A. Email your teacher right away. It would be the safest option.
Answer:
(d) 2 pF
Explanation: the charge on capacitor is given by the expression
Q=CV
where Q=charge
C=capacitance
V=voltage across the plate of the capacitor
here we have given Q=500 pF, V=250 volt
using this formula C=
=500××
=2×
=2 pF
Answer:
first step here is to substitute the 3 of your two equations into the second;
3 Ne^(-Q_v/k(1293)) = Ne^(-Q_v/k(1566))
Since 'N' is a constant, we can remove it from both sides.
We also want to combine our two Q_v values, so we can solve for Q_v, so we should put them both on the same side:
3 = e^(-Q_v/k(1293)) / e^(-Q_v/k(1566))
3 = e^(-Q_v/k(1293) + Q_v/k(1566) ) (index laws)
ln (3) = -Q_v/k(1293) + Q_v/k(1566) (log laws)
ln (3) = -0.13Q_v / k(1566) (addition of fractions)
Q_v = ln (3)* k * 1566 / -0.13 (rearranging the equation)
Now, as long as you know Boltzmann's constant it's just a matter of substituting it for k and plugging everything into a calculator.
Answer:
I feel as if the phrase quantity over quality is the correct answer.