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Brilliant_brown [7]
2 years ago
12

What fitness components can impact a soccer players performance?

Physics
1 answer:
NemiM [27]2 years ago
6 0

Answer:

C) Cardiovascular, muscular strength/power, muscular endurance

Explanation:

I did take the quiz

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Agility is the
olga_2 [115]

Answer:

combination of strength and speed

Explanation:

please like and Mark as brainliest

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3 years ago
What is polygamy? Explain please:)
Mariulka [41]

I always thought that 'polygamy' refers to the social system in which a man
may have more than one wife, and I thought there's some other word for
where a woman is married to more than one man. 

But when I went to look it up just now to answer your question, (which by the
way you could also have done very easily), I found a definition that says it's a
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how the genders may be represented or distributed within it.

5 0
3 years ago
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Is the earth clean...? <br> A. hell no! <br> B. YES<br> C. I dont know.
Tamiku [17]

Answer:

A

Explanation:

7 0
3 years ago
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You are designing a new home for a cold climate. You want one room in the house to be warmed by the Sun's energy. Which material
solniwko [45]

Answer:

glass if for a good view, but for absorbing heat concrete and brick is the best

Explanation:

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8 0
3 years ago
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You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t
Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

W=-Fd

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

F=\mu mg

where

\mu is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

W=-\mu mg d (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

-\frac{1}{2}mu^2 = -\mu mg d

And solving for \mu, we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767

7 0
3 years ago
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