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Rudiy27
4 years ago
9

A wave will "break" because ________. a wave will "break" because ________. the sediment load of water is greater near the shore

of random molecular motion in wave forms the density of water increases closer to shore the bottom interferes with its oscillatory motion the wind increases its speed near shorelines
Physics
1 answer:
vodka [1.7K]4 years ago
5 0
A wave will "break" because the bottom interferes with its oscillatory motion. Breaking of waves may occur anywhere that the amplitude is sufficient, including in mid-ocean. When waves enter shallow water they break because the motion of water in lower part of the wave nearest the bottom is slowed by friction so that their oscillation is faster than its supporting portion at the bottom. Thus, the wave collapses forward and breaks.

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Two children want to balance horizontally on a seesaw. The first child is sitting one meter to the left of the pivot point locat
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4 0
3 years ago
Davisson and Germer performed their experiment with a nickel target for several electron bombarding energies. At what angles wou
pochemuha

Answer:

The angle of diffraction are 67.75 deg and 53.57 deg.

Explanation:

Given:

Davisson and Germer  experiment with nickel target for electrons bombarding.

Voltages : 38\ eV and 54\ eV

We have to find the angles that is  \phi_3_8 and \phi_5_4 .

Concept:

  • Davison Germer experiment is based on de Broglie hypothesis where it says matter has both wave and particle nature.
  • When electrons get reflected from the surface of a metal target with an atomic spacing of D, they form diffraction patterns.
  • The positions of diffraction maxima are given by Dsin(\phi) = n\lambda .
  • An atomic spacing is D = 0.215\ nm, when  the principal maximum corresponds to n=1
  • The wavelength is \lambda, and   \lambda =\frac{1.227 \sqrt{V} } {\sqrt{V_o}}\ nm .

Solution:

Finding the wavelength at V_o=38\ eV .

⇒ \lambda_3_8 =\frac{1.227 \sqrt{V} } {\sqrt{38V}}\ nm

⇒ \lambda_3_8 =0.199 nm

    Plugging the values of wavelength.

⇒  sin(\phi)=\frac{\lambda}{D}

⇒  \phi_3_8=sin^-1(\frac{0.199}{0.215} )

⇒ \phi_3_8 =67.75 degrees.

Now

For for the electrons with energy 54\ eV, V_0=54V the wavelength is.

⇒ \lambda_5_4 =\frac{1.227 \sqrt{V} } {\sqrt{54V}} = 0.173 nm

And

⇒ \phi_5_4=sin^-1(\frac{0.173}{0.215} ) = 53.57 degrees.

So,

The angles of diffraction maxima are 67.75 deg and 53.57 deg.

6 0
3 years ago
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