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Reika [66]
3 years ago
9

While attempting to score a goal, a player from the opposing team accidentally kicked Vance in the shin, causing an injury. What

is one way that Vance might have been able to prevent this injury?
Physics
2 answers:
hjlf3 years ago
7 0

Answer:

Explanation:

The shin bone is the tibia. It is located in the lower leg bone. It is larger then it's neighboring bone that is fibula. It supports the weight of the human body. It provides stability while standing. It is sensitive to fracture. Therefore, it is advice to wear the leg guard or shin guard to protect oneself from injury.

Therefore, Vance should wear leg guard or shin guard to prevent injury.

sergiy2304 [10]3 years ago
4 0
To prevent this injury, Vance could have protected his shins using a shin guard. Hope this helps! :)
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Travels 11,000 feet along a dark desert highway if the car averages 84 mph find the amount of time to cover this distance
laila [671]

The time taken by traveler to cover the distance is,

t=\frac{d}{v}

Substitute the known values,

\begin{gathered} t=\frac{(11000\text{ ft)}}{(84\text{ mph)(}\frac{1.46667\text{ ft/s}}{1\text{ mph}})_{}} \\ \approx89.3\text{ s} \end{gathered}

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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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