Given that Q = -3.35*10^-6 C, d = 0.0477 m, m = 9.6*10^-3 kg, q'
= -8.29*10^-6 C, v = 57.3 m/s
find the distance the particle travel before its speed is
zero.
initial energy = kqq'/d + mv^2/2
final energy = kqq'/x
kqq'/d + mv^2/2 = kqq'/x
1/d + mv^2/(2kqq') = 1/x
so x = 1/[1/d + mv^2/(2kqq')]
x = 1/[1/0.0477 +
(9.6*10^-3*(57.3)^2)/(2*9*10^9*3.35*10^-6*8.29*10^-6)] = 0.01056
m
the required distance = d - x
= 0.0477-0.01056
= 0.03714 m
7 m, a particle of mass 9.60 x 103 kg and charge -8.29 uC is fred...
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Two particles are fixed to an x axis: particle 1 of charge
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could you please solve it with Explaining & circle on the
final answer
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