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2 years ago

As the pressure on a sample of gas increases, the volume of the sample _____. decreases stays the same increases

Physics

2 answers:

TEA [102]2 years ago

5 0

Answer : As the pressure on a sample of gas increases, the volume of the sample decreases.

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

$P\propto \frac{1}{V}$ (At constant temperature and moles of gas)

From the we conclude that the higher the pressure of gas, lower will be the volume and lower the pressure of gas, higher will be the volume of gas.

Hence, as the pressure on a sample of gas increases, the volume of the sample decreases.

Allisa [31]2 years ago

3 0

By Boyle's law the volume of the sample decreases, provided temperature is constant.

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scZoUnD [109]

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Explanation:

Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.

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This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.

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ε0 is the electric constant.

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A body of mass 2 kg is moving in the positive X-Direction with a speed of 4 m/s collides head on with an another body of mass 3

Inga [223]

$m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.$

The momentul of the system preserves:

$m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.$

Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with $E$ .

Now, we write the energy conservation law:

$\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E$

From the above equation, you find $E$ , and then conclude that the sound energy can certainly not be greater than this.

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2 years ago