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4 years ago

Choose all the answers that apply.

Physics

2 answers:

kozerog [31]4 years ago

8 0

Answer:

Skipping a cool-down can cause them.

They take rest to heal.

A broken bone is one type.

Tendonitis is one type.

Explanation:

Olenka [21]4 years ago

5 0

Answer:

<h3>Skipping a cool-down can cause them.</h3><h3>They take rest to heal.</h3><h3>A broken bone is one type.</h3><h3>Tendonitis is one type.</h3>

Explanation:

<em>Overuse injuries represent any type of muscle or bone injury caused by repetitive trauma as tendinitis or stress fractures.</em> <u>The most common overuse injuries are from muscle and bone overuse,</u> when the athlete don't rest properly, the excess of activity push muscles and bones out of the limit. Also, training errors conduct to these injuries.

Overuse injuries can be caused by skipping cool-down periods, or rest periods, that's what get the athlete in a overuse state. After these injuries occur, to rest is mandatory, it's the only way to heal, because muscles and bone need to stop to recover from the wide time use.

In addition, as we said before, tendinitis and stress fractures are the most common overuse injuries, due to the overuse of tendons and bones, at that point, they stress and become vulnerable.

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A plastic rod and a piece of cloth are both uncharged. A student rubs the plastic rod with the cloth. The plastic rod becomes ne

Molodets [167]

Answer: a sock and a Pepe when the Pepe is rubbed by the sock the energy is enormous

Explanation:

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3 years ago

You drop a 0.375 kg ball from a height of 1.37 m. It hits the ground and bounces up again to a height of 0.67 m. How much energy

Radda [10]

2.57 joule energy lose in the bounce .

<u>Explanation</u>:

when ball is the height of 1.37 m from the ground it has some gravitational potential energy with respect to hits the ground

Formula for gravitational potential energy given by

Potential Energy = mgh

Where ,

m = mass

g = acceleration due to gravity

h = height

Potential energy when ball hits the ground

m= 0.375 kg

h = 1.37 m

g = 9.8 m/s²

$Potential Energy = 0.375\times9.8\times1.37$

Potential Energy = 5.03 joule

Potential energy when ball bounces up again

h= 0.67 m

$Potential Energy = 0.375\times0.67\times9.8$

Potential Energy = 2.46 joule

Energy loss = 5.03 - 2.46 = 2.57 joule

2.57 joule energy lose in the bounce

6 0

3 years ago

5. What is the force of gravity on a

zvonat [6]

Answer:

32.40 N

Explanation:

Acceleration due to gravity on the moon is 1.62 m/s².

W = mg

W = (20 kg) (1.62 m/s²)

W = 32.40 N

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4 years ago

What is the most important part when passing a volleyball?

Bezzdna [24]

Answer: ur form (how u hit it)

Explanation: if it’s not what ur looking for lmk i’ll try to answer differently

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4 years ago

Read 2 more answers

Please solve this asap

brilliants [131]

Let $\vec u$ and $\vec v$ be the vectors, and let $x=\|\vec u\|=\|\vec v\|$ be their common magnitude.

The resultant $\vec u + \vec v$ is $\sqrt 2$ times larger in magnitude than either vector alone, so $\|\vec u+\vec v\| = \sqrt2\,x$ .

Recall the dot product identity

$\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos(\theta)$

where $\theta$ is the angle between the vectors $\vec a$ and $\vec b$ . In the special case of $\vec a=\vec b$ , we get

$\vec a \cdot \vec a = \|\vec a\|^2 \cos(0^\circ) \implies \|\vec a\| = \sqrt{\vec a\cdot\vec a}$

Now, to get the angle between $\vec u$ and $\vec v$ , we have

$\vec u \cdot \vec v = \|\vec u\| \|\vec v\| \cos(\theta) \implies \cos(\theta) = \dfrac{\vec u \cdot \vec v}{x^2}$

To compute the dot product, we take the dot product of the resultant with itself.

$(\vec u+\vec v) \cdot (\vec u + \vec v) = \|\vec u + \vec v\|^2$

Solve for $\vec u\cdot\vec v$ .

$(\vec u\cdot\vec u) + 2(\vec u\cdot\vec v) + (\vec v\cdot\vec v) = \|\vec u + \vec v\|^2$

$\|\vec u\|^2 + 2(\vec u\cdot \vec v) + \|\vec v\|^2 = \|\vec u+\vec v\|^2$

$x^2 + 2(\vec u \cdot \vec v) + x^2 = (\sqrt2\,x)^2$

$2(\vec u\cdot\vec v) + 2x^2 = 2x^2$

$2(\vec u\cdot\vec v) = 0$

$\vec u\cdot\vec v = 0$

Since their dot product is zero, $\vec u$ and $\vec v$ are perpendicular, so $\theta=90^\circ$ .

7 0

2 years ago