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stealth61 [152]
2 years ago
14

In the modern quantum-based atomic theory, what is the name given to a particular space around the nucleus in which an electron

moves?
Physics
1 answer:
Anastaziya [24]2 years ago
8 0
In the modern quantum based atomic theory, the name given to a particular space around the nucleus in which an electron moves is called orbital. You might be interested to know a little more about the term orbital. Atomic orbital is actually a mathematical function that describes the wave like nature of a single electron or a pair of electrons within an atom.
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Electron grops could be considered which of the following
Maurinko [17]

Answer:

Electron groups could be considered as  Lone pair electrons and bonded pairs of electrons.

Answer: Option D & B

Explanation:

The two or more electrons can be bonded by single bond, double bond, covalent bond of electrons can simply be lone pair of electrons. Unshared pair of electrons are generally termed as lone pair of electrons in an atom which are generally present in the outermost shell of atoms. Hence electron groups can be determined by bonded pairs and lone pairs of electrons.

I got this from another brainly user

5 0
1 year ago
I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block
borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

8 0
2 years ago
Brainlist nd 20 points
nadezda [96]

Answer:

I'm not sure but I think it's 35-39

4 0
2 years ago
Read 2 more answers
What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to
Vadim26 [7]

Answer:

The correct answer to the question is

B. It always decreases

Explanation:

To solve the question, we note that the foce of gravity is given by

F_G=\frac{Gm_1m_2}{r^2} where

G= Gravitational constant

m₁ = mass of first object

m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2} = \frac{2}{4} \frac{Gm_1m_2}{r^2}

Therefore the gravitational force is halved. That is it will always decrease

4 0
3 years ago
SI is considered a consistent system because it what
NemiM [27]

Answer:

because it is a worldwide system....

Explanation:

5 0
2 years ago
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