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Crazy boy [7]
3 years ago
15

A frog jumps up at time t = 0.

Physics
2 answers:
valina [46]3 years ago
6 0
Is there a picture that I can see
vesna_86 [32]3 years ago
4 0

Answer:

Decreasing

Explanation:

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This way the person experimenting can use several variables and make observations. Upon making observations, the person can be able to gather as much data as he can in order to answer the original question that he asked.
7 0
3 years ago
Read 2 more answers
What is the bouncing of light when it changes media
matrenka [14]

When light passes from one medium to another, part of it continues on
into the new medium, while the rest of it bounces away from the boundary,
back into the first medium.

The part of the light that continues on into the new medium is <em>transmitted</em>
light.  Its forward progress at any point in its journey is <em>transmission</em>. 

Its direction usually changes as it crosses the boundary.  The bending is <em>
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5 0
3 years ago
Please please help
Feliz [49]

Answer:

D

Explanation:

5 0
2 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
Conclusions _____.
PIT_PIT [208]

Answer:

hjk

Explanation:

hjkhjk

5 0
2 years ago
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