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What is the de Broglie wavelength (in meters) of a 455 g football when it is kicked for an extra point at a velocity of 39.3 meters per second?

Answer :

Final answer:

The de Broglie wavelength of the football is approximately 3.706 × 10^-36 meters.

Explanation:

The de Broglie wavelength of a particle can be calculated using the equation λ = h / p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle.

In order to find the de Broglie wavelength of a 455 g football kicked at a velocity of 39.3 m/s, we first need to calculate the momentum of the football.

The momentum of an object can be calculated by multiplying its mass by its velocity: p = m * v. Plugging in the values, we get p = 0.455 kg * 39.3 m/s = 17.8115 kg·m/s.

Now, we can calculate the de Broglie wavelength using the equation λ = h / p.

Since the question provided the value of Planck's constant, we can directly substitute the values to find the wavelength.

Therefore, the de Broglie wavelength of the football is λ ≈ 6.626 × 10^-34 kg·m²/s / 17.8115 kg·m/s = 3.706 × 10^-36 meters.

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