Answer:i think they do i dont understand your question it does not make sence
Explanation:
Answer:
(a)20.65g
(b)0.19m
Explanation:
(a) The total mass would be it's mass per length multiplied by the total lenght
0.355(50 + 23*0.355) = 20.65 g
(b) The center of mass would be at point c where the mass on the left and on the right of c is the same
Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g



We have that the values for F north,
F east,
F up are
From the Question we are told that
electric force 
electric force , 
electric force , 
charge on this ball one 
charge on this ball two 
Generally the equation for the F north is mathematically given as


For F East


For F UP


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The momentum would increase assuming the velocity stays the same. P=Mv