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3 years ago

In this case, lithium would be classified as a(n)

Physics

2 answers:

atroni [7]3 years ago

8 0

Answer: C) ion.

Explanation:

An atom is the smallest entity of any matter which may or may not have independent existence.

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.

Ion is formed by a neutral atom or loss or gain of electron. Loss of electron results into positively charged ion called as cation and gain of electron results in negatively charged ion called as anion.

Lithium is an atom with atomic number 3 and thus lose 1 electron to attain stable configuration.

$Li:3:1s622s^1$

$Li^+:2:1s^2$

Molecule can be formed by combination of similar elements or combination of different elements.

Black_prince [1.1K]3 years ago

5 0

Lithium is to be considered as an ION

Atom is defined as its natural state where it will have all its electrons present in its shell

Since in this case Lithium is giving its electron so after giving the electron it will become an ion

Compound is the combination of two or more different atoms

Molecules is the combination of same type of atoms

So here answer must be

<em>C) ion</em>

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garik1379 [7]

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3 years ago

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Andru [333]

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3 years ago

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3 years ago

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$

Where in this case:

$m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m$

Therefore:

$E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J$

8 0

3 years ago