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nataly862011 [7]
3 years ago
14

What grade of sprain is a completely torn ligament?

Physics
1 answer:
Alex Ar [27]3 years ago
6 0
Grade 1: Stretching or slight tearing of the ligament with mild tenderness, swelling and stiffness. The ankle feels stable and it is usually possible to walk with minimal pain.

Grade 2: A more severe sprain, but incomplete tear with moderate pain, swelling and bruising. Although it feels somewhat stable, the damaged areas are tender to the touch and walking is painful.

Grade 3: This is a complete tear of the affected ligament(s) with severe swelling and bruising. The ankle is unstable and walking is likely not possible because the ankle gives out and there is intense pain.

source - https://www.rushcopley.com/health/physician-articles/varying-degrees-of-ankle-sprains/
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Answer:

The vertical distance is  d = \frac{2}{k} *[mg + f]

Explanation:

From the question we are told that

   The mass of the cylinder is  m

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Generally from the work energy theorem

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Here E the the energy of the spring which is increasing and this is mathematically represented as

       E =  \frac{1}{2} * k  *  d^2

Here k is the spring constant

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     P  = mgd

And

     W_f  is the workdone by friction which is mathematically represented as

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So

    \frac{1}{2} * k  *  d^2 =  mgd +  f *  d

=>    \frac{1}{2} * k  *  d^2 =  d[mg +  f    ]

=>  \frac{1}{2} * k  *  d =  [mg +  f    ]

=> d = \frac{2}{k} *[mg + f]

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.200 rev/s . The magnitude
blagie [28]

Answer:

(A). the angular velocity of fan is 0.380 rev/s.

(B). The number of revolution is 0.059 rad.

(C). The tangential speed of the blade is 0.907 m/s.

(D). The tangential acceleration of the blade is 2.10 m/s²

Explanation:

Given that,

Initial angular velocity = 0.200 rev/s

Angular acceleration = 0.883 rev/s²

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Time = 0.204 s

(A). We need to calculate the angular velocity of fan

Using equation of angular motion

\omega_{f}=\omega_{i}+\alpha t

Put the value into the formula

\omega_{f}=0.200+0.883\times0.204

\omega_{f}=0.380\ rev/s

(B). We need to calculate the number of revolution

Using formula of angular displacement

\theta=\omega_{i}t+\dfrac{1}{2}\alpha t^2

Put the value into the formula

\theta=0.200\times0.204+\dfrac{1}{2}\times0.883\times(0.204)^2

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(C). We need to calculate the tangential speed of the blade

Using formula of speed

v=\dfrac{d}{2}\omega_{f}

Put the value into the formula

v=\dfrac{0.760}{2}\times0.380\times2\pi

v=0.907\ m/s

(D). We need to calculate the tangential acceleration of the blade

Using formula of tangential acceleration

a_{t}=\alpha r

Put the value into the formula

a_{t}=0.883\times0.38\times2\pi

a_{t}=2.10\ m/s^2

Hence, (A). the angular velocity of fan is 0.380 rev/s.

(B). The number of revolution is 0.059 rad.

(C). The tangential speed of the blade is 0.907 m/s.

(D). The tangential acceleration of the blade is 2.10 m/s²

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