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zhenek [66]
3 years ago
7

Not sure on this I know it’s not D please help

Physics
1 answer:
svet-max [94.6K]3 years ago
7 0
This is the amplitude I believe.
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4. If the length of a cube is 2000 um calcula<br>its volume​
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A 2.2-kg block slides on a horizontal surface with a speed of v=0.80m/s and an
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Answer:

μ = 0.33

Equal to 3.2 m/s²

Explanation:

Draw a free body diagram of the block.  There are three forces:

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3 years ago
In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is
meriva

The spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

<h3>What is double slit experiment?</h3>

The double-slit experiment serves as a proof in current physics that both light and matter may exhibit properties of classically defined waves and particles. It also illustrates the inherently probabilistic nature of quantum mechanical events. Thomas Young carried out the first experiment of this kind employing light in 1802, illustrating how light behaves like a wave. It was formerly believed that light was made up of either waves or particles. About a century later, with the advent of modern physics, it was discovered that light may in fact exhibit behaviour like that of both waves and particles. The identical behaviour of electrons was first shown by Davisson and Germer in 1927, and it was later extended to atoms and molecules.

The separation between the slits, d = 0.05mm = 5×10⁻⁵ m

The distance from the slits to a screen, D = 2.5 m

Let x is the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits,

λ = 600nm = 6× 10⁻⁷ m  

We know that the bright fringe is given by :

y = nλD/d

So, the spacing between the first-order and second-order bright fringes is :

x = 2λD/d - λD/d

x =  λD/d

x = 6 × 10⁻⁷ × 2.5/5 × 10⁻⁵

x = 0.03 m

or

x = 3 cm

So, the spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

to learn more about double slit experiment go to -

brainly.com/question/28108126

#SPJ4

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What is the primary difference between an ideal emf device and a real emf device?.
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A real emf device has an internal resistance, but an ideal emf device does not.

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