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3 years ago

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an athlete whirls an 8.71 kg hammer tied to the end of a 1.5 m chain in a simple horizontal circle where you should ignore any v

ertical deviations the hammer moves at the rate of 1.16 rev/s what is the tension in the chain

1 answer:

-BARSIC- [3]3 years ago

5 0

Answer:

T = 692.42 N

Explanation:

Given that,

Mass of hammer, m = 8.71 kg

Length of the chain to which an athlete whirls the hammer, r = 1.5 m

The angular sped of the hammer, $\omega=1.16\ rev/s=7.28\ rad/s$

We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :

$F=m\omega^2r\\\\=8.71\times (7.28)^2\times 1.5\\\\=692.42\ N$

So, the tension in the chain is 692.42 N.

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Answer:

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Given:

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coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

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towards positive Y axis.

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$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

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$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

<em>For angle in counterclockwise direction from the +x-axis</em>

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

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