Answer:
If an inhibitory synapse fires at the same time and at the same distance from the initial segment as an excitatory synapse of the same intensity there will be no changes in the potential in the firing zone.
Explanation:
Under normal conditions, the transmembrane potential depends on the ionic charges present in the intracellular and extracellular spaces. The extracellular space load is usually positive and in the cytoplasm is negative.
- <u>Depolarization</u> occurs by opening ion channels that allow sodium to enter the cell, making the intracellular space more positive.
- An opening of potassium channels releases this ion to the extracellular space, leading to <u>hyperpolarization</u>.
An excitatory synapse is one capable of depolarizing a cell and boosting the production of action potential, provided it is capable of reaching the threshold of said potential.
On the other hand, an inhibitory synapse is able to hyperpolarize the cell membrane and prevent an action potential from originating, so that they can inhibit the action of an excitatory synapse.
The interaction between two synapses, one excitatory and one inhibitory, -called synapse summation- will depend on the strength that each of them possesses. In this case, the intensity of both synapses being the same, there will be no changes in the membrane potential in the firing zone.
Learn more:
Excitatory and inhibitory postsynaptic potentials brainly.com/question/3521553
I think your answer should be B. an organism. Hope this helps. =^-^=
Don't trust those link, my guy its a scam or virus
Answer:
26- Zone of aeration- Soil and rock that both contain water
27- Groundwater- Water held underground
28- Zone of saturation- Saturated with water
29- Sinkhole- Hole in the ground caused by erosion
30- Sediment- Material that is broken down by weathering or erosion
31- Floodplain- Low-laying ground next to the river or water area
32- Caverns- A large cave or chamber
Answer:
0.104M
Explanation:
The following equation is given in this question:
mava=mbvb
Where;
ma = molarity of acid (M)
mb = molarity of base (M)
va = volume of acid (ml)
vb = volume of base (ml)
According to this inputted values;
max5.0ml=5.2mlx0.10m
ma = unknown molarity of acid
mb = 0.10M
va = 5.0ml
vb = 5.2ml
Hence, ma x 5.0ml = 5.2ml x 0.10m
5ma = 0.52
ma = 0.52 ÷ 5
ma = 0.104M
