We know that,
KE = (1/2)mv^2
F = mv^2 /r
Solving both, F = 2KE/r
F = 2*50/3.5 = 28.57 N
The line breaks when the tension is 28.57 N
F' = 2*52/3.5 = 29.71 N
Now, Resultant force F'
F' = F + Fa
Where, Fa is attraction force. So it is negative
Fa = 29.71 - 28.57 = 1.14 N
By using Columbs law,
F = Kq1*q2/d^2
- 1.14 = 9*10^9*(+q)*(-q)/(3.5)^2
q^2 = 1.14*(3.5)^2 /(9*10^9)
q = 4*10^-5 C
An electrically neutral model airplane is flying in a horizontal cirele on a 3.5-m guideline, which...
An electrically neutral model airplane is flying in a horizontal circle on a 3.5-m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is 50 J. Reconsider the same situation, except that now there is a point charge of +q on the plane and a point charge of -q at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is 53.5...
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Chapter 18, Problem 22 GO An electrically neutral model airplane is flying in a horizontal circle on a 3.0-m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is 50 J. Reconsider the same situation, except that now there is a point charge of +q on the plane and a point charge of -q at the other end of the guideline In this case, the line breaks when the kinetic energy...
please show steps and formulas when working answer An electrically neutral model airplane is flying in a horizontal circle on a 3.0 m guideline, which is nearly parallel to the groun. The line breaks when the kinetic energy of the plane is 50.0 J. Reconsider the same situation, except that now there is a point charge of +q on the plane and a point charge of - q at the other end of the guideline. In this case, the line...
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