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vovangra [49]
3 years ago
5

A ball rolling down a hill was displaced 21.9 m while uniformly accelerating from rest. If the final velocity was 7.14 m/s, what

was the rate of acceleration?
Physics
1 answer:
belka [17]3 years ago
4 0

Answer:

a = 1.16 m/s²

Explanation:

In order to find the acceleration of the ball we will use 3rd equation of motion.

2as = Vf² - Vi²

where,

a = acceleration = ?

s = displacement = 21.9 m

Vf = Final Velocity = 7.14 m/s

Vi = Initial Velocity = 0 m/s (Since, ball starts from rest)

Therefore, using the values, we get:

2a(21.9 m) = (7.14 m/s)² - (0 m/s)²

a = (50.97 m²/s²)/(43.8 m)

<u>a = 1.16 m/s²</u>

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Define
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A point P1 is located by the vector A = (3.74)î + (1.64)ĵ and a point P2 is located by the vector B = (1.60)î + (3.66)ĵ. The vec
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