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:
The "pressure" of the electricity is electric potential. Electric potential is the amount of energy available to push each unit of charge through an electric circuit. The unit of electric potential is the volt. ... A volt is the force needed to move one amp through a conductor that has 1 ohm of resistance
Answer:
The electric flux is zero because charge is zero.
Explanation:
Given that,
Positive charge 
Negative charge 
We need to calculate the total charged
Using formula of charge
Put the value into the formula
We need to calculate the electric flux
Using formula of electric flux
Put the value into the formula
Hence, The electric flux is zero because charge is zero.
46.6666 that is the mass number
Explanation:
14 divided 3.0
Answer:
The starting velocity.
Explanation:
We must understand that this equation comes from the following equation of kinematics.
where:
Vf = final velocity = 33 [m/s]
Vo = starting velocity [m/s]
a = acceleration = 3 [m/s²]
t = time = 30 [s]
So, these values can be assembly in the following way:
