Explain how the extension of a spring is determined
1 answer:
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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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Answer:
Q = 5 L/s
Explanation:
To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):
V: Volume (volumen) = 200L
t: time (tiempo) = 40 s
you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):
Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)