Una bala calibre se deja caer desde el piso 84 (a una altura de 250 metros) La bala tiene una masa de 0,008 kg. ¿A qué velocidad

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.