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OLEGan [10]
3 years ago
9

The ball and socket joint in the hip is similar to the ball and socket joint of the shoulder but is designed more for __________

than ___________.
Physics
2 answers:
Viktor [21]3 years ago
8 0

Answer:

a. power

b. precision

Explanation:

The hip is a spherical joint that allows the upper leg to move from front to back and from side to side. The largest joint that supports weight in the body, the hip joint is surrounded by strong ligaments and muscles.

It differs mainly with the circular articulation of the shoulder since it allows it to perform movement of greater power and with greater precision

kondaur [170]3 years ago
5 0

Answer:

The ball and socket joint in the hip is similar to the ball and socket joint in the shoulder, but is designed <em>more</em> for bio-mechanical strength, stability and ability to carry weight than a wide range of movement.

Explanation:

The kind of weights that the hips carry vary in nature.

Some weights are better described as stationary and others as ,constantly changing.  

Besides carrying it's own weight (that is the upper torso), the hip is also active and instrumental in the ability of mankind to lift and carry additional weights.

The study of how the hip manages these weights effortlessly is important in the design of prosthetic appendages and replacement hips for accident victims and geriatric patients who have one hip issue or the other.

Cheers!

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dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

Therefore the precession period is 5.4s

7 0
3 years ago
Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 81 mN when sepa
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Answer:

Explanation:

Check the attachment for solution

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3 years ago
The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be
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By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

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But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

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Non examples of gravity ?
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