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Coherent light of frequency 6.37×1014 Hz passes through 2 thin slits and also drops on a display screen 88.0 cm ameans. You observe that the

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The distance between the two slits is 40.11 μm.

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Explanation:

Given that,

Frequency Distance of the screen l = 88.0 cm

Position of the 3rd order y =3.10 cm

We must calculate the wavelength

Using formula of wavelength wright here, c = rate of light

f = frequency

Put the value right into the formula  We must calculate the distance between the two slits  Where, m = number of fringe

d = distance in between the two slits

Here, Put the value into the formula   Hence, The distance in between the 2 slits is 40.11 μm.

A parallel combicountry of a 1.13-μF capacitor and a 2.85-μF one is associated in series to a 4.25-μF capacitor. This three-capaci

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(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are associated to the battery having actually voltage V. According to the difficulty, C₁ and C₂ are linked in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the over equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Because, C₄ and C₃ are associated in series, tright here tantamount capacitance is:

C₅ = Substitute 4.25 μF for C₃ and also 3.98 μF for C₄ in the above equation.

C₅ = C₅ = 2.05 μF

The charge on the tantamount capacitance is recognize by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and also 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3 = 35.46 μC

Because, the capacitors C₃ and also C₄ are linked in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

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Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = = = 8.90 volts

Since, C₁ and also C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.