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bagirrra123 [75]
2 years ago
5

Please help! Kiki makes a table to compare the particles in a magnesium atom to those in a magnesium ion. She knows that a magne

sium ion has an electric field that points away from the ion. What values should she use to complete her table?
(The table looks something like this:
Magnesium Atom  Magnesium Ion
Protons (Magnesium Atom)12  (and Magnesium Ion) X
Neutrons (MA)12  (MI)Y
Electrons (MA) 12  (MI) Z
A. X: 14 Y: 12 Z: 12
B.X: 12 Y: 12 Z: 10
C. X: 12 Y: 10 Z: 12 X: 12
D. Y: 12 Z: 14
Physics
2 answers:
Liono4ka [1.6K]2 years ago
7 0
The answer would be D
Alex2 years ago
5 0

Answer:

B.X: 12 Y: 12 Z: 10  

Explanation:

<u>Electric field points away from a positive ion and towards for a negative ion. An atom becomes an ion when it loses or gains electrons. </u>

A positive ion is formed when the electrons are lost to another atom and remaining atom has more number of protons than electrons.

Atomic number of magnesium is 12 which means it has 12 protons and 12 electrons in a neutral atom. atomic mass of magnesium is 24 which means it has 12 neutrons.

A positive magnesium ion must have electrons less than protons. There would be no change in number of protons and neutrons.

<u>From the given options, in option B, number of protons and neutrons is same i.e. 12 where as number electrons is 10 which is less than the number of protons. This means these are the correct options for magnesium ion. </u>

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given,

net charge = +2.00 μC

we know,

1 coulomb charge =  6.28 x 10¹⁸electrons

1 micro coulomb  charge =  6.28 x 10¹⁸ x 10⁻⁶ electron

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2.00 μC = 2 x 6.28 x 10¹² electrons

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since net charge is positive.

The number of protons should be 1.256 x 10¹³ more than electrons.

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A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

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m= 20 kg

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Here when the mass when hits at spring its speed is

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using

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Now we know;

F = ma

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Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750
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Answer:

Part a)

f = 1911.5 Hz

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\lambda = 0.186 m

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Here the source and observer both are moving towards each other

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f_0 = 1750 Hz

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now we will have

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