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3 years ago

What is the first step of the STOP procedure for assessing acute sports injuries?

Physics

2 answers:

IrinaVladis [17]3 years ago

4 0

Answer:

B. Stop the player from further aggravating the injury.

Explanation:

Troyanec [42]3 years ago

3 0

The first step of the STOP procedure is S, which stands for "stop." This indicates that you should stop what you're doing to place your full attention on the situation at hand. This will allow you to stay focused on what's specifically happened to an athlete.

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A client is receiving an IV solution of sodium chloride 0.9% (Normal Saline) 250 ml with amiodarone (Cordarone) 1 gram at 17 ml/

tester [92]

Answer:

1.1mg/min

Explanation:

We are given that

Volume of solution=250 ml

Mass of amiodarone=1 g

Infusion rate=17 ml/hr

We know that

1 g= 1000 mg

Ratio of 1000g:250 ml= $\frac{1000}{25}=4 mg/ml$

The concentration of solution=4mg/ml

Amiodarone infusing (mg/min)= $\frac{infusion \;rate(ml/hr)\times concentration}{60}$

Because 1 hr= 60 minute

Amiodarone infusing=1.1mg/m

Hence, 1.1 mg/min of amiodarone is infusing.

6 0

3 years ago

Which of the following types of energy originates below Earth's surface?

faltersainse [42]

Geothermal energy. Electromagnetic energy originates at the sun (think UV rays; they are part of the electromagnetic spectrum). Electrical energy technically can be anywhere, as long as there are positive and negative charges. Mechanical energy is like a generator or an engine; there is always something working to get things done/made. I hope this answer and these explanations helped!

6 0

3 years ago

A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep

salantis [7]

The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force $F_f$ which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
$F-F_f=0$
so
$F=F_f$

The frictional force can be rewritten as
$F_f = \mu m g$
where $m=50 kg$ , $g=9.81 m/s^2$ . Re-arranging, we can solve this equation to find $\mu$ , the coefficient of dynamic friction:
$\mu = \frac{F}{mg}= \frac{99 N}{(50 kg)(9.81 m/s^2)} =0.20$

4 0

3 years ago

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

3 years ago

Shelley is in an elevator that is traveling downward and slowing down at a rate of

liraira [26]

Answer:

N = 648.55[N]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

∑F = Forces applied [N]

m = mass = 73.2 [kg]

a = acceleration = 0.950 [m/s²]

Let's assume the direction of the upward forces as positive, just as if the movement of the box is upward the acceleration will be positive.

By performing a summation of forces on the vertical axis we obtain all the required forces and other magnitudes to be determined.

$-m*g + N = -m*a\\$

where:

g = gravity acceleration = 9.81 [m/s²]

N = normal force (or weight) measured by the scale = 83.4 [N]

Now replacing:

$-(73.2*9.81)+N=-73.2*0.950\\-718.092+N=-69.54\\N = -69.54+718.092\\N = 648.55[N]$

The acceleration has a negative sign, this means that the elevator is descending at that very moment.

8 0

3 years ago