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omeli
18 days ago
13

THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A

RE PERFORMED, WHAT IS THE TOTAL ANGULAR DISTANCE AND THE TOTAL ANGULAR DISPLACEMENT UNDERGONE AT THE KNEE?
Physics
1 answer:
ValentinkaMS [2.4K]18 days ago
5 0

Answer: total angular distance = 1700° and 29.7 rad

   the total angular displacement = 0

Explanation:

This is a breakdown of the calculations needed.

The task is to determine both the total angular distance and displacement experienced by the knee.

To calculate the distance traveled by the knee, consider that when squatting, the knee bends 85° to lower, and then another 85° to return to standing (upright). Thus, the cumulative angular movement during the squat totals 170°.

For 10 squats, the knee must undergo 170° motion multiplied by 10, resulting in:

10 * 170° = 1700°

As such, the total angular distance reached is 1700°.

Now converting this to radians since both degree and radian outputs are required:

Since 2π equals 360°, it follows that one degree equates to about 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1700° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

For the second part, remember that angular displacement is determined by the angular distance divided by time, leading to a displacement of zero because the knee's ending position is the same as its starting position.

I hope this is helpful!!!!

π

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Thus,

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22 days ago
-. What is the acceleration of 4 kg trolling bag pulled by a girl with a<br> force of 3 N?
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1 month ago
Two planets having equal masses are in circular orbit around a star. Planet A has a smaller orbital radius than planet B. Which
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Answer:

Explanation:

To approach this problem, we need to understand two key concepts.

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Kinetic energy for each planet is computed as:

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