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3 years ago

After removing a TLC plate from the developing chamber, what is the first thing to do with the TLC plate?

Physics

1 answer:

postnew [5]3 years ago

4 0

Answer:

The solvent front line should be drawn.

Explanation:

TLC is thin layer chromatography, a techniques used for the separation of substances based on their mobility on their solvents. The silver or thin metal films are used for the absorption.

The TLC plate must be removed from the chamber. After its removal the solvent front line should be drawn to known the absorbance of the material. This technique is quiet useful for the distinction of molecules.

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3 years ago

(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation

icang [17]

Explanation:

(a) Given that,

Area of a parallel plate capacitor, $A=1.8\ cm^2=1.8\times 10^{-4}\ m^2$

The separation between the plates of a capacitor, $d=0.01\ mm = 10^{-5}\ m$

The dielectric constant of, k = 2.1

When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :

$C=\dfrac{k\epsilon_o A}{d}$

Putting all the values we get :

$C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF$

(b) We know that the Teflon has dielectric strength of 60 MV/m, $E=60\times 10^6\ V/m$

The voltage difference between the plates at this critical voltage is given by :

$V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V$

V = 0.6 kV

3 0

3 years ago

Two point charges are on the y axis. A 4.50-σC charge is located at y=1.25 cm , and a -2.24-σC charge is located at y=-1.80 cm.

kvasek [131]

Two point charges are on the y axis. A 4.50-σC charge ( $q_1$ ) is located at y=1.25 cm ( $r_1$ ) , and a -2.24-σC charge ( $q_2$ ) is located at y=-1.80 cm ( $r_2$ ).

The total electric potential at the origin is $2.12\times10^6V$ .

How to calculate the total electric potential at the origin?

The total electric potential at some point due to several point charges is represented by:

$V=k_e \sum _i \frac{q_i}{r_i}$

Here,

$k _e= 8.988 \times 10^9 N \cdot m^2/C^2$ (Columbic constant)

The total electric potential at point $O$ is the sum of the potentials due to $q_1$ and $q_2$ will be:

$V_O=k_e(\frac{q_1}{r_1} +\frac{q_2}{r_2} )$

Substituting the values, we get:

$V_O=8.988\times10^9(\frac{4.5 \times 10^{-6}}{0.0125} +\frac{-2.24\times10^{-6}}{0.018} )=2.12\times10^6V$

Hence, the total electric potential at the origin is $2.12\times10^6V$ .4

To learn more about total electric potential, refer to:

brainly.com/question/16843018

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2 years ago

The________is a measure of the number of waves that pass a point in a given amount of time

cricket20 [7]

Frequency measures how many waves pass a certain point in a given amount of time.
High frequency means the waves are closer together and more frequent, making for a lower wave length.
Low frequency means the waves are farther apart and less frequent, making for a higher wave length.

So the answer is frequency.

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4 years ago

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3 years ago