Answer:
The upper motor neurons synapse in the spinal cord connect with anterior horn cells of lower motor neurons, usually via interneurons. The anterior horn cells are the cell bodies of the lower motor neurons and are located in the grey matter of the spinal cord.
Explanation:
Interneurons are the central nodes of neural circuits, enabling communication between the upper motor neurons, sensory or motor neurons located in the brain and spinal cord and they send signals to lower motor neurons or central nervous system (CNS) in the brain stem and spinal cord . When they get a signal from the upper motor neurons, they send another signal to your muscles to make them contract. They play vital roles in reflexes, neuronal oscillations, and neurogenesis in the adult mammalian brain.
Renshaw cells are among the very first identified interneurons. They are excited by the axon collaterals of the motor neurons. In addition, Renshaw cells make inhibitory connections to several groups of motor neurons.
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The three types are alpha beta and gamma
Answer:
A) R = (200 i ^ + 100 j ^ + 30k ^) m
, B) L = 223.61 m
, C) R = 225.61 m
Explanation:
Part A
This is a vector summing exercise, let's take a Reference System where the z axis corresponds to the height (flights), the x axis is the East - West and the y axis corresponds to the North - South.
Let's write the displacements
Descending from the apartment
10 flights of 3 m each, the total descent is 30 m
Z = 30 k ^ m
Offset at street level
L1 = 0.2 i ^ km
L2 = 0.1 j ^ km
Let's reduce everything to the SI system
L1 = 0.2 * 1000 = 200 i ^ m
L2 = 100 j ^ m
The distance traveled is
R = (200 i ^ + 100 j ^ + 30k ^) m
Part B
The horizontal distance traveled can be found with the Pythagorean theorem for the coordinates in the plane
L² = x² + y²
L = √ (200² + 100²)
L = 223.61 m
Part C
The magnitude of travel, let's use the Pythagorean theorem for the sum
R² = x² + y² + z²
R = √ (30² + 200² + 100²)
R = 225.61 m
Answer:
a. E = 122.4 N/C
b. E = 58.2 N/C
c. E = 0
Explanation:
The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.
In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.
A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.
B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.
C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.
