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brasherfamily14
05.09.2020 •
Physics
A 0.22-caliber handgun fires a 28-g bullet at a velocity of 765 m/s. Calculate the de Broglie wavelength of the bullet.
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Ответ:
de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m
Explanation:
Given;
mass of bullet, m = 28 g = 0.028 kg
velocity of the bullet, v = 765 m/s
de Broglie wavelength of the bullet is given by;
λ = h / mv
where;
λ is de Broglie wavelength of the bullet
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
λ = h / mv
λ = (6.626 x 10⁻³⁴ ) / (0.028 x 765)
λ = 3.093 x 10⁻³⁵ m
Therefore, de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m
Ответ:
c and f
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The moon will come between the earth and the moon...which also means the moons shadow will be on the earth's surface.Hope it helps:)