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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Explanation:
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Battery capacity (AH) is defined as a product of the current that is drawn from the battery while the battery is able to supply the load until its voltage is dropped to lower than a certain value for each cell.
Answer: The electrons flowing through the wire are referred to as a quantity of electricity, and the flow of electricity is referred to as “an electric current.”
Explanation: Hope it Helps have a blessed day
Answer
Time period T = 1.50 s
time t = 40 s
r = 6.2 m
a)
Angular speed ω = 2π/T
= 
= 4.189 rad/s
Angular acceleration α = 
= 
= 0.105 rad/s²
Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²
b)The maximum speed.
v = 2πr/T
= 
= 25.97 m/s
So centripetal acceleration.
a = 
= 
= 108.781 m/s^2
= 11.1 g
in combination with the gravitation acceleration.
