Witam,
proszę o pomoc w rozwiązaniu następujących zadań (będę wdzięczny za jakiekolwiek naprowadzenie albo najlepiej o wzory, z których mogę skorzystać, gdyż nie wiem nawet, od jakiej strony zabrać się za te zadania).
1. To define the change of thermal action in the same dielectrics, which have placed at
first in an electrical field by frequency 50 kHz, and then - 15 MHz. To consider, that
tan \delta accordingly will decrease by 4 times, and E=const.
2. To define the change of capacitive resistance of a biological tissue owing to change
of current frequency, if at a voltage 25 V and frequency 1000 Hz the value of a
current is 0.04 A, and at frequency 5000 Hz – 0.045 A. A biological tissue to
consider as model with consecutive connection of active (R = 420 om) and capacitive
resistances.
3. What is the lifetime τ of the molecule in the excited state if the fluorescence quantum
yield decreased by 4 times when adding extinguishing material with concentration
6 mM? Constant of extinguishing k=10^9 M^-1 s^-1.
4. What is the quantum yield of luminescence matter if optical density is 0.06, and the
luminescence intensity is 3 times less than the intensity of the excitation light?
5. What is the concentration of extinguishing if the fluorescence quantum yield decrees
by four times? The lifetime of a molecule in an excited state t = 4,2×10^-3. Constant
extinguishing k=10^9 M^-1 s^-1.
Zadania akademickie z fizyki
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Re: Zadania akademickie z fizyki
Problem 1
Data:
\( f_{1} = 50\cdot 10^3 Hz \)
\( f_{2} = 15\cdot 10^6 Hz \)
\( \frac{\tg \delta_{1}} {\tg \delta_{2}} = \frac{1}{4} \ \ (*) \)
\( E= const.\)
Calculate:
\( \frac{C_{2}}{C_{1}} \)- change of thermal capicity.
Solution
\( \tg \delta_{1} = \frac{1}{2\pi \cdot f_{1}\cdot R \cdot C_{1}},\)
\( \tg \delta_{2} = \frac{1}{2\pi \cdot f_{2}\cdot R \cdot C_{2}},\)
From the loss tangent quotient \( (*)\)
\( \frac{C_{2}}{C_{1}} = \ \ ... \)
Data:
\( f_{1} = 50\cdot 10^3 Hz \)
\( f_{2} = 15\cdot 10^6 Hz \)
\( \frac{\tg \delta_{1}} {\tg \delta_{2}} = \frac{1}{4} \ \ (*) \)
\( E= const.\)
Calculate:
\( \frac{C_{2}}{C_{1}} \)- change of thermal capicity.
Solution
\( \tg \delta_{1} = \frac{1}{2\pi \cdot f_{1}\cdot R \cdot C_{1}},\)
\( \tg \delta_{2} = \frac{1}{2\pi \cdot f_{2}\cdot R \cdot C_{2}},\)
From the loss tangent quotient \( (*)\)
\( \frac{C_{2}}{C_{1}} = \ \ ... \)