A person walks 20.5° north of east for 4.90 km. If a person would walk first due north and then due east to arrive at the same location, what total distance would he have walked?

D) the event horizon, a large planet and the chute located between the two.

A) the moment of inertia is given by i = mr^2 here m = 0.25 and r = 0.5. i = 0.25 * (0.5)^2 = 0.0625. b) i = 9.74 * 10^(37) and mass = 5.98 * 10^(24) so r^2 = i/m = 9.74 * 10^(37)/5.98 * 10^(24) = 1.628 * 10^(37 - 24) = 1.628 * 10^(13) r = √ 1.628 * 10^(13) = 4034848.200 c) since angular momentum is the product of moment of inertia and angular velocity, a = i * v; but i = mr^2 = 2 * 0.6^2 = 0.72 so a = 0.72 * 1.1 = 0.792. if she pulls her hand in 0.15 m. the new radius = 0.6 - 0.15 = 0.45. v = angular momentum/ radius = 0.792/ 0.45 = 1.76