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Ksenya-84 [330]
3 years ago
8

learning task 2: assess your health status by using the question below try to determine if you are capable to do the physical fi

tness test .write your answer in your answer sheet​
Physics
1 answer:
Arlecino [84]3 years ago
7 0

To assess your health status, it is necessary to take the physical fitness test, which will determine how healthy you are, according to some questions, such as how often you have seen a doctor, how is your heart, etc.

Some sample questions to assess your health status are:

  • Have you seen a pshysician?

  • Whats your heart condition according to your physician?

  • Are you recommended to do some physical activity?

  • Do you feel any chest pain if you are not doing physical activity?

  • This past few days, do you feel any chest pain if you are not doing phsysical activity?

  • Do you ever lose consciusness or lose your balance because of dizziness?

  • Do you have any joint or bone injury or problem that could be made worse by performing physical activity?

Therefore, the physical fitness test is necessary so that a person can perform physical activities safely, avoiding injuries and other associated risks.

Learn more here:

brainly.com/question/24757304

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She could look for an online job or a job she can work at from home.

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A constant horizontal F force began to act on the initially immovable body placed on a horizontal surface. After t time the forc
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Answer:

The coefficient of friction is (F/(19.6·m)

Explanation:

The given parameters are;

The force applied to the immovable body = F

The time duration the force acts = t

The time the body spends in motion = 3·t

The acceleration due to gravity, g = 9.8 m/s²

From Newton's second law of motion, we have;

The impulse of the force = F × t = m × Δv₁

Where;

Δv₁ = v₁ - 0 = v₁

The impulse applied by the force of friction, F_f is F_f × (3·t - t) =  F_f × (2·t)

Given that the motion of the object is stopped by the frictional force, we have;

The impulse due to the frictional force = Momentum change = m × Δv₂ = F_f × (2·t)

Where;

Δv₂ = v₂ - 0 = v₂

Given that the velocity, v₂, at the start of the deceleration = The velocity at the point the force ceased to  act, v₁, we have;

m × Δv₂ = F_f × (2·t) = m × Δv₁ = F × t

∴ F_f × (2·t) = F × t

F_f = F × t/(2·t) = F/2

The coefficient of dynamic friction, \mu _k = Frictional force/(The weight of the body) = (F/2)/(9.8 × m)

\mu _k = (F/(19.6·m)

The coefficient of friction, \mu _k = (F/(19.6·m)

5 0
3 years ago
If an airplane were traveling westward with a thrust force of 450 N and there was a headwind (drag) of 200 N, what would the res
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Answer:

The resulting net force on the airplane would be 250N.

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10 newtons

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A uniform thin circular rubber band of mass M and spring constant k has an original radius R?
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