation:
In accordance with Columb's law
E = k Q1 Q2 / r.r = 8.99 x 10^9 x 5.0 x 10^-6 x 5.8 x 10^-6 / 0.084 x 0.084
= 36948.6961 x 10^-3 = 36.848 N/C
Answer:
add 44m/s and 22m/s then multiply it by 11
Explanation:
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%