All categories

2 years ago

Why are overuse injuries particularly frustrating set-backs?

Physics

2 answers:

olganol [36]2 years ago

7 0

Because they are caused by your exercise

anastassius [24]2 years ago

4 0

Answer:

Because they are caused by the excess exercise and lack of rest to recover the body.

Explanation:

Over use injuries can be say that the inflammation of the connecting tissues which attaches the bones to the muscles. They can be occur if someone does not modify his or her workout after a break.

These injuries particularly are frustrating set backs because you are not give a much time of rest to your body to recover from these injuries.

You might be interested in

A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig

trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

$V=\frac{MU+mu}{M+m}$

Substituting values:

$V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s$

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

$V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134$

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0

2 years ago

At a distance of 0.208 cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485 n/c . wh

Anna35 [415]

We have the equation for electric field E = kQ/ $d^{2}$

Where k is a constant, Q is the charge of source and d is the distance from center.

In this case E is inversely proportional to $d^{2}$

So, $\frac{E_{1} }{E_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}$

$E_{1}$ = 485 N/C

$d_{1}$ = 0.208 cm

$d_{2}$ = 0.620 cm

$E_{2}$ = ?

$\frac{485 }{E_{2}} = \frac{0.620^{2}}{0.208^{2}}$

$E_{2}$ = $\frac{485*0.208^{2}}{0.628^{2}}$

$E_{2}$ = 53.20 N/C

7 0

2 years ago

Read 2 more answers

Rudiy27

Okay so yeah u have to minus then subtract then decide it it’s a method i was taught to do

5 0

3 years ago

Where are the magnetic<br> fields<br> strongest near a bar magnet?

mixer [17]

<h2>ANSWER:</h2>

<u>The north pole</u>

The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole when compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center.

CarryOnLearning(◍•ᴗ•◍)

6 0

2 years ago

Which list places different units of matter in the correct sequence from largest to smallest

weqwewe [10]

Answer:

What do you mean?

6 0

2 years ago

Read 2 more answers