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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
a. 5.3 m/s/s.
b. 9.3 m/s/s.
c. 4.0 m/s/s.
d. 26 m/s/s.
e. 6.6 m/s/s.

Respuesta :

The net acceleration of the car at the top of the hill is most nearly is option A. . 5.3 m/s²

Acceleration is the rate of trade of the velocity of an object with appreciation to time. Accelerations are vector portions. The orientation of an object's acceleration is given by using the orientation of the internet pressure performing on that item.

Calculation:-

V² = U² - 2aS

a = U²- V² / 2S

  = 25² - 10² / 2 × 50

 = 625 -100/100

 = 525/100

 = 5.25 m/s² ≈  5.3 m/s²

Acceleration is the price at which speed modifications with time, in terms of every velocity and course. A thing or an item shifting in a instantly line is increased if it quickens or slows down. motion on a circle is extended despite the truth that the price is constant due to the fact the direction is usually changing.

Learn more about acceleration here:- https://brainly.com/question/29110429

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