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(B) Since the particles are moving in circular orbits,the centripetal force is provided by the given force $F(r)$.$\frac{mv^2}{r} = |F(r)| = \frac{16}

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$6 	imes 10^{-2}$

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(B) Since the particles are moving in circular orbits,the centripetal force is provided by the given force $F(r)$.$\frac{mv^2}{r} = |F(r)| = \frac{16}{r} + r^3$Multiplying both sides by $\frac{r}{2}$,we get the kinetic energy $K = \frac{1}{2}mv^2 = \frac{1}{2}(16 + r^4)$.For the first particle at $r = 1$:$K_1 = \frac{1}{2}(16 + 1^4) = \frac{17}{2} = 8.5$.For the second particle at $r = 4$:$K_2 = \frac{1}{2}(16 + 4^4) = \frac{1}{2}(16 + 256) = \frac{272}{2} = 136$.The ratio of kinetic energies is:$\frac{K_1}{K_2} = \frac{17/2}{272/2} = \frac{17}{272} = \frac{1}{16} = 0.0625$.Thus,the ratio is approximately $6 	imes 10^{-2}$.

Two particles of the same mass are moving in circular orbits because of a force given by $F(r) = -\frac{16}{r} - r^3$. The first particle is at a distance $r = 1$,and the second is at $r = 4$. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to:

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