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Allisa [31]
3 years ago
8

As altitude increases, atmospheric pressure decreases, which means less oxygen is diffused into the blood from the lungs. for in

stance, at 18,000 feet you would obtain 29% less oxygen than you would at sea level. this can lead to altitude sickness, or hypoxia as it is called. imagine a mountain climber, who lives at sea level, trying to climb mount kilimanjaro for the first time. the climber is very fit so he plans to summit the 19,341 foot peak in four days, though it usually takes first-time climbers five or six days. use your knowledge of equilibrium to predict the climberâs reaction to this speedy climb, and the reaction needed to restore equilibrium.
Physics
2 answers:
Anni [7]3 years ago
8 0

 

I believe this question has the following sub questions:

1. When the climber ascends the mountain before acclimating, ___ to restore homeostasis within the body. <span>
2. As the climber rehydrates and rests at a basecamp, hemoglobin (Hb) is produced, which is the body's response to the stress. As a result, ___ </span>
3. When the climber returns to sea level before reacclimating, ___ to restore homeostasis within the body. 
4. As the climber reacclimates to sea level, hemoglobin is replaced at a lower rate, decreasing hemoglobin concentrations in the blood. As a result, ___

 

The answers on each sub question are:

1) Equilibrium shifts to release O2<span>
<span>2) Equilibrium shifts to formation of HbO2</span>
<span>3) Equilibrium shifts to formation of HbO2</span>
<span>4) Equilibrium shifts to release O2</span></span>

Elina [12.6K]3 years ago
5 0

Question subitems:

1. When the climber ascends the mountain before acclimating, __to restore homeostasis within the body.

2. As the climber rehydrates and rests at a basecamp, hemoglobin (Hb) is produced, which is the body's response to the stress. As a result, __

3. When the climber returns to sea level before reacclimating, __ to restore homeostasis within the body.

4. As the climber reacclimates to sea level, hemoglobin is replaced at a lower rate, decreasing hemoglobin concentrations in the blood. As a result, __

Explanation:

1) The body will try to compensate the concentration of oxygen so the equilibrium will shift to release O2

2) The body will try to restore the ammount of oxygen in blood so it will producec Hb and shift to the production of HbO2

3) The higher oxygen concentration will trend to increase the formation of HbO2

4) To restorre the homeostasis the equilibrium will shift to the release of O2

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A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?
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Answer:

118\; \rm Hz.

Explanation:

The frequency f of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of v= 295\; \rm m\cdot s^{-1}. In other words, the wave would have traveled 295\; \rm m in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that 295\; \rm m? The wavelength of this wave\lambda = 2.50\; \rm m gives the length of one wave cycle. Therefore:

\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118.

That is: there are 118 wave cycles in 295\; \rm m of this wave.

On the other hand, Because that 295\; \rm m of this wave goes through that point in each second, that 118 wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

f = 118\; \rm s^{-1} = 118\; \rm Hz.

The calculations above can be expressed with the formula:

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Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

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The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

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