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vlada-n [284]
2 years ago
9

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.

Learn more in:

brainly.com/question/4439815?referrer=searchResults

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Select the statements that are TRUE. Electrophilic aromatic substitution reactions have energies of activation that are very low
NikAS [45]

Answer:

Addition reactions with benzenes lead to the loss of aromaticity.

Benzene and its derivatives undergo a type of substitution reaction in which a hydrogen atom is replaced by a substituent, but the stable aromatic benzene ring is regenerated at the end of the mechanism.

Benzene and its derivatives tend to undergo electrophilic aromatic substitution reactions.

Explanation:

5 0
3 years ago
When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
A train travelling at 50km/h approaches another train moving towards the first at 90km/h. If they are 35km apart (on a straight
Ede4ka [16]

it will take 36 seconds


8 0
3 years ago
Calculate the force of attraction between
vagabundo [1.1K]

Answer:

So the force of attraction between the two objects is 3.3365*10^-6

Explanation:

m1=10kg

m2=50kg

d=10cm=0.1m

G=6.673*10^-11Nm^2kg^2

We have to find the force of attraction between them

F=Gm1m2/d^2

F=6.673*10^-11*10*50/0.1^2

F=3.3365*10^-8/0.01

F=3.3365*10^-6

4 0
3 years ago
- In Einstein's famous equation E = me?, describing the
mel-nik [20]

Answer:

option C is correct

Explanation:

8 0
3 years ago
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