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Gelneren [198K]
3 years ago
13

Important nuclei of the indirect (multineuronal) system that receive impulses from the equilibrium apparatus of the inner ear an

d help to maintain balance by varying muscle tone of postural muscles are the ________.A. Red nuclei.B. Vestibular nuclei.C. Reticular nuclei.D. Superior Colliculi.
Physics
1 answer:
Lera25 [3.4K]3 years ago
6 0

Answer:

B. Vestibular nuclei

Explanation:

The nerve information generated by the vestibular receptors travels through the vestibular portion of the eighth pair that penetrates the brain stem at the level of the brain stem bridge. At this level there are four vestibular nuclei, which receive the synapses of these axons, coming from the ridges and macules. The semicircular ducts predominantly terminate in the superior and medial nuclei. While the fibers coming from the macules end on the lateral, medial and inferior nuclei. Some fibers of the eighth pair end in the flocculonodular lobe of the cerebellum, <u>these connections play an important role in controlling posture and balance.</u>

From the vestibular nuclei, two bundles of fibers descending to the spinal cord originate from the medial and lateral vestibular spinal bundles and a bundle of fiber that rises in the brain stem that participates in the coordination of eye movements, the medial longitudinal fascicle, which participates in Rotational nystagmus This system also participates in an important way in the control of some ocular movements by the fibers that it contributes to the medial longitudinal fascicle, which is a structure that interconnects the motor nuclei of the extrinsic muscles of the eyeballs VI or abdicens nucleus (abductor) on one side and IV or pathetic nucleus (trochlear) and III or nucleus of the common ocular motor (oculomotor) on the opposite side.

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In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 58 N to the input pi
AfilCa [17]

Answer:

The correct answer is "3899.92 N".

Explanation:

The given values are:

Force,

F_{app}=58 N

Ratio,

\frac{R_2}{R_1}=8.2

As we know,

Area, A=\pi r^2

or,

⇒  \frac{F_2}{F_1} =\frac{A_2}{A_1}

On substituting the value of "A", we get

⇒  \frac{F_2}{F_1} =\frac{\pi r_2^2}{\pi r_1^2}

⇒  \frac{F_2}{F_1} =\frac{r_2^2}{r_1^2}

On applying cross-multiplication, we get

⇒  F_2=F_1\times (\frac{r_2}{r_1} )^2

On substituting the given values, we get

⇒       =58\times (8.2)^2

⇒       =58\times 67.2

⇒       =3899.92 \ N

6 0
3 years ago
Calculate the speed (in m/sec) of a wave with a wavelength of 2.1 meters and a period of 9.4 second.
Lorico [155]

Wave speed = (wavelength) x (frequency)

We know the wavelength, but we don't know the frequency. How can we find the frequency ?  "Here frequency frequency."

We know the period, and frequency is just (1 / period).  So . . .

Wave speed = (wavelength) / (period)

Wave speed = (2.1 meters) / (9.4 seconds)

Wave speed = (2.1 / 9.4) m/s

<em>Wave speed = 0.223 m/s</em>

8 0
3 years ago
Which statement is true about the speed of heat transfer by conduction?
antiseptic1488 [7]

Answer:

Conduction is fastest in solid

Explanation:

A basic property of conduction is that for heat to transfer there must be contact. From kinetic theory, configuration of solid is such that the molecules are held together and they move or vibrate about their mean position. Because of this arrangement heat transfer by conduction is faster in solid that any other state of matter

3 0
3 years ago
Help pls lol and tysm
Zanzabum

Answer:

umm section 2 and 4

Explanation:

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7 0
3 years ago
A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth
andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

F_{D}=\frac{kq^{2}}{DC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

F_{B}=\frac{kq^{2}}{BC^{2}}

F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

F_{A}=\frac{kq^{2}}{AC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N

The direction of FA is along A to C.

The net force along +X axis

F_{x}=F_{A}Cos45-F_{D}

F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N

The net force along +Y axis

F_{y}=F_{B}-F_{A}Sin45

F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N

The resultant force is given by

F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}

F = 4.03\times10^{7}N

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0
3 years ago
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