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ozzi
3 years ago
13

Why are overuse injuries particularly frustrating set-backs?

Physics
2 answers:
olganol [36]3 years ago
7 0
Because they are caused by your exercise 
anastassius [24]3 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.

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What number is between 2000 and 2500
Kitty [74]
If its just a number 2,300 is an answer 
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3 years ago
A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 22.0 m/s and returns the shot with the ball
morpeh [17]
I need an answer choice
8 0
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The amount of the lighted side of the moon you can see is the same during
attashe74 [19]

<u>Answer:</u>

The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".

<u>Explanation:</u>

The Moon is in sequential rotation with Earth, and thus displays the Sun, the close side, always on the same side. Thanks to libration, Earth can display slightly greater than half (nearly 59 per cent) of the entire lunar surface.

The side of the Moon facing Earth is considered the near side, and the far side is called the reverse. The far side is often referred to as the "dark side" inaccurately but it is actually highlighted as often as the near side: once every 29.5 Earth days. During the New Moon the near side becomes blurred.

5 0
3 years ago
A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o
inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) =  \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_i^2) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2  ) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 -  \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 -  \frac{3}{4}v_f^2\\\\

h = \frac{3}{4g}(v_1^2 -v_f^2)

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0
2 years ago
A 15 W electric shaver is used for 3 minutes. How much energy does it use in joules?​
borishaifa [10]
P=change in E/t
Change in E=p*t
=15*3
=45

The answer is 45J.
3 0
2 years ago
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