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1 year ago

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A low partial pressure of oxygen promotes hemoglobin binding to carbon dioxide. This is an example of the?

1 answer:

soldier1979 [14.2K]1 year ago

4 0

A low partial pressure of oxygen promotes hemoglobin binding to carbon dioxide. This is an example of the Haldane effect

The less saturated hemoglobin is and the lower the partial pressure of oxygen in the blood is, the more readily hemoglobin binds to carbon dioxide. This is an example of the Haldane effect.

The Haldane effect is a property of hemoglobin first described by John Scott Haldane, within which oxygenation of blood in the lungs displaces carbon dioxide from hemoglobin, increasing the removal of carbon dioxide. Consequently, oxygenated blood has a reduced affinity for carbon dioxide.

The Haldane effect is a property of hemoglobin first described by John Scott Haldane, within which oxygenation of blood in the lungs displaces carbon dioxide from hemoglobin, increasing the removal of carbon dioxide. Consequently, oxygenated blood has a reduced affinity for carbon dioxide.

To learn more about Haldane effect here

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if you have a mass of 55 kg and you are standing 3 meters away from your car, which has a mass of 1234 kg, how strong is the for

bagirrra123 [75]

Gravitational force between two masses is given by formula

$F = \frac{Gm_1m_2}{r^2}$

here we know that

$m_1 = 55 kg$

$m_2 = 1234 kg$

$r = 3 m$

$G = 6.67 \times 10^{-11} Nm^2/kg^2$

now from the above equation we will have

$F = \frac{(6.67 \times 10^{-11})(55)(1234)}{3^2}$

$F = 5.03 \times 10^{-7}N$

so above is the gravitational force between car and the person

5 0

3 years ago

12. Which of the following statements about

aliya0001 [1]

A seems to be correct

8 0

3 years ago

two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

3 0

3 years ago

Read 2 more answers

An electron at Earth's surface experiences a gravitational force of magnitude F=(9.11×10−31 kg)⋅(9.8 m/s2). Part A How far away

katrin [286]

Answer:

r = 5,085 m

Explanation:

The force exerted by on the surface of the Earth on an electron is its weight

W = F = 9.11 10⁻³¹ 9.8

W = 8.9 10⁻³⁰ N

The electric force between an electron and a proton is given by Coulomb's Law

Fe = k q₁ q₂ / r²

Fe = - k q² / r²

They ask us that W = Fe

W = k q² / r²

r = √ k q² / W

Let's calculate

r = √ 8.99 10⁹ (1.6 10⁻¹⁹)² /8.9 10⁻³⁰

r = √ 25.86

r = 5,085 m

Let's look for the relationship of this distance with the harmonic distance

R / R_atomic = 5,085 / 10⁻¹⁰

R / R_Atomic = 5 10¹⁰

We see that this distance is 10¹⁰ times the interatomic distance, so the gravitational attraction force is very small at atomic scale

8 0

3 years ago

The structures shown are primarily involved in

VLD [36.1K]

Removing waste.
_____

6 0

3 years ago

Read 2 more answers