Question

W=\dfrac{1}{2}CV^2W= 2 1 ​ CV 2 W, equals, start fraction, 1, divided by, 2, end fraction, C, V, squared The equation above describes the amount of energy stored in a capacitor in joules, WWW, in terms CCC, the capacitance of the capacitor in farads, and VVV, the voltage in volts. Which of the following correctly gives CCC in terms of WWW and VVV ?

Answer 1

Answer:

We Should Know Some Trignometric Identities For Solving This Question.

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$\begin{gathered}1. \: \: \sin(2 \alpha ) = 2 \sin( \alpha ) \cos( \alpha ) \\\end{gathered} </p><p>1.sin(2α)=2sin(α)cos(α)$

$2. \: \: \sin(180 - \alpha ) = \sin( \alpha )2.sin(180−α)=sin(α)$

$3. \: \: \sin(90 - \alpha ) = \cos( \alpha )3.sin(90−α)=cos(α)$

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$L.H.S :- Sin10. Sin30. Sin50. Sin70$

$\begin{gathered}= \cos(90 - 10) \sin(30) \cos(90 - 40) \cos(90 - 70) \\ = \cos(80) \: \frac{1}{2} \: \cos(40) \: \cos(20) \\ = \frac{1}{4 \sin(20) } \cos(8 0 ) \cos(40) . \: 2 \sin(20) \cos(20) \\ = \frac{1}{8 \sin(20) } \cos(80) 2. \cos(40) \sin(40) \\ = \frac{1}{16 \sin(20) } 2 \cos(80) \sin(80 ) \\ = \frac{1}{16 \sin(20) } \sin(160) \\ = \frac{1}{16 \sin(20) } \sin(180 - 20) \\ = \frac{1}{16 \sin(20) } \sin(20) \\ = \frac{1}{16} \: \: \: \: \: \: \: \: .........R.H.S\end{gathered}$

$=cos(90−10)sin(30)cos(90−40)cos(90−70)$

$=cos(80) 2/1$

$cos(40)cos(20)= 4sin(20)1$

$cos(80)cos(40).2sin(20)cos(20)= 8sin(20)1$

$cos(80)2.cos(40)sin(40)= 16sin(20)1$

$2cos(80)sin(80)= 16sin(20)1$

$sin(160)= 16sin(20)1$

$sin(180−20)= 16sin(20)1$

$sin(20)= 16/1$

.........R.H.S,

Answer 2

Answer:

idk

Step-by-step explanation:

when i did this i got confused