Answer:
THE BOHR SHIFT ON THE OXYGEN-HEMOGLOBIN DISSOCIATION CURVE IS PRODUCED BY CHANGES IN THE CONCENTRATION OF CARBON IV OXIDE.
Explanation:
The oxygen-hemoglobin dissociation curve shows the relationship between the saturated hemoglobin concentration and oxygen. It shows how the blood hold on to and releases oxygen. The Bohr shift can occur as a result of changes in concentration of carbon iv oxide and other factors such as acidity or pH, 2,3-bisphosphoglycerate, exercise, also temperature of the body. These factors contributes to the right or left shift on the curve. Carbon iv oxide prevents the binding of oxygen to the hemoglobin. The is because hemoglobin has the same binding site for both oxygen and carbon iv oxide. Carbon iv oxide increase also leads to a change in the pH of the blood through the formation of bicarbonate ion. Bicarbonate ion formation causes reduced acidity and therefore lead a shift in the dissociation curve for more of the carbon iv oxide to be excreted as hemoglobin's affinity for oxygen reduces. And when the concentration of carbon iv oxide is low in the plasma, acidity increases and this provides more affinity for oxygen by the hemoglobin.
Answer:
A. Is the one that the experimenter manipulates directly
Explanation:
The independent variable is the one that is manipulated during an experiment by the experimenter.
The dependent variable is the one that is effected by the independent variable in an experiment.
Answer:
a) a = 4.57 m/s², b) a = 6.48 m / s²
, c) a = 1.42 m / s²,d) r = 82.3 m
Explanation:
The centripetal acceleration is the acceleration responsible for the change of direction of the acceleration vector and occurs in circular movements, the expression is
a = v² / r
let's apply this precaution to our cases
a) let's calculate
a = 8²/14
a = 4.57 m/s²
b) an automobile at v = 65 km / h (1000 m / 1km) (1 h / 3600 s) =18,055 m/s
let's reduce feet to meters
1 ft = 0.3048 m
r = 165 ft (0.3048 m / 1 ft) = 50.292 m
a = 18,055 2 / 50,292
a = 6.48 m / s²
c) we calculate
a = 1.25²2 / 1.1
a = 1.42 m / s²
d) we look for the radius
a = v² / r
r = v² / a
we reduce
v = 80 km / h (1000 m / 1km) (1h / 3600s) = 22.22 ms
r = 22.22²/6
r = 82.3 m
e) the cenripeta acceleration is used to take the curves on the highway,
Used in centrifuges to separate compounds
It is used in the games of the park of atraccio
Used in CD players and computer hard drives
