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gtnhenbr [62]
3 years ago
11

g Describe how an electron can possess the properties of both a particle and a wave. How do the wave properties of electrons rel

ate to the Heisenberg uncertainty principle? Electrons possess the properties of particles in that they ---Select--- . However, Louis de Broglie theorized that matter can also show wave properties, specifically characteristic wavelengths based on its mass and velocity. The wave properties of electrons were observed by their ability to ---Select--- when passed through a crystal, similar to the diffraction of electromagnetic waves. The Heisenberg uncertainty principle states that wave properties of electrons ---Select--- the exact location of an electron in space.
Physics
1 answer:
nalin [4]3 years ago
8 0

Answer:

See explanation

Explanation:

According to Louis de Broglie, matter has an associated wavelength. Hence, there exist no clear cut difference between matter and wave. Matter may be regarded as a wave and vice versa depending on the behavior of each under the given circumstances.

According to Heisenberg uncertainty principle, the position and momentum of matter can not be simultaneously determined with precision. This further reinforces the wave-particle concept of the electron.

When electrons are passed through crystals, they are diffracted just like electromagnetic waves. This further reinforces the wave-particle paradox.

According to The Heisenberg uncertainty principle, the wave property of electrons determine their exact location in space

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Answer:

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Approximately, Angular acceleration =

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Explanation:

Length of the rod = 2.0m long

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Rotational Inertia of the Rod(I) = 1/3ML²

Angular Acceleration = ?

There is an equation that shows us the relationship between Torque and Angular acceleration.

The equation is :

Torque(T) = Inertia × Angular Acceleration

Angular acceleration = Torque ÷ Inertia

Where:

Torque = L/2(MgCosθ)

Where M = Mass

L = Length = 2.0m

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g = Acceleration due to gravity = 9.81m/s²

Inertia = 1/3ML²

Angular Acceleration =  (Mass × g × Cos (30°) × (L÷2)) ÷ 1/3ML²

Angular Acceleration =

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Angular Acceleration = (3 × 9.81m/s² × cos 30°) ÷ 2× L

Angular Acceleration = 3 × 9.81m/s² × cos 30°) ÷ 2× 2.0m

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Approximately Angular Acceleration =

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