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3 years ago

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A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3

.43 m/s and a centripetal acceleration of magnitude 2.32 m/s2. Position vector locates him relative to the rotation axis. (a) What is the magnitude of___________.

1 answer:

Advocard [28]3 years ago

5 0

Answer:

r = 5.07 m

Explanation:

given,

velocity of the man , v = 3.43 m/s

centripetal acceleration, a = 2.32 m/s²

magnitude of position of = ?

using centripetal acceleration formula

$a_c =\dfrac{v^2}{r}$

$2.32 =\dfrac{3.43^2}{r}$

$r =\dfrac{3.43^2}{2.32}$

r = 5.07 m

The magnitude of the position vector relative to rotational axis is equal to 5.07 m.

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