F 13 (on q 3 due to q 1) = [k *q 3 *q 1 ]/ r 2 =[(9*10 9 )(8*10 -9 )(8*10 -9 )]/

Question

F13 (on q3 due to q1) = [k *q3*q1]/ r2
                               =[(9*109)(8*10-9)(8*10-9)]/(.035)2
   =4.702*10-4 N
F23 (on q3 due to q2) = [k *q3*q2]/ r2
      =[(9*109)(8*10-9)(16*10-9)]/(.035)2
                                  = 9.404*10-4 N
F13x = F13 * cos(60) = +2.351*10-4 N
F13y = F13 * sin(60) = +4.072*10-4 N
F23x = F23 * cos(0) = + 9.404*10-4 N
F23y = 0

Fx = 2.351*10-4 +9.404*10-4 = 1.176*10-3 N
Fy = 4.072*10-4 N

F = 10-3(1.176)2 +(0.4072)2 = 1.2445*10-3 N
F gives the direction of acceleration a a = (change in velocity ) /time a=(v-0)/t a is directed towards F but a, v, and t are unknowns

How do I draw the velocity vector?? Please help.

Explanation / Answer

so a=(v-0)/t=v/t so the direction of acceleration is the direction ofvelocity now direction of acceleration is the direction offorce so direction of velocity is the direction of force

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