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A 1600kg car is traveling over a hill that has a radius of curvature of 25m. The car is slowing down as it goes over the hill. It slows down at a constant rate from a speed of 25ms to a speed of 10ms over a distance of 50m ending at the top of the hill. The net acceleration of the car at the top of the hill is most nearly.

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The  correct answer is net acceleration of the car at the top of the hill is 6.6 m/s²

distance traveled by the car, d = 50 m mass of the car, m = 1600 kg radius of the hill's curvature, r = 25 m beginning speed of the car, u = 25 m/s

The third kinematic equation is applied to get the car's tangential acceleration, as shown below;

v2 = u2 + 2as, where a represents the vehicle's acceleration and 2as = v2 - u2 equals a=5.25 m/s

A net equals a2 + aca net = (5.25) 2 + 42a net = 6.6 m/s. 2

Consequently, the vehicle's net acceleration at the summit of the hill is 6.6 m/s2.

to learn about acceleration click the link below:

brainly.com/question/12550364

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