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Step2247 [10]
3 years ago
5

Explain the process of synaptic transmission, beginning with the neurotransmitters at the axon terminal of the presynaptic cell

and ending with the impulse being sent down the postsynaptic cell.
Physics
2 answers:
nydimaria [60]3 years ago
8 0

The synapse is actually the link between 2 neurons. Now when an action potential contacts the synaptic knob of a neuron, the voltage-gate calcium channels are unlocked, resulting in an influx of positively charged calcium ions into the cell. This makes the vesicles containing neurotransmitters, for example acetylcholine, to travel towards the pre-synaptic membrane. When the vesicle arrives at the membrane, the contents are released into the synaptic cleft by exocytosis. Neurotransmitters disperse across the space, down to its concentration gradient, up until it reaches the post-synaptic membrane, where it connects to the correct neuroreceptors. Connecting to the neuroreceptors results in depolarisation in the post-syanaptic neuron as voltage-gated sodium channels are also opened, and the positively charged sodium ions travel into the cell. When adequate neurotransmitters bind to neuroreceptors, the post-synaptic membrane overcame the threshold level of depolarisation and an action potential is made and the impulse is transmitted.

skelet666 [1.2K]3 years ago
5 0

Answer:

Synaptic transmission, beginning with the neurotransmitters at the axon terminal of the presynaptic cell and ending with the impulse being sent down to the postsynaptic cell isn't very complicated as it sounds. When an action potential contacts the synaptic knob of a neuron, the voltagegate calcium channels are unlocked, resulting in an influx of positively charged calcium ions into the cell. So, that makes the vesicles containing neurotransmitters to travel towards the presynaptic membrane. Then when the vesicle arrives at the membrane, the contents are released into the synaptic cleft by exocytosis. Neurotransmitters disperse across the space, down to its concentration gradient, up until it reaches the postsynaptic membrane, where it connects to the correct neuro receptors. Connecting to the neuro receptors results in depolarization in the post synaptic neuron as voltagegated sodium channels are also opened, and the positively charged sodium ions travel into the cell. When adequate neurotransmitters bind to neuroreceptors, the postsynaptic membrane overcame the threshold level of depolarisation and an action potential is made and the impulse is transmitted.

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A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

F_f=-\mu mg, the force of friction, with \mu=0.25 being the coefficient of friction, m=30.0 kg being the mass of the crate, and g=9.8 m/s^2. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

where

F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N is the magnitude of the force of friction

d = 4.5 m is the displacement

\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

W=F_f d cos \theta

We notice that both gravity and normal force are perpendicular to the displacement: therefore, \theta=90^{circ}, and so

cos \theta=0

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0
2 years ago
Dante is leading a parade across the main street in front of city hall. Starting at city hall, he marches the parade 4 blocks ea
Ipatiy [6.2K]

Answer:

The correct option is A)

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

Given that Dante is leading a parade across the main street in front of city hall.

Let, Initial location of parade is 0i+0j

One block of city is one units on the XY- graph

Statement 1: Parade marches the parade 4 blocks east, then 3 blocks south

New location of parade is 4i-3j

Statement 2: The parade marches 1 block west and 9 blocks north and finally stops.

Final location of parade is (4i-3j)+(-1i+9j)=3i+6j

Displacement is given by

Displacement = (Final destination)-(Initial destination)

Displacement = (3i+6j)-(0i+0j)=3i+6j

Thus,

Magnitude of displacement = \sqrt{3^{2}+6^{2}}

                                              = 6.71 m

Direction of displacement =  tan^{-1}(\frac{Y}{X} )

                                           =  tan^{-1}(\frac{6}{3} )

                                           = 63.43 NE

Therefore, the correct option is A) Displacement: 6.71 m, Direction: 63.4 degrees north of east

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Answer:

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