Question

Question I have to answer.

Consider Problem 21.66 in the book. What would the charge 42 have to be to produce a net force of 6.00 N in the +x direction on 93? 92 = 59.6 nc Submit Answer Incorrect. Tries 6/99 Previous Tries What would the charge 92 have to be to produce a net force of 6.00 N in the -x direction on 93? 92 = -11.6 nC Submit Answer Incorrect. Tries 1/99 Previous Tries Assuming 92 has the value you determined in the first question above (the one that yielded a force of 6.00 N in the +x direction), at what new x location could you place 43 in order to yield zero net force on 93? (In other words, we can now move 43 away from the origin in an attempt to get to zero net force. Where should we put it if that is our goal?) Assume 41 and 42 do not change their locations. X3 = Submit Answer Tries 0/99

Question that is being referenced.

Any help would be appreciated.

Answer 1

о: 3 м у чин, 55 (-0 • 2 ) , (о: Ит). на ч, 2 - 4 xio - с ч, =3x6:4c а о.y-o: а – о. Чт н, я, - -о-2-0-0 = -0. Эт

а): не ква, - кл. : 6 хо и . о. 54 xjp4 27xo° | - хе"] - *, - [ + ] холь 4,7 o-ose X -3 он Saxio - вс Paut (b); я хло" ххк“Га“ |-- – 1 = Aь- -, - o-1-2 к/s° - з хв°] 4, = 54 xio-*

Banyeres 22,95 Karo 22 (04) * = Ob -1 = 3:14. Tube Than ** (hi 0.144 mi alle