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andreyandreev [35.5K]
3 years ago
8

Suppose there is a sample of xenon in a rectangular container. The gas exerts a total force of 4.47 N perpendicular to one of th

e container walls, whose dimensions are 0.125 m by 0.209 m . Calculate the pressure of the sample.
Physics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

The pressure is P=170.61 Nm^{-2}.

Explanation:

Calculate the area of the rectangular container.

A=LB

Here, L is the length and B is the breadth.

Put L=0.125 m  and B=  0.209 m.

A=(0.125)(0.209)

A = 0.0262 m^{2}

The expression for the pressure in terms of area and force is as follows;

P=\frac{F}{A}

Here, P is the pressure and F is the force.

Put  A = 0.0262 m^{2} and F= 4.47 N.

P=\frac{4.47}{0.0262}

P=170.61 Nm^{-2}

Therefore, the pressure of the given sample is P=170.61 Nm^{-2}.

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(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5
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Answer:

The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Given that,

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Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

a=\dfrac{v^2}{r}

Put the value into the formula

a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}

Put the value into the formula

a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}

a=2.32\times10^{15}\ m/s^2

We need to calculate the rate at which it emits energy because of its acceleration is

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Put the value into the formula

\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}

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\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s

\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s

We need to calculate the fraction of its energy that it radiates every second

\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}

\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}

Hence, The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

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Answer:

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