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Airida [17]
3 years ago
7

The following are steps involved in transmission at the cholinergic synapse:

Physics
1 answer:
Westkost [7]3 years ago
6 0

Answer:

3. Step 1; An action potential depolarizes the axon terminal at the presynaptic membrane

2. Step 2; Calcium ions enter the axon terminal

4. Step 3; Acetylcholine is released from storage vesicles by exocytosis

5. Step 4; Acetylcholine binds to receptors on the postsynaptic membrane

1. Step 5; Chemically gated ion channels on the postsynaptic membrane are opened

Explanation:

3. The cholinergic synapse starts at the point of arrival of an electrochemical impulse or action potentials at the synaptic knob of the axon terminal of a presynaptic neuron membrane

2. The arrival of the action potential at the axon terminal causes the calcium ion Ca²⁺ channels to open and Ca²⁺ enters into the synaptic knob, resulting in the fusion of the presynaptic membrane and synaptic vesicles

4. The fusion enables the release into the synaptic cleft of many acetylcholine (ACh) transmitter molecules by exocytosis

5. Some of the ACh are transported across the synaptic cleft and bind to postsynaptic neuron membrane embedded ACh receptors

1. The binding of the ACh neurotransmitter molecules to receptors on the membrane of the dendrites of a neuron it leads to the opening of ion channels

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Read 2 more answers
An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec
jolli1 [7]

Answer:

v_o=39\ m/s\\t_m=4\ s

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

\displaystyle h_m=H+\frac{v_o^2}{2g}

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.

(a)

To find the initial speed we solve for vo:

\displaystyle v_o=\sqrt{2gh_m}

v_o=\sqrt{2\cdot 9.8\cdot 77.5}

v_o=39\ m/s

(b)

The maximum time or the time taken by the object to reach its highest  point is calculated as follows:

\displaystyle t_m=\frac{v_o}{g}

\displaystyle t_m=\frac{39}{9.8}

t_m=4\ s

7 0
2 years ago
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