Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
Answer:
<em>The amount of electric charge transported = 0.192 C</em>
Explanation:
Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)
Q = It..................... Equation 1
Where Q = Electric charge, I = electric current, t = time.
<em>Given:</em> I = 285 mA, t = 674 milliseconds.
<em>Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A</em>
<em> (ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s </em>
Substituting these values into equation 1
Q = 0.285 × 0.674
<em>Q = 0.192 C</em>
<em>Therefore the amount of electric charge transported = 0.192 C</em>
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One is chemo. Chemo is a special magnetic field like to treat cancer
I think Option B, i’m sorry if that’s wrong.
