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4 years ago

14

Which is a characteristic of a mixture? Its components retain their properties. It can only be separated using electricity. It i

s the result of a chemical a reaction.

2 answers:

kakasveta [241]4 years ago

6 0

The first answer choice is correct

kifflom [539]4 years ago

5 0

The second one; its components retain their properties.

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The newton is described in units of force!

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How can you degauss a magnet temporarily?

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3 years ago

(Equation 17.6) Write the equation for the path-length difference at a bright fringe (constructive interference). Define all var

Andrews [41]

Answer:

d sin tea = m λ

Explanation:

When we have a two-slit system, the optical path difference determines whether the intensity reaching an observation screen is maximum or zero.

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OP = d sin θ

now to have constructive interference and see a bright line this leg must be an integer number of wavelengths, ose

d sin tea = m λ

where

d is the distance between the two slits

θ complexion the angle sea the point hold it between the two slits

λ the wavelength of the coherent light used

m an integer, which counts the number of lines of interference

Units in the SI system

d, lam in meters

θ degrees

m an integer

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3 years ago

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

Anni [7]

Answer:

The phase difference is $\Delta \phi = 180^o$

Explanation:

From the question we are told that

The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

The wavelength is $\lambda = 400nm$

The distance of the wave from the central maximum is $L = 5mm = 5*10^{-3} m$

Generally the path difference of this waves is mathematically represented as

$y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum

This implies that

$tan \theta = \frac{L}{D}$

=> $\theta = tan ^{-1} \frac{L}{D}$

$\theta = tan ^{-1} [\frac{5*10^{-3}}{1}]$

$\theta =0.2865$

Substituting values into the formula for path difference

$y = 0.2 *10^{-3} sin(0.2864)$

$y = 9.997*10^{-7} \ m$

The phase difference is mathematically represented as

$\Delta \phi = \frac{2 \pi }{\lambda } * y$

Substituting values

$\Delta \phi = \frac{2 \pi }{400 *10^{-9} } \ * 9.997*10^{-7}$

$\Delta \phi =5 \pi$

Converting to degree

$\Delta \phi =5 \pi radians = 5 (180^o) = 180^o$

the solution is subtracted by 360° in order to get the actual angle

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3 years ago