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Dennis_Churaev [7]
3 years ago
7

The potential difference between two parallel conducting plates in vacuum is 165 V. An alpha particle with mass of 6.50×10-27 kg

and charge of 3.20×10-19 C is released from rest near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 40.0 cm.
Physics
1 answer:
shepuryov [24]3 years ago
4 0

Answer:

kinetic energy (K.E) = 5.28 ×10⁻¹⁷            

Explanation:

Given:

Mass of  α particle (m) = 6.50 × 10⁻²⁷ kg

Charge of  α particle (q) = 3.20 × 10⁻¹⁹ C

Potential difference ΔV = 165 V

Find:

kinetic energy (K.E)

Computation:

kinetic energy (K.E) = (ΔV)(q)

kinetic energy (K.E) = (165)(3.20×10⁻¹⁹)

kinetic energy (K.E) = 528 (10⁻¹⁹)

kinetic energy (K.E) = 5.28 ×10⁻¹⁷              

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57. Estimate Potential Energy A boulder with a
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A .025 kg bullet is moving at 350 m/s when it hits and then passes through a 7 kg block of wood. After the bullet passes through
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