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Drupady [299]
3 years ago
11

The diagram shows a spectrometer being used to analyze a sample of a solution.

Physics
2 answers:
fredd [130]3 years ago
4 0

Answer

B. X

A plate of glass or metal ruled with very close parallel lines, producing a spectrum by diffraction and interference of light.

diffraction grating is an optical component that splits and diffracts light into several beams travelling in different directions. The light emerges with its colours separated.

joja [24]3 years ago
3 0

Answer:

X

Explanation:

I just answered this question on my quiz

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1. bleus

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Just add an -s to make it plural

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Technical skills: knowledge base of sport rearrange our coaching skills from best (at the top) to worst (at the bottom) by dragg
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The steps which enhance strengths and address limitations as a coach include:

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1 year ago
Begin any simulation, and turn on Gravity Force in the central menu on the right. The gravity force arrow shows the direction an
Katyanochek1 [597]

Answer:

The gravity arrow for each body rotates, always pointing toward the other body. Both arrows grow longer when the bodies come closer to one another and shorter when they move farther apart. This change shows that the gravitational force is stronger the closer together the bodies are.

Explanation:

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2 years ago
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6 0
3 years ago
A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this
wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

r=r_o\times A^{\frac{1}{3}}

r_o=1.25 \times 10^{-15} m = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}

r=4.6656\times 10^{-15} m=4.6656 fm

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) F=k\times \frac{q_1q_2}{a^2}

k=9\times 10^9 N m^2/C^2 = Coulombs constant

q_1,q_2 = charges kept at distance 'a' from each other

F = electrostatic force between charges

q_1=+1.602\times 10^{-19} C

q_2=+1.602\times 10^{-19} C

Force of repulsion between two protons on opposite sides of the diameter

a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m

F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}

F=2.6527 N

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0
3 years ago
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