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3 years ago

Where are alkaline earth metals found on the periodic table? Group 1 Group 2 Groups 3–12 Group 17'

Physics

2 answers:

anzhelika [568]3 years ago

8 0

Answer:

B- Group 2... It says my answer is too short so how ya'll doing today?

kogti [31]3 years ago

8 0

Answer:

Elements of this group are beryllium, magnesium, calcium, strontium, barium and radium. Elements of group 1 are called alkali metals and elements of group 17 are called halogens. Thus, we can conclude that out of the given options alkaline earth metals are found on the periodic table in group 2

Explanation:

Hope this helps:)

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3 years ago

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2 years ago

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Answer:

Explanation:

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3 years ago

If a ball rolls off a 4m high table at 7m/s, how long will it take until it hits the floor

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3 years ago

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,

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

5 0

3 years ago