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KIM [24]
3 years ago
12

An Object, Start from rest w Confront Aiceleration 8m/s2 along a

Physics
1 answer:
shutvik [7]3 years ago
6 0

Answer:

A)   v = 40 m / s, B)   v_average = 20 m / s

Explanation:

For this exercise we will use the kinematics relations

         

A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero

         v = vo + a t

         v = 0 + 8 5

         v = 40 m / s

B) the average velocity can be found with the relation

         v_average = vf + vo / 2

         v-average = 0+ 40/2

          v_average = 20 m / s

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2 years ago
A golf ball is dropped from rest from a height of 9.5m.  It hits the pavement then bounces back up rising  just 5.7 m before fal
Oksanka [162]

Answer: 3.4s

Explanation:

There are three stages in the motion of the ball, so you have to calculate the times for every stage.

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d = Vo + gt² / 2

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⇒ t = √ (1.94 s²) = 1.39s

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i) Determine the initial speed:

Vf² = Vo² - 2gd

Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²

⇒ Vo = 10.6 m/s

ii) time rising

Vf = Vo - gt

Vf = 0 ⇒ Vo = gt ⇒

t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s

3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)

i) d = 5.7m - 1.20m = 4.5m

ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²

⇒ t = √ (0.92 s²) = 0.96s

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t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s

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3 years ago
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Answer:

Situation one:

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