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neonofarm [45]
3 years ago
12

How is the pressure of a gas related to its concentration of particles?A) Pressure will expand a gas, enlarging its volume and r

educing its density and concentration of particles. B) Pressure will magnify a gas, developing its volume and multiplying its density and concentration of particles. C) Pressure will compress a gas, reducing its volume and giving it a greater density and concentration of particles. Eliminate D) Pressure will accelerate a gas, extending its volume and allowing a smaller density and concentration of particles.
Physics
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

C) Pressure will compress a gas, reducing its volume and giving it a greater density and concentration of particles.

Explanation:

At constant temperature, pressure and volume are inversely related.

P V = constant

\Rightarrow P \propto \frac{1}{V}

As the pressure increases, the gas compresses, the particles come closer reducing the volume of gas.

As we know, with decrease in volume, density increases.

Density = \frac{Mass}{Volume}

Density \propto \frac{1}{Volume}

Thus, the pressure of a gas is directly related to concentration of particles. Increase in pressure causes increase in concentration of the particles.

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a = 0.159 m/s²

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Answer:

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(b) 24150268.34 N/C

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Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

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E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

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