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2 years ago

Reuptake refers to the...

Physics

1 answer:

Arada [10]2 years ago

4 0

Reuptake refers to the REABSORPTION of excess neurotransmitter molecules by a sending neuron (Option b).

Reuptake is the mechanism by which cells reabsorb chemical messengers produced and secreted by them. In nerve terminals, reuptake is used to reabsorb released neurotransmitters.

The reuptake mechanism is exploited in therapeutics for the development of target drugs and treatments.

Serotonin is a neurotransmitter that acts to stabilize different emotions such as mood, feelings of well-being, appetite and happiness.

For example, serotonin reuptake inhibitors which are capable of blocking the reuptake of serotonin to modulate serotonin brain levels have recently been developed.

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Identify the wavelength of this wave.

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Answer:

Explanation:

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Read 2 more answers

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$ --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a - 420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

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3 years ago

The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 60 mm?

jok3333 [9.3K]

V = [4/3]π r^3 => [dV / dr ] = 4π r^2

[dV/dt] = [dV/dr] * [dr/dt]

[dV/dt] = [4π r^2] * [ dr/ dt]

r = 60 mm, [dr / dt] = 4 mm/s

[dV / dt ] = [4π(60mm)^2] * 4mm/s = 180,955.7 mm/s

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