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A 4.0-g bead carries a charge of 20 μC. The bead is accelerated from rest through a potential difference V, and afterward the bead is moving at 3.0 m/s. What is the magnitude of the potential difference V? *
1 point
900 V
400 V
200 V
400 kV

Respuesta :

onyebuchinnaji

Answer:

The magnitude of the potential difference is 900 V.

Explanation:

Given;

mass of the bead, m = 4.0 g = 0.004 kg

charge of the bead, Q = 20 μC = 20 x 10⁻⁶ C

final velocity of the bead, v = 3 m/s

What is the magnitude of the potential difference V?

Apply the principle of conservation of energy;

The electric potential energy at the beginning is equal to kinetic energy of the bead  at the end of the journey.

qV = ¹/₂mv²

[tex]V = \frac{mv^2}{2q} \\\\V = \frac{0.004 \ \times \ (3)^2}{2(20 \times 10^{-6})}\\\\V = 900 \ V[/tex]

Therefore, the magnitude of the potential difference is 900 V.

Lanuel

The magnitude of the potential difference (V) is equal to 900 Volts.

Given the following data:

  • Mass of bead = 4.0 g to kg = 0.004 kg
  • Charge of bead = 20 μC = [tex]20 \times 10^{-6} \;C[/tex]
  • Final velocity of bead = 3 m/s

To determine the magnitude of the potential difference (V):

How to calculate the potential difference (V).

We would apply the law of conservation of energy, which states that the electric potential energy possessed by the bead at the beginning is equal to the kinetic energy possessed by the bead at the end of the journey:

[tex]qV = \frac{1}{2} mv^2\\\\V = \frac{\frac{1}{2} mv^2}{q} \\\\V = \frac{\frac{1}{2} \times 0.004 \times 3.0^2}{20 \times 10^{-6}} \\\\V = \frac{ 0.002 \times 9}{20 \times 10^{-6}}[/tex]

V = 900 V.

Read more on kinetic energy here: https://brainly.com/question/1242059

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