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OleMash [197]
2 years ago
13

What is the electron dot diagram for magnesium oxide?

Physics
2 answers:
Luda [366]2 years ago
7 0

Hey there mate,

The electron dot diagram of magnesium oxide, as per question, is attached as picture. Click on the picture.

Please check.

Answered by Benjemin360

timofeeve [1]2 years ago
3 0

Answer:

The electron dot diagram for magnesium oxide consists of magnesium in Mg2+form, and oxygen showed with 8 valence electrons (dots) with a negative charge of 2.

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A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti
aleksley [76]

Magnitude of normal force acting on the block is 7 N

Explanation:

10N = 1.02kg

Mass of the block = m = 1.02 kg

Angle of incline Θ =  30°

Normal force acting on the block = N

From the free body diagram,

N = mgCos Θ

N = (1.02)(9.81)Cos(30)

N = 8.66 N

Rounding off to nearest whole number,

N = 7 N

Magnitude of normal force acting on the block = 7 N

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3 years ago
A boy accidentally touch an electric cable and quickly withdrew his hand. Narrate the impulses involved in such action.
Lelechka [254]

Answer:

Quick action was in this as a thought of he knew he could die from this

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3 years ago
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On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi
Lostsunrise [7]

Answer:

0.06\Omega/m

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (I=\frac{V}{R}). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

\frac{13}{20z+r} =7.6   .......(1)

\frac{13}{40z+r} =4.5     ......(2)

Substituting the value of r from (2) in (1), we have,

13=152z+7.6\times\frac{13-180z}{4.5}

which simplifying gives us, z=0.0589\Omega/m\approx0.06\Omega/m (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 \Omega

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3 years ago
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Answer:2 amperes

Explanation:

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