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andrezito [222]
3 years ago
10

Light is described as having a dual wave-particle nature. Which piece of evidence provides support for the model of light as a w

ave?
Light causes electrons to be released when it strikes a metal surface.
Light reflects when it hits a surface.
Packets of energy called photons.
Light travels fastest through a vacuum.
Physics
2 answers:
S_A_V [24]3 years ago
8 0

Answer:

The evidence that supports the wave particle nature of light as a wave is:

"Light reflects when it hits a surface."

Explanation:

Light exhibit wave-particle duality. This means, in some cases the properties of light are just like a wave and in other cases it behaves as particles.

The properties like reflection or refraction from a surface are attributed as properties of waves. The phenomena of interference and diffraction also fall in this category.

Therefore, from the given options, the only option that supports the evidence of light as a wave is "Light reflects when it hits a surface", as reflection is entirely the property of a wave.

Oksi-84 [34.3K]3 years ago
7 0

Answer:

Light reflects when it hits a surface    

Explanation:

Light has wave-particle dual nature.

Particle nature:

The light is composed of packets of energy known as photons. The photoelectric effect shows that light has particle nature. When light strikes a metal surface, it ejects electrons if the energy of the light matches the work function of the metal.

Wave nature:

Light is an electromagnetic wave. Reflection, refraction, interference and Diffraction are phenomenons that are described by its wave nature.

Light reflects from a surface-provides an evidence for light as a wave.

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A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of
Rudiy27

Answer:

Explanation:

Given

mass of spring m=100\ gm

extension in spring x=5\ cm

downward velocity v=70\ cm/s

Position in undamped free vibration is given by

u(t)=A\cos \omega _0t+B\sin \omega _0t

where \omega _0^2=\frac{k}{m}

also \frac{k}{m}=\frac{g}{L}

\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}

\omega _0=14

u(t)=A\cos(14t)+B\sin(14t)

it is given

u(0)=0

u'(0)=70\ cm/s

substituting values we get

A=0

u(t)=B\sin (14t)

u'(t)=14B\cos (14t)

70=14B

B=\frac{10}{2}

B=5

u(t)=5\sin (14t)

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