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3 years ago

What must happen to an atom of magnesium in order to become a magnesium ion

Physics

1 answer:

Zielflug [23.3K]3 years ago

6 0

I would assume that the magnesium would have to come into contact with an ion in order to become a magnesium ion.

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Which particles account for most of the mass of an atom?

BartSMP [9]

Protons and Neutrons account for most of an atoms mass

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3 years ago

In the following description, QUOTE the portion that addresses weather in the area.

olga nikolaevna [1]

Answer:

Portland normally had rainy, but mild winters. Yesterday it rained 0.75 inches in Portland.

Explanation:

This talks about the weather

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3 years ago

A 19 nC charge is moved in a uniform electric field. The electric field does 5.3 μJ of work as the charge moves from point A to

Marizza181 [45]

Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

Work done to move a charge from point B to C, W₂ = -12.2 μJ

Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

W = qV

V = W/q ....(1)

Using equation (1), the potential difference between points A and B is:

$V_{1}=\frac{W_{1} }{q}$

Substitute the suitable values in the above equation.

$V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }$

V₁ = 278.95 V

Using equation (1), the potential difference between points B and C is:

$V_{2}=\frac{W_{2} }{q}$

Substitute the suitable values in the above equation.

$V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }$

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

8 0

3 years ago

A QL = -26 μC charge is placed on the x-axis at x = - 0.2 m. A QR = 26 μC charge is placed at x = +0.2 m. (for all answers below

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The y-component of the electric field at x = 0 m and y = 0.2 m is 2.

</span>

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3 years ago

Click on the ray diagram that shows the proper positioning of an image formed by a concave lens.

Reika [66]

I don't think so

Answer:

Explanation:

beautiful answer

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2 years ago