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Рефераты - Физика - I must do my duty
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I must do my duty

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(continue)

I.G. Goriachko, St. Petersburg, in February, 23.

Forms of neutral and changed body’s trajectories in Nature are very differently and depended on a pressure (p) and temperature (T) of surroundings. With reference to microcosm, the neutral bodies (photons, roentgen-rays, gamma-rays, free neutrons, and etc.) for lack of influence to them coulomb’s forces always moving merely by the parabolical trajectories. The changed orbital bodies (electrons, protons, I must do my duty-particles, and etc., being on the steady atom’s and nucleus’s orbits) always moving merely by elliptical orbits. These changed particles by their crossing from the one to the next steady elliptical orbit (and a free changed ones) always moving merely by hyperbolical trajectories. As a result of these crossings eccentricities of trajectories suddenly changed from the rangeI must do my dutyto the I must do my dutyI must do my duty (and back). This signified that eccentricity (e) of the orbit is appeared as the quantum parameter depended on (p, T). Into the atom and into its nucleus acted only exactly balanced coulomb’s and gravity forces of differently (I must do my duty) orientations. Therefore, so-called «nuclei ‘s forces» are not existing. The circle or the straight-line motions in Nature are impossible in order to a really curvature in space and time. But as natural indicator of this curvature may be namely the form of body’s trajectory.

3.The real body’s trajectories in Nature

The explanation 2. It was shown in the exp. 1 for the orbital body: W < 0, I must do my dutyI must do my dutyfor the non-orbital body: W>0, I must do my duty

Write down the conical equation (for the circle, ellipse, parabola, hyperbola) in the non-dimensional forms

- for the ellipse

I must do my duty , where 0 <I must do my duty, (2)

- for the hyperbola I must do my duty , where 1I must do my duty. (3)

By means of (2), (3),(A) we obtain

I must do my duty I must do my duty, where I must do my duty, (a)

I must do my dutyI must do my duty I must do my duty, where I must do my duty (b)

For the straight-line: I must do my duty Therefore,I must do my duty (c). For the circle: I must do my duty . Therefore,I must do my duty(d)

For the parabola:I must do my duty Therefore,I must do my duty(e). But for the elliptical and hyperbolical trajectories parametersI must do my dutyare periodical ones.


4.Some useful formulas

Cephler’s the Third law has the next modern form (where f – the gravity constant, M – mass of the Sun, I must do my duty - the orbit ‘s semi-axis, I must do my duty – period) I must do my duty But as w=I must do my dutyI must do my duty (where m – the planet’s mass) we obtain from (1) I must do my dutyI must do my duty From these equations we obtain the correlation (A), where I must do my duty. Also from the (1) we obtain: a)I must do my duty For the circle (I must do my duty) motion I must do my duty - the first cosmic velocity. For the parabolic (I must do my duty) motion I must do my duty - the second (or so-called - the «parabolic») cosmic velocity, and etc. b) I must do my duty (where M – the planet’s mass, r–the planet’s radius, I must do my dutythe basis vector). Therefore, the body’s weight (I must do my duty) on the planet depended on the form its orbit. For the circle (I must do my duty) motion, we obtain: I must do my duty - the acceleration of gravity. From the (B’) for the photon (I must do my dutyI must do my duty) , we obtain : I must do my duty- the full photon’s energy. And etc. You will be certained soon about the parameter I must do my dutyI must do my duty fundamental significance in thermomechanics of macro- and microcosm.


Thank you. To the next report in March, 08. G.I.G.

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