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olga55 [171]
4 years ago
8

On a particle level, what happens when thermal conduction occurs within a liquid?

Physics
2 answers:
Kay [80]4 years ago
5 0

Answer:

a) When fast particles rise to the top.

Explanation:

Heat is a form of energy that can be studied through the thermal agitation of the molecules that a material involves. When heat is delivered to a body, its temperature increases, the mobility of its molecules increases. In this way, the system is not in thermal equilibrium: the temperature at each point of the body is different and the variations over time.

There are three forms of heat transmission: conduction, convection and radiation. In conduction, heat is transferred only because of molecular movement and shocks between fast and slow molecules, without displacement of matter. Convection is due to the global movement of matter and is only important in liquids and gases. Radiation is an electromagnetic interaction between bodies and does not require the existence of a material means to transmit heat from one to another. The conduction of heat in a medium can be more or less favorable according to the material.

Elodia [21]4 years ago
4 0

B sounds perfect to me

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A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel
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If the particle is an electron E_y = 3.311 * 10^3 N/C

If the particle is a proton, E_y = 6.08 * 10^6 N/C

Explanation:

Initial speed at the origin, u = 3 * 10^6 m/s

\theta = 38^0 to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

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Mass of an electron, m_e = 9.1 * 10^{-31} kg

Mass of a proton, m_p = 1.67 * 10^{-27} kg

The electric field intensity along the positive y axis E_y, can be given by the formula:

E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\

If the particle is an electron:

E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C

If the particle is a proton:

E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

E_y = 6.08 * 10^6 N/C

8 0
3 years ago
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