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sergiy2304 [10]
3 years ago
8

The bohr Effect states that the lower the ph,?

Physics
1 answer:
Oksana_A [137]3 years ago
5 0

Answer:

Explanation:

The Bohr effect describes how low pH (acidity) lowers the affinity of hemoglobin for oxygen, making hemoglobin more likely to offload oxygen in areas of low pH, which for reasons I'll get into, tissues in need of oxygen tend to have.

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What is the time period of a simple harmonic oscillator with a mass of 0.3kg and a force constant of 5N/m?
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Why does Alpha Centauri appear to be yellow-white?
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Room's become bright at the day time although there is no bright sunlight in the rooms why
Aneli [31]

Answer:

The room become bright in the day by a process call scattering of light or Rayleigh scattering.

Explanation:

It is called Rayleigh scattering or scattering of light because Rayleigh scattering is the scattering if light or electromagnetic radiation by smaller particles which have radius of less than 110 nanometer in a medium and the wavelengths of the electromagnetic. Wave or light remain unchanged. The scattering of light occur in the day time in the room and this bring of brighten up of the room.

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3 years ago
A flat loop of wire consisting of a single turn ofcross-sectional area 7.90 cm2is perpendicular to a magnetic field that increas
Westkost [7]

To develop this problem, it is necessary to apply the concepts related to Faraday's law and Magnetic Flow, which is defined as the change that the magnetic field has in a given area. In other words

\Phi = BA Cos\theta

Where

B= Magnetic Field

A = Area

\theta = Angle between magnetic field lines and normal to the area

The differentiation of this value allows us to obtain in turn the induced emf or electromotive force.

In this case we have that the flat loop of wire is perpendicular to the magnetic field, therefore the angle is 0 degrees, since its magnetic field acts parallel to the area:

\theta = 0 then our expression can be written as

\Phi = BA

From the same value of the electromotive force we have to

\epsilon = -\frac{d\Phi}{dt}

Replacing we have

\epsilon = -A\frac{B}{dt}

Replacing with our values we have that

\epsilon = -(7.9*10^{-4}m)\frac{(3.5-0.5)}{0.1}

\epsilon = -0.0237V

Therefore the magnitude of the induced emf in the loop is 0.0237V

On the other hand we have that the current by Ohm's Law can be defined as

I = \frac{\epsilon}{R}

For the given value of the resistance and the previously found potential we have to

I = \frac{0.237}{1.3}

I= 0.0182A

6 0
3 years ago
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