11:33A moodle.nct.edu.omQuestion 12Not yet answeredMarked out of 1.00Flag questionAccording to the kinetic theory of gases, whencollisions occur the molecules did not lose theirO a. Kinetic EnergyO b. Potential EnergyO c. Gravitational EnergyO d. Pressure and temperature of the gasQuestion 13Not yet answeredMarked out of 1.00Flag questionA magnetic field of 7.2 T passes through a singleloop square coil of side length 5.2 x 10-2 m whichmakes 32.9° angle with the area vector of thecoil. Find the magnetic flux associated with the ccin Wb/m2?个

13. Magnetic field, B = 7.2 T side, s = 5.2 x 10-2 m Angle, A = 32.9 degree

A magnetic field of 7.2 T passes through a single loop square coil of side length 5.2 x 10-2 m which makes 32.9° angle with the area vector of the coil. Find the magnetic flux associated with the cc in Wh/m2?

A magnetic field of 7.2 T passes through a single loop square coil of side length 5.2 x 10-2 m which makes 32.9° angle with the area vector of the coil. Find the magnetic flux associated with the cc in Wh/m2?

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter20: Kinetic Theory Of Gases

Section: Chapter Questions

Problem 44PQ

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Consider the Maxwell-Boltzmann distribution function plotted in Problem 28. For those parameters, determine the rms velocity and the most probable speed, as well as the values of f(v) for each of these values. Compare these values with the graph in Problem 28. 28. Plot the Maxwell-Boltzmann distribution function for a gas composed of nitrogen molecules (N2) at a temperature of 295 K. Identify the points on the curve that have a value of half the maximum value. Estimate these speeds, which represent the range of speeds most of the molecules are likely to have. The mass of a nitrogen molecule is 4.68 1026 kg. Equation 20.18 can be used to find the rms velocity given the temperature, Boltzmanns constant, and the mass of the atom or molecule. The mass of a nitrogen molecule is 4.68 1026 kg. vrms=3kBTm=3(1.381023J/K)4.681026kg=511m/s Using the results of Problem 28 and the rms velocity, we can calculate the value of f(v). f(vrms) = (3.11 108)(511)2 e(5.75106(511)2) = 0.00181 The most probable speed, for which this function has its maximum value, is given by Equation 20.20. vmp=2kBTm=2(1.381023J/K)(295K)4.681026kg=417m/s f(vmp) = (3.11108)(417)2 e(5.75106(417)2) = 0.00199 We plot these points on the speed distribution. The most probable speed is indeed at the peak of the distribution function. Since the function is not symmetric, the rms velocity is somewhat higher than the most probable speed. Figure P20.29ANS
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