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Debora [2.8K]
2 years ago
10

What determines the state of matter for any substance

Physics
1 answer:
Andrews [41]2 years ago
7 0

Amount of molecules present in the substance

Explanation:

In general state of matter is classified into three types based on the amount of molecules present in the matter. They are widely classified as

  • SOLID
  • LIQUID
  • GAS

SOLID

In solid the molecules are tightly packed without any free space to move inside. Therefore Solids  have strong Inter-Molecular Force of attraction and have definite shape and volume.  

LIQUID      

In Liquid the molecules are  not tightly packed. So that molecules move  freely   inside the matter. Therefore Liquids  have Weaker Inter-Molecular Force of attraction and have no definite shape and  no definite volume.

GAS

In Gas the molecules are loosely packed with  more  free space to move inside. Therefore Gas  have the weakest Inter-Molecular Force of attraction and don't have definite shape and  definite volume.  

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A 0.60 kg rubber ball has a speed of 2.0 m/s at point A, and kinetic energy of 7.5 J at point
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A race car has a maximum speed of 0.104 km/s .What is this speed in miles per hour ?
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Explanation:

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The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

3 0
3 years ago
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