Physics, asked by aaryangoyal4168, 11 months ago

Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, T_(A)//T_(B), is :

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Answered by Anonymous

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Answered by Theopekaaleader

Explanation:

$Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1$

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