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timofeeve [1]
3 years ago
8

An athlete in the early weeks of an aerobic training program develops a decreased hemoglobin concentration but does not report d

isproportionate fatigue. Which of the following statements is true of the athlete's decreased hemoglobin levels? a. Vitamin and mineral supplements are necessary to reverse this condition. b. Decreased hemoglobin concentrations are never a cause for concern. c. Vitamin and mineral supplements are necessary to reverse this condition d. The athlete has a low blood volume. e. The athlete has an iron deficiency.
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0

Answer: B

Explanation: Hemoglobin is a protein in red blood cells that acts as the vehicle for oxygen and carbon dioxide from the lungs to tissues and vice-versa.

What is considered normal Hemoglobin levels vary throughout one's life, from childhood to adult life and in older adults. It even varies between women and men. In women, it also varies depending on whether they are pregnant or not.

When hemoglobin does become less than normal, it results in anemia, a condition accompanied by fatigue. Because the athlete does not have this symptom, he is not suffering from anemia, and so there is no cause for concern. Although the athlete's hemoglobin level has reduced, the absence of fatigue indicates that it is not lower than normal.

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A boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of 5.9 m/s. if the coefficient of fricti
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Frictional force = vertical force x coeff of friction = 590N x 0.045 = 26.55N

Mass of boy and sled = 590N / g = 590N / 9.8m/s^2 = 60.20 kg

Deceleration due to friction = 26.55N / 60.20kg = 0.44 m/s^2.

For constant acceleration we have:

v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and ti is time. In this case v = 0, u = 5.9 m/s, a = -0.44 m/s^2. So we have
0 = 5.9 - 0.44t
which gives t = 5.9 / 0.44 = 13.409 s.

Distance traveled in this time d = ut + 0.5at^2 = 5.9 x 13.409 - 0.5 x 0.44 x 13.409^2 = 39.56 m

<span>Answer: </span>40 meters
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The amount of matter in an object
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Water drips from the nozzle of a shower onto the floor 193 cm below. The drops fall at regular (equal) intervals of time, the fi
Law Incorporation [45]

Answer: 108.81 cm and 48.66 cm

Explanation:

In this, we have to make sure to keep in mind the Gravity effects on the drops. The drops will accelerate when they fall making them travel faster. This means, the velocity is not constant.

What is know:

Height (h) = 193 cm

Gravity (g) =  981 cm/s^{2}

Initial Velocity = 0

First, we can know how long it take to the drop to travel to the floor. It can be done with the following equation:

x = V_{0} t + \frac{1}{2} at^{2}    (1)

Where:

x is the distance which is 193cm

Vo is the Initial Velocity  which is zero

t is the time the time it takes the drop to travel from the shower to the floor

a is the aceleration, which in this case is the gravity.

With the Initial Velocity equals zero the equations simply:

193 cm = \frac{1}{2}gt^{2}

To search for the time:

t =\sqrt{\frac{2*193cm}{981cm/s^{2} } }

t = 0.627 s

This is the time it takes a drop to fall to the floor, with this time and knowing other 3 drops have driped from the shower by this time. We can calculate how much time it takes the shower to drip each drop.

Time for Drip = t/4

Time for Drip = 0.156

This time is the difference between each drop, using the same equation we can calculate where was each drop, because now it is know how much time had each drop after being drip from the shower.

Our first is already on the floor (193 cm) with 0.627 s, The second drops have been falling for (0.627s - 0.156) 0.471 s and our third drop for (0.627s - 0.156 - 0.156) 0.315 s

We can use (1) to know how far have each drop traveled on these times. We know the Initials Velocity are 0, know we need ot know the distances.

For the second drop:

x = \frac{1}{2} (981cm/s^{2})(0.471s)^{2}

x =108.81 cm

For the third drop:

x = \frac{1}{2} (981cm/s^{2})(0.315s)^{2}

x = 48.66 cm

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Answer:

Explanation:

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