as a mercury atom absorbs a photon of energy as electron in the atom changes from energy level B to energy level E. calculate the frequency of the absorb photon.

Explanation:

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Answer:

Answer:

A) F = V²E_o•πr²/2d²

B) U = E_o•Aπr²V²/2d

Explanation:

A) Since we have two circular plates, the formula for the electric field is expressed as;

E = V/d

Where;

V is voltage

d is distance

However, the net electric field produced is given by;

E' = V/2d

The tension in the cable can then be expressed as;

F = qE'

Where q is charge

Thus;

F = qV/2d - - - (eq 1)

We also know that;

C = q/V = E_o•A/d

A is area = πr²

Thus;

q/V = E_o•πr²/d

q = VE_o•πr²/d

Let's put VE_o•πr²/d for q in eq 1 to get;

F = V²E_o•πr²/2d²

B) formula for the energy stored in the electric field is;

U = ½CV²

From earlier, we saw that; C = E_o•A/d

Thus;

U = ½E_o•AV²/d

A = πr²

Thus;

U = E_o•Aπr²V²/2d

Answer:

Explanation:

It is a case of adiabatic expansion .

[tex]T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}[/tex]

T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.

Given ,

V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .

[tex]( 3 )^{\gamma-1}= \frac{292}{ T_2}[/tex]

[tex]( 3 )^{1.4-1}= \frac{292}{ T_2}[/tex]

1.552 = 292 / T₂

T₂ = 188 K .