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Solve for v...?
p = 1 / [ √(1 - (v^2/c^2)) ]
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3 Answers
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- :-)Lv 41 decade agoFavorite Answer
p = 1 / [ √(1 - (v^2/c^2)) ]
p [ √(1 - (v^2/c^2)) ] = 1
squaring on both sides to get rid of the sq.root
p^2 (1- (v^2/c^2)) = 1
1 - (v^2/c^2) = 1 / p^2
v^2/c^2 = 1 - (1/p^2)
v^2 = c^2 (1 - (1/p^2))
Apply sq.root on both sides to get v
v = √ (c^2 (1 - (1/p^2)))
v = c √ (1 - (1/p^2))
- lenpol7Lv 71 decade ago
p = 1 / [ â(1 - (v^2/c^2)) ]
1/p = [ â(1 - (v^2/c^2)) ]
(1/p)^2 = 1- (v^2/c^2)
(1/p)^2 - 1 = -v^2/c^2
c^2[(1/p)^2 - 1] = -v^2
v^2 = [c^2 - c^2/p^2]
v = +/- sqrt[c^2 - c^2/p^2]
- 1 decade ago
p = 1 / [ â(1 - (v²/c²)) ]
Square this equation,
p² = 1 / (1-v²/c²)
Cross multiply
1-v²/c² = (1/p²)
v²/c² = 1 - (1/p²)
v² = c² [1 - (1/p²)]
take square root
v = c â[1 - (1/p²)]
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