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
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Answer:
30 kilometers is a reasonable measurement
Answer:
4. total energy
Explanation:
According to Bernoulli's principle at any two points along a streamline flow The total energy that is sum of pressure energy , Kinetic energy and potential energy of the liquid all taken in per unit volume remains constant. Therefore,
for ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any given point in the pipe, the fluid has a constant Total Energy.