MasteringPhysics: Homework 2a- Google Chrome Secure https//session.masteringphys
ID: 1575259 • Letter: M
Question
Explanation / Answer
using law of conservation of energy
Total energy at the seperation of 1 mm = Total energy at the seperation of 1.0 cm
K1+U1 = K2+U2
0+(k*q1*q2/r1) = (0.5*m*v^2)+(k*q1*q2/r2)
0+((9*10^9*1*10^-6*1*10^-6)/(1*10^-3)) = (0.5*35*10^-3*v^2)+((9*10^9*1*10^-6*1*10^-6)/(1*10^-2))
v = 21.5 m/s
b) when r2 = 1 km = 1000 m
0+(k*q1*q2/r1) = (0.5*m*v^2)+(k*q1*q2/r2)
0+((9*10^9*1*10^-6*1*10^-6)/(1*10^-3)) = (0.5*35*10^-3*v^2)+((9*10^9*1*10^-6*1*10^-6)/(1000))
v = 22.6 m/sec
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