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adell [148]
3 years ago
7

A student constructs a simple constant volume gas thermometer and calibrates it using the boiling point of water, 100°C, and the

freezing point of a specific brine solution, -19°C. The pressures measured at the calibration points are 1.366 atm and 0.9267 atm, respectively. From these data, what temperature will the student calculate, in degrees Celsius, for the value of absolute zero?
Physics
1 answer:
just olya [345]3 years ago
4 0

Answer:

The pressure corresponding to the absolute zero temperature is 0.997atm.

Explanation:

To solve this question, you draw a straight vertical line with the boiling point temperature and pressure on top of the line and the freezing point temperature and pressure on the lower part. The absolute temperature somewhere in the middle of the line with the pressure to be obtained.

So, we have;

0- (-19) / 100 - (-19) = P - 0.9267 / 1.366 - 0.9267

19 / 119 = P - 0.9267 / 0.4393

Cross multiply, we have

19 * 0.4393 = 119(P -0.9267)

8.3467 = 119P - 110.2773

119P = 118.624

P = 0.997 atm

So at 0°C, the pressure of the thermometer is 0.997atm.

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<h2>Answer:</h2>

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Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o
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Answer:

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For the point outside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

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Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

<u>x = -x0:</u>

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dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}

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