Finding time constants of a circuit? So this is a homework question and I am having trouble figuring out what they are asking. The potential difference (voltage) across the capacitor at time t > 0 is given by V_C(t)=q(t)/C. The quantity RC has the dimensions of time and is often called the time constant for the circuit. How many time constants does it take for a capacitor to charge to 90% of the applied voltage, V0? Justify your answer' So change in V, or V_C , or delta V is 90%. In other words we have 0.9V=q(t)/C? I have found in a previous question that q(t)=V0C(1−e^(−t/CR)) so 0.9V=(V_0C(1-e^((-t)/(CR))))/(C)

$\frac{20b}{{\left(4b^3\right)}^3}$

Which operation could we perform in order to find the number of milliseconds in a year??
${\displaystyle 60\cdot 60\cdot 24\cdot 7\cdot 365}$
${\displaystyle 1000\cdot 60\cdot 60\cdot 24\cdot 365}$
${\displaystyle 24\cdot 60\cdot 100\cdot 7\cdot 52}$
${\displaystyle 1000\cdot 60\cdot 24\cdot 7\cdot 52?}$

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A) No
B) 0
C) Yes
D) 6

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149600000000 is equal to
A)$1.496\times <!--\; \times \; -->{10}^{11}$
B)$1.496\times <!--\; \times \; -->{10}^{10}$
C)$1.496\times <!--\; \times \; -->{10}^{12}$
D)$1.496\times <!--\; \times \; -->{10}^8$

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