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 answer is b
Explanation:
Second and third class levers are differentiated by <u>the location of the </u><u>load.</u>
<em>Hope</em><em> </em><em>this</em><em> </em><em>help</em><em> </em><em>you</em><em> </em><em>out </em><em>and have</em><em> </em><em>a </em><em>nice</em><em> </em><em>day </em><em>=</em><em>)</em>
Answer:
2156 J
Explanation:
From the question,
Work done = Combined mass of the bucket and water×height×gravity.
W = (M+m)hg............................. Equation 1
Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.
Given: M = 20 kg, m = 2 kg, h = 10 m
Constant: g = 9.8 m/s²
Substitute these value into equation 1
W = (20+2)×10×9.8
W = 22×98
W = 2156 J
Answer:
c. Light energy to thermal energy
Explanation:
The energy from the sun comes in the form of light energy but is converted to thermal energy.