Answer: D) All of the above
Explanation:
voltage across 2.0μf capacitor is 5.32v
Given:
C1=2.0μf
C2=4.0μf
since two capacitors are in series there equivalent capacitance will be
[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]
=1.33μf
As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.
Q=CV
given,V=8v
charge on 2.0μf capacitor is
=5.32v
learn more about series capacitance from here: brainly.com/question/28166078
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The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1
Temperature that will be my answer number 1
Answer:
in the parallel connection the light bulbs shine less than in the series connection
Explanation:
In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is
P = i² R
where R is the resistance of each bulb and i the current of the circuit.
If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,
i = i₁ + i₂
if the two light bulbs are the same
i = 2 i₁
i₁ = i / 2
so the power of each bulb is is
P = i₁² R
P = R i² / 4
P = ¼ P_initial
Therefore we see that in the parallel connection the light bulbs shine less than in the series connection
