Given G = 20a r + 50a + 40a , at (1, /2, /6) the
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Given G = 20ar + 50aθ + 40aφ, at (1, π/2, π/6) the component of G perpendicular to surface θ = π/2 is
(a) 20ar
(b) 50aθ
(c) 40aφ
(d) 20ar + 40aθ
(e) - 40aθ + 20aφ
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The correct answer is e 40a 20a This is because at 1 2 6 the component o...View the full answer
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