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
In the current brainly version, unfortunately you could not change your username.
The only way is to create a new account, but of course all of your points will be gone
<span>
A. The closet point in the Moon's orbit to Earth . . . . . perigee
B. The farthest point in the Moon's orbit to Earth . . . . . apogee
C. The Sun's orbit that is closest to the Moon . . . . . a meaningless description
D. The closest point in Earth's orbit of the Sun . . . . . perihelion
-- The farthest point in Earth's orbit of the Sun . . . . . aphelion
</span>
Answer:
Explanation:
Given
mass of archer 
Average force 
extension in arrow 
Work done to stretch the bow with arrow
This work done is converted into kinetic Energy of arrow
where v= velocity of arrow
(b)if arrow is thrown vertically upward then this energy is converted to Potential energy
Answer:
a) 145.6kgm^2
b) 158.4kg-m^2/s
c) 0.76rads/s
Explanation:
Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.
a) From I = MK^2
I = (160Kg)(0.91m)^2
I = 145.6kgm^2
b) The magnitude of the angular momentum is given by:
L= r × p The raduis and momentum are perpendicular.
L = r × mc
L = (1.20m)(44.0kg)(3.0m/s)
L = 158.4kg-m^2/s
c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:
L = Iw
158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]
w = 158.6/208.96
w = 0.76rad/s
