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2 years ago

10

when two capacitor 3muF and 6muF are connected in a parallel and combination is charged to a potential of 120 volt the potential

difference across the 3muF capacitor is

1 answer:

Lunna [17]2 years ago

4 0

Answer:

V₁ = V = 120 V

Explanation:

Such a combination of capacitors in which;

1- Potential difference across each capacitor is the same

2- Total charge is distributed amongst the capacitors

; is called Parallel Combination.

Therefore, in this case, the potential difference across each capacitor will also be the same. Because the capacitors are connected in parallel here. So the voltage across 3 μF capacitor will be the same as the voltage across the 6 μF capacitor and they both will be equal to the total potential difference.

<u>V₁ = V = 120 V</u>

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asambeis [7]

By using the Plancks-Einstein equation, we can find the energy;
E = hf
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3 years ago

A 0.10 g honeybee acquires a charge of +23 pC while flying.

kari74 [83]

Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

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Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

a.

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

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2 years ago

imagine you are going on a rid in a spacecraft next to earth. Your trip takes one whole year. Describe earth's tilt in the north

mylen [45]

You will have to fly around the whole earth to get to your landing station

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2 years ago

Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and fi

steposvetlana [31]

Answer:

$W=K_f-K_i$

Explanation:

The work done on a particle by external forces is defined as:

$W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,$

According to Newton's second law $F=ma$ . Thus:

$W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\$

Acceleration is defined as the derivative of the speed with respect to time:

$W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,$

Speed is defined as the derivative of the position with respect to time:

$W=m\int\limits^{v_f}_{v_i} v \cdot dv \,$

Kinetic energy is defined as $K=\frac{mv^2}{2}$ :

$W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i$

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2 years ago

6) If a mass of an object is decreased to half and acting force is reduced by quarter the acceleration of its motion

Zepler [3.9K]

Answer:

Decreases to half.

Explanation:

From the question given above, the following data were obtained:

Initial mass (m₁) = m

Initial force (F₁) = F

Initial acceleration (a₁) =?

Final mass (m₂) = ½m

Final force (F₂) = ¼F

Final acceleration (a₂) =?

Next, we shall determine a₁. This can be obtained as follow:

F₁ = m₁a₁

F = ma₁

Divide both side by m

a₁ = F / m

Next, we shall determine a₂.

F₂ = m₂a₂

¼F = ½ma₂

2F = 4ma₂

Divide both side by 4m

a₂ = 2F / 4m

a₂ = F / 2m

Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:

a₁ = F / m

a₂ = F / 2m

a₂ : a₁ = a₂ / a₁

a₂ / a₁ = F/2m ÷ F/m

a₂ / a₁ = F/2m × m/F

a₂ / a₁ = ½

Cross multiply

a₂ = ½a₁

From the illustrations made above, the acceleration of the car will decrease to half the original acceleration

7 0

2 years ago