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Daniel [21]
3 years ago
12

A He-Ne laser (wavelength λ = 600 nm) shines through a double slit of unknown separation d onto a screen 1.44 m away from the sl

it. The distance on the screen between the m = 4 maxima and the central maximum of the two-slit diffract
Chemistry
1 answer:
ludmilkaskok [199]3 years ago
7 0

Answer:

1.28 ×10^-4m

Explanation:

Interference is the superimposition of the wavefront. The superimposition causes bright and dark fringes.

The dark fringe is formed when the additional distance traveled is equal to the multiple wavelength integrals.

From the question, the parameters given are; Wavelength of the light,λ= 600nm, Distance of the screen (D) = 1.44m, order of the bright fringe (m) = 4, Distance of the 4 order fringe (y) = 2.7 cm.

Therefore, for the maximum bright fringe;

dy/D=mλ-----------------------------------------------------------------------------------------------------------------------(1)

Therefore, slotting in equation (1);

d×0.027/1.44=4×600×10^-9

d= 1.28×10^-4 m.

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PLEAZE HELP!!!
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Answer:

V= 12mL

Explanation:

you had the right idea with your Significant figures however, when we divide we see that it requires 2 significant figures as our least amount. this is because when looking at our division, 62 has 2 sig. fig. while 5.35 has a total 3. when looking at your answer we see that you had a total of 3 sig. figures. so in actuakity you had to round up to 12 and not to the tenths because the decimal makes .6 count as your third sig fig.

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What is the particle that is labeled with a question mark (?) in the diagram?
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(Link; https://www.chemteam.info/SigFigs/SigFigsFable.html)
Anna007 [38]

1. The measurement turned out to be so expensive since the student did not rely on the significant figures to calculate the edge of the cube and approximate it to the next value (2.1 cm possibly) that would allow a simpler construction and therefore its cost was much lower .

2. The student had to take all the significant figures in his calculations, with which he would have:

Volume = \frac{mass}{density}

Volume = \frac{80g}{8.67g/mL}

Volume = 9.2272203 mL

Since the figure to be constructed is a cube, he had to calculate the cube root of the volume to find the value of the edge of the cube:

Edge of the cube = \sqrt[3]{9.2272203 cm^{3} } (Taking into account that cm^{3} is proportional to mL)

Edge of the cube = 2.097443624 cm

Because the cube edge value is so specific, in order to manage his budget, he was able to order a 2.1 cm cube, which would bring the mass up to 80.29287 g, and in the lab reduce one of the faces to the appropriate weight. .

On the other hand, the main thing he had to do was ask how much it would cost to make a cube with those specifications, especially when they mentioned that it would be "expensive" and he only had $50.

The significant figures guarantee the correct operation of a machinery, a gear, a team in general, for which the accuracy will not only be taken to the millimeter, but sometimes microns or much more specific, as in the case of computer components, Therefore, it is very important, if not, and if arbitrary measurements are taken that do not consider significant figures, the components could not function properly, which would cause a loss in time, effort and manpower.

If you want to learn more about exercises with significant figures, you can see the next link: brainly.com/question/11904364?referrer=searchResults

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