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Zielflug [23.3K]
2 years ago
13

If the valence electrons were removed, what would be the ion charge of the element?

Physics
1 answer:
kherson [118]2 years ago
6 0

Answer:

I think C.

Explanation:

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An airplane flew 1000 m in 400 seconds. What is the airplane's speed?
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Answer:

speed =  \frac{distance}{time}  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   =  \frac{1000}{400}  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = 2.5m {s}^{ - 1}

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The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these
Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

F = \frac{kq_{1}q_{2}}{r^2}\\

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\

<u>F = 3.86 x 10⁻⁶ N</u>

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2 years ago
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The Greek philosopher Aristotle and the Roman Catholic Church also believed the sun revolved around the earth. In 1543, Nicolaus Copernicus<span> published a new theory stating the earth revolves around the sun. This is known as the Copernican theory.</span>
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3 years ago
Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is m
Genrish500 [490]

Answer:

the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

Explanation:

Given the data in the question;

Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J

Mass of proton = 1.673 × 10⁻²⁷ kg

Charge of proton = 1.602 × 10⁻¹⁹ C

distance d = 2 m

we know that

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so

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

Potential difference ΔV = 20287.14 V

Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

E = Potential difference ΔV / distance d

we substitute

E = 20287.14 V / 2 m

E = 10143.57 V/m or 1.01 × 10⁴ V/m

Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

3 0
2 years ago
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