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WITCHER [35]
2 years ago
11

an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco

nd. The radius of the hammer's path is 2.0 meters.
Physics
1 answer:
Semenov [28]2 years ago
6 0

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

a_c = \frac{v^2}{r}

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

a_c = \frac{(12\ m/s)^2}{2\ m}

<u>ac = 72 m/s²</u>

<u></u>

Now, the centripetal force applied by the athlete on the hammer will be:

F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)

<u>Fc = 504 N</u>

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As given in the problem we have to calculate the Kinetic energy of the Jemima,

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