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Alexxx [7]
3 years ago
6

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

in 6.73s. what is the magnitiude of the girls average acceleration.
Physics
1 answer:
Mashutka [201]3 years ago
8 0

Answer:

a = 0.53 m/s^2

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

\omega = 2\pi f

\omega = 2\pi ( \frac{20}{60})

\omega = 2.10 rad/s

so final tangential speed is given as

v = r\omega

v = 1.71 (2.10) = 3.58 m/s

now average acceleration of the girl is given as

a = \frac{v_f - v_i}{\Delta t}

a = \frac{3.58 - 0}{6.73}

a = 0.53 m/s^2

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The value of the c will be (0 i+0 j+0 k). c is a vector that is along the positive x-axis and has the condition a(b+c)=0.

<h3>What is a vector?</h3>

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Given ;

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