A scooter is traveling at a constant speed v when it encounters a circular hill of radius r = 480 m. The driver and scooter together have mass m = 159 kg. (a) What speed in m/s does the scooter have if the driver feels weightlessness (i. e., has an apparent weight of zero) at the top of the hill?
(b) If the driver is traveling at the speed above and encounters a hill with a radius 2r,

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

so easy. just google the electromagnetic waves and it’ll pop up.

well clustered together.

points close on a scatter diagram