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Alborosie
2 years ago
13

Binding of a signaling molecule to which type of receptor leads directly to a change in the distribution of substances on opposi

te sides of the membrane?.
Physics
1 answer:
Mrac [35]2 years ago
3 0

When two sides of a membrane are in contact with each other, the distribution of ions will alter as a result of the binding of a signal molecule to a ligand-gated ion channel.

<h3>What is a ligand-gated ion channel?</h3>

Ligand-gated ion channels (LGICs) are membrane proteins that are structurally integral and feature a pore that permits the controlled passage of particular ions across the plasma membrane. The electrochemical gradient for the permeant ions drives the passive ion flux.

When a chemical ligand, such as a neurotransmitter, attaches to the protein, ligand-gated ion channels open. Changes in membrane potential cause voltage channels to open and close. When a receptor physically deforms, as in the case of pressure and touch receptors, mechanically-gated channels open.

Learn more about ligand-gated ion channel here:

brainly.com/question/15215628

#SPJ4

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In a tug of war, when one team is pulling with a force of 85 N and the other 40 N, what is the net
olya-2409 [2.1K]
85 N - 40 N = 45 N
And depending on direction the greater force is being pulled towards
4 0
3 years ago
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

7 0
3 years ago
Ex 10: My dog runs at 6 m/s for 18 meters. How long did she run for?
Hunter-Best [27]
She ran for 3s

Put 18/6 because in order to find how long she ran for you need to divide the distance by the meters ran, once you do that you will get 3.
7 0
3 years ago
An airplanes lands with a velocity of +75 m/s and comes to a stop in 20 seconds. What is the acceleration of the airplane while
erma4kov [3.2K]

Answer:

The acceleration of the car, a = -3.75 m/s²

Explanation:

Given data,  

The initial velocity of the airplane, u = 75 m/s

The final velocity of the plane, v = 0 m/s

The time period of motion, t = 20 s

Using the I equations of motion

                    v = u + at

                     a = (v - u) / t

                        = (0 - 75) / 20

                        = -3.75 m/s²

The negative sign indicates that the plane is decelerating

Hence, the acceleration of the car, a = -3.75 m/s²

7 0
3 years ago
1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant
Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction (v_j) = 550 mph

Velocity of jet stream from west to east direction (v_s)=80\ mph

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph

Similarly, the velocity of the stream is, \vec{v_s}=80\vec{i}

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle \theta with the x axis in the third quadrant.

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

5 0
3 years ago
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