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miv72 [106K]
3 years ago
10

Neurons that relay information within the cns and are the location of information processing are called __________.

Physics
1 answer:
emmainna [20.7K]3 years ago
7 0

Answer:

Interneurons

Explanation:

An interneuron or integrative neuron is a central nervous system neuron, usually small and short axon, that interconnects with other neurons; but never with sensory receptors or muscle fibers, allowing more complex functions.

The interneuron, also called the association neuron, has the function of analyzing sensory information and storing part of it. It also acts on reflex acts, transforming a stimulus in response at the level of the spinal cord. They are located between sensory and motor neurons and are located in the upper nerve centers. Interneurons are multipolar neurons, which connect afferent neurons with efferent neurons in the neuronal or nerve tracts. In other words, they function as a communicational bridge, intercommunicating sensory neurons with motor neurons. Like motor cells, interneurons are only found in the central nervous system. In contrast to the peripheral nervous system, all CNS neurons appear to be interneurons, as they are in communication with many other neurons. However, the term "interneuron" refers to neurons that have axon and dendritic extensions of local extension and not distant, that is, short.

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umka21 [38]

Answer:it would be C

Explanation:

8 0
3 years ago
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Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along  positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, \overrightarrow{v_1}=-120\ \vec{i}

Velocity in north direction is, \overrightarrow{v_2}=120\ \vec{j}

Now, since v_1\ and\ v_2 are perpendicular to each other, their resultant magnitude is given as:

|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}

Plug in the given values and solve for the magnitude of the resultant.This gives,

|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

4 0
2 years ago
What are the three stages of matter and how?​
n200080 [17]

Answer:

Solid, Liquid and Gas

Explanation:

They are the 3 states or stages of Matter

7 0
3 years ago
The climate of the earth throughout history has always _____.
Vlad [161]

Answer:

The climate of the earth throughout history has always been <em><u>fluctuated between hot and cold periods. </u></em>

7 0
3 years ago
Read 2 more answers
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
2 years ago
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