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Johannes Kepler revealed three empirical laws by using the data on planetary motion, pertaining to the orbit of planets. These laws are contained here because of the huge importance they have regarding the motion of planets.
1. First law (Law of orbit): Every planet moves in an elliptical orbit, with the light at single focus of the ellipse. This is known as law of orbits. For simplification, the orbit can be supposed to be nearly circular because circle is special case of ellipse. Every planet has a definite orbit and dissimilar planets have different orbits.
2. Second law (Law of area): A line joining any planet to the sun (i.e. radius vector of the planet from sun) sweeps equivalent areas in equal interval of time i.e. the linear speed of the planet changes such that the areal velocity of planet is constant. The linear speed is maximum when the planet is nearest to sun, whereas it is minimum when the planet is farthest from the sun.
3. Third law (Law of period): The Square of the time periods of the planet is proportional to the cube of semi-major axis of ellipse. If T is time period of revolution of a planet around the sun and a is the semi-major axis of the orbit of ellipse, then
types of equilibriam
Kepler's 1-2-3 law Another formulation of Kepler's third law, that relates to the mass m of the primary to a secondary's angular velocity omega & semimajor axis a: m o= ome
A coil of resistance 1.5k Ω and 0.25H inductance is connected in parallel with a variable capacitance across a 10V, 8kHz supply. Calculate (a) the capacitance of the capacitor
Sodium iodide crystal (NaI) It scintillate with good efficiency when gamma ray is absorbed, these crystals where attached directly to the PTM to detect a weak flashes of light
When unsaturated compounds undergo addition reactions with free radicals, it is known as free radical addition reaction.
Volume resistance: The resistance provided to the current which flows through the material is called volume resistance. For a cube of unit dimensions this is called volume resisti
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A stone drops from the edge of the roof. It passes the window 2 m high in 0.1s . How far is the roof above the top of the window? Solution) Height of the window (h) = 2m Acce
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