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iragen [17]
3 years ago
9

A man in a gym is holding an 8.0-kg weight at arm's length, a distance of 0.55 m from his shoulder joint. What is the torque abo

ut his shoulder joint due to the weight if his arm is held at 30° below the horizontal?
Physics
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

  \tau =37.34\ N m

Explanation:

given,

mass of the weight = 8 Kg

distance = 0.55 m

angle below horizontal = 30°

torque about shoulder

  \tau = \vec{r} \times \vec{F}

  \tau = r \times F cos \theta

  \tau = 0.55 \times 8 \times 9.8 \times cos 30^0

  \tau =37.34\ N m

torque about his shoulder join is equal to   \tau =37.34\ N m

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goblinko [34]

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Explanation:

Because I got it right on my quiz :D

also because you can use the impulse momentum formula, Ft=m(triangle)v

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3 years ago
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Answer:

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Now we have to convert from light-years to seconds in order to get the distance in meters.

t = 6.45 [light-years]*365[\frac{days}{1light-year}]*24[\frac{hr}{1day}] *60[\frac{min}{1hr}]*60[\frac{seg}{1min} ] =203407200 [s]

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