Find the magnitude of the average force ⟨Fx⟩⟨Fx⟩ in the x direction that the particle exerts on the right-hand wall of the conta
iner as it bounces back and forth. Assume that collisions between the wall and particle are elastic and that the position of the container is fixed. Be careful of the sign of your answer. Express the magnitude of the average force in terms of mmm, vxvxv_x, and LxLxL_x.
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Answer:
λ = 451.7 nm
Explanation:
The expression for the constructive interference of the double diffraction experiment is
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the experiment occurs at very small angles
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
d y / L = m λ
λ =
let's calculate
λ =
λ = 4.51699 10⁻⁷ m
λ = 4.517 10⁻⁷ m (109 nm / 1m)
λ = 451.7 nm