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Margaret [11]
3 years ago
11

A body accelerates uniformly from rest at 2m/s square. calculate it's velocity after traveling 9m.​

Physics
1 answer:
Talja [164]3 years ago
3 0

Answer:

6m/s

Explanation:

Given parameters:

Initial velocity  = 0m/s

Acceleration  = 2m/s²

Distance  = 9m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we use the expression below:

         v²  = u²  + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

        v²   = 0² + (2 x 2 x 9) = 36

        v  = 6m/s

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a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?

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