Answer this
a) which ink is likely to be pure? Why?
b) What does the chromatography tell us about ink Y
c) Why are the three different spots separated out from ink Y found at different heights?​

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:

8760 N

Explanation:

think this is the right answer :)

Explanation:

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4) encourage them to go school regularly.

Answer:

This question is incomplete, the complete question is;

People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm.

a) What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.50 cm? (Assume the lens-to-retina distance is 2.00 cm.)

b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)  __ mm

Answer:

1) the power in D of the eyes of a woman is 61.7647 D

2) the size in mm of an image of a 8.00 mm object is -1.882 mm

Explanation:

Given the data in the question;

a) power in D of the eyes of woman who can see an object clearly at a distance of only 8.5 cm and the lens-to-retina distance is 2.00 cm,

so

u = 8.5 cm = ( 8.5 / 100 )m = 0.085 m

v = 2.00 cm = ( 2 / 100 )m =  0.02 m

Now, we know that power of lens p = 1 / u + 1 / v

so we substitute

p = ( 1 / 0.085 ) + ( 1 / 0.02 )

p = 11.7647 + 50

p = 61.7647 D

Therefore,  the power in D of the eyes of a woman is 61.7647 D

b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)

we know that;

m = -v / u

we substitute

m = -0.02 / 0.085

m = -0.2353

since H₀ = 8.0 mm

H[tex]_i[/tex] = m × H₀

H[tex]_i[/tex] = -0.2353 × 8.0

H[tex]_i[/tex] = -1.882 mm

the size in mm of an image of a 8.00 mm object is -1.882 mm

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 .