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How do they derive at the 202.6 degrees on the vector table below (first column, second cell)? This is for F 21 (3.83, 202.6 degrees)
How do they derive at the 202.6 degrees on the vector table below (first column, second cell)? This is for F21 (3.83, 202.6 degrees). How do they get 202.6 degrees from 3.83?
Example 16.3 Suppose three point charges are arranged as shown in the figure. A charge q1 = + 1.2 \"C is located at the origin of an (x, y) coordinate system; a second charge q2 = 0.60 [AC is located at (1.20 m, 0.50 m) and the third charge q3 = + 0.20 uC is located at (1.20 m, 0). What is the force on q2 due to the other two charges? .\\~ 0.5 m \fVectors (mN, () X - components Y - components F21 (3.83, 202.6.) 3.83 cos 202.6 = -3.53 3.83 sin 202.6 = -1.47 F23 (4.32, -90' ) 4.32 cos (-90 ) = 0 4.32 sin (-90 ) = -4.32 F2 -3.53 -5.79 F, = V3.53 2 + 5.792 = 6.78 mN y tan O = Ry / RX tan O = -5.79/-3.53 = 1.64 92 0 = tan-1 (1.64) = 58.6 F 2 ( = 90 - 58.60 = 31 a = inv tan (1.2/0.5 ) 91 + X + 93 (The correct angle is 270 - 31 = 239) 38Step by Step Solution
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