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djyliett [7]
3 years ago
15

An insulated container with a divider in the middle contains two separated gases. Gas 1 is initially at a higher temperature tha

n gas 2. The divider is then removed. Which of the following observations might be made over a period of time as the two gases mix together, and why
(A) Gas 1 remains at a higher temperature than gas 2 because gas 1 started at a higher temperature.
(B) Gas 1 remains at a higher temperature than gas 2 because gas 1 started with a higher kinetic energy.
(C) On average, the molecules of gas 1 lose all of their kinetic energy to the molecules of gas 2 through collision, resulting in gas 2 eventually having a higher temperature than gas 1.
(D) On average, the molecules of gas 1 lose some of their kinetic energy to the molecules of gas 2 through collisions, resulting in the two gases eventually having the same temperature.
Physics
2 answers:
12345 [234]3 years ago
4 0

Answer:

Option (D) On average, the molecules of gas 1 lose some of their kinetic energy to the molecules of gas 2 through collisions, resulting in the two gases eventually having the same temperature.

Explanation:

From the question given, Gas 1 was initially at a higher temperature than Gas 2.

As the two gas mixes together, there will be a transfer of heat from Gas 1 molecules to Gas 2 molecules. Now, as this continues over a period of time, the two gas will eventually have the same temperature.

marissa [1.9K]3 years ago
3 0

Answer:

The correct answer is <u><em>(D) On average, the molecules of gas 1 lose some of their kinetic energy to the molecules of gas 2 through collisions, resulting in the two gases eventually having the same temperature.</em></u>

Explanation:

Molecular kinetic theory is a theory that postulates that matter is made up of a set of particles known as atoms or molecules, which are constantly in motion.  These particles collide continuously with each other and with the walls of the container that contains it. These collisions are carried out kinetically, in other words, energy is transferred without loss.

The kinetic energy of the particles increases with increasing temperature. In this case, gas 1 is initially at a higher temperature than gas 2.  So gas 1 has higher kinetic energy than gas 2.  Then, to reach an equilibrium when the mixture of both gases occurs, gas 1 must lose part of its kinetic energy, transferring it to gas 2 through particle collisions to reach the same temperature.

<u><em>The correct answer is (D) On average, the molecules of gas 1 lose some of their kinetic energy to the molecules of gas 2 through collisions, resulting in the two gases eventually having the same temperature.</em></u>

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A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m
Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

  • OA = range of initial position of the airplane from the point of observation = 375 m
  • OB = range of the final position of the airplane from the point of observation = 797 m
  • \theta = angle of the initial position vector from the observation point = 43^\circ
  • \alpha = angle of the final position vector from the observation point = 123^\circ
  • \vec{AB} = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} =  (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} =  (-708.34\ \hat{i}+412.67\ \hat{j})\ m

The magnitude of the displacement vector = \sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

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