Answer:
The consecutive charge configuration has a more intense field than alternating
Explanation:
In each corner we place a different account there are only two different settings, see attached.
In the case of alternating charging (+ - + -) see diagram 1, the electric field in the center is canceled in pairs, resulting in a zero field
In the case of consecutive loads (+ + - -) in this case we have a result between the two charges, therefore the total field is
E = 2 k q / ra2 a cos 45
The consecutive charge configuration has a more intense field than alternating
<span>The process of making alloys involves ( Heating ) pure metals to remove impurities. Then the pure metals are(mixed) with other components. </span><span>An </span>alloy<span> is a mixture of metals or a mixture of a metal and another element. </span>Alloys<span> are defined by a metallic bonding character.</span>
Answer:
Angle θ = 30.82°
Explanation:
From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ
where;
I_o is the intensity of the polarized wave before passing through the filter.
In this question,
I is 0.708 W/m²
While I_o is 0.960 W/m²
Thus, plugging in these values into the equation, we have;
0.708 W/m² = 0.960 W/m² •cos²θ
Thus, cos²θ = 0.708 W/m²/0.960 W/m²
cos²θ = 0.7375
Cos θ = √0.7375
Cos θ = 0.8588
θ = Cos^(-1)0.8588
θ = 30.82°
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:
The relationship between the initial stored energy
and the stored energy after the dielectric is inserted
is:
c) 
Explanation:
A parallel plate capacitor with
that is connected to a voltage source
holds a charge of
. Then we disconnect the voltage source and keep the charge
constant . If we insert a dielectric of
between the plates while we keep the charge constant, we found that the potential decreases as:

The capacitance is modified as:

The stored energy without the dielectric is
The stored energy after the dielectric is inserted is:

If we replace in the above equation the values of V and C we get that


Finally
