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About Indian National Physics Olympiad - 2017: Solutions
1. The power radiated through gravitational waves by a massive star orbiting a black hole is given by LG = Kcx5y-1M2R4ω6, where x = -5 and y = 1. The total mechanical energy of the star is E = -GMbM/2R. The rate of decrease in the orbital period is dT/dt = -3KG5/3MbM(2π)8/3c5T5/3. 2. For a cylinder partly filled with mercury rotating at 78 rpm, the radius of curvature of the mercury surface at the lowest point is ρx=0 = 15 cm. The image position of a point object at
Detailed Information
| Author: | ['Gwen Angira'] |
|---|---|
| Publication Year: | 2019 |
| Pages: | 8 |
| Language: | English |
| Format: | |
| Price: | FREE |
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