When an object is in the air, and there is no air resistance acting on the object, the only force that will act on the object is the gravitational force (on the vertical axis).

Then, if the only force acting on the object is the gravitational force, the acceleration of the object will be equal to the gravitational acceleration.

We know that the gravitational acceleration is equal to:

g = 9.8m/s^2

Then the acceleration on the vertical axis will be equal to:

a(t) = 9.8m/s^2

The correct option is the first one:

A. 9.8 m/s^2

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3 years ago

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

Nata [24]

Answer:

(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 connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

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

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = $\frac{C_{3}C_{4} }{C_{3} + C_{4} }$

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

C₅ = $\frac{4.25\times3.98 }{4.25 + 3.98 }$

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

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

Q = 2.05 μF x 17.3 = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = $\frac{Q}{C_{4} }$ = $\frac{35.46\times10^{-6} }{3.98\times10^{-6}}$ = 8.90 volts