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maxonik [38]
3 years ago
13

The 'little brain' attached to the rear of the brainstem is called the: (2 points)

Physics
2 answers:
11111nata11111 [884]3 years ago
6 0

Answer:

The 'little brain' attached to the rear of the brainstem is called the cerebellum. See the explanation below, please.

Explanation:

The cerebellum is in the back of the brain, it is small in size. It is responsible for functions such as integrations of sensory and motor pathways, balance, coordination of movements, posture.

Dimas [21]3 years ago
6 0

<u>Answer: </u>

Option(d)

"The 'little brain' is said to be present in the rear part of the brainstem known as the pons.  

<u> Explanation</u>:

Pons are small structures present in the  brainstem. It  connects both the cerebral cortex and other part called medulla oblongata. It controls the communication between the two hemisphere of the brain.  

It is the largest part of  brainstem and is packed with  nerves fibers. It also controls the breathing and the sensation of taste, seeing and hearing.

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The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can
Sonja [21]

Answer:

Explanation:

a ) speed of passenger = circumference / time

= 2π R / Time

= 2 x 3.14 x 50 / 60

= 5.23 m /s

b )

centrifugal force = m v² /R

= (882 /9.8 ) x 5.23² / 50

= 77.47 N

Apparent weight at the highest point

real weight  - centrifugal force

= 882 - 77.47

= 804.53 N

Apparent weight at the lowest point

real weight  + centrifugal force

= 882 +77.47

= 959.47 N

c )  if the passenger’s apparent weight at the highest point were zero

centrifugal force = weight

mv² /R = mg

v² = gR

= 9.8 X 50

v = 22.13 m /s

d )

apparent weight

mg - mv² / R

= 882 - (882 / 9.8 )x 22.13²/50

= 882 +  882

= 1764 N  

=

6 0
3 years ago
A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a
AVprozaik [17]
<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}</h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x 10^{24} Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = \frac{GM}{r^{2} }

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x 10^{-11} m^{3} kg^{-1} s^{-2}

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g =  \frac{(6.67 * 10^{-11}) (5.972 * 10^{24})  }{5733^{2} }

g = 12.12 m/s^{2}

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}

6 0
4 years ago
A magnetic field would be produced by a beam of
kumpel [21]
3 protons should be your answer
4 0
3 years ago
You drag a suitcase of mass 8.2 kg with a force of f at an angle 41.9 ◦ with respect to the horizontal along a surface with kine
DedPeter [7]

Answer:

35.6 N

Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

There are two forces acting along the horizontal direction:

- The horizontal component of the pushing force, which is given by

F_x = F cos \theta

with \theta=41.9^{\circ}

- The frictional force, whose magnitude is

F_f = \mu mg

where \mu=0.33, m=8.2 kg and g=9.8 m/s^2.

The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:

F_x - F_f=0\\F_x = F_f\\F cos \theta = \mu mg\\F=\frac{\mu mg}{cos \theta}=\frac{(0.33)(8.2 kg)(9.8 m/s^2)}{cos(41.9^{\circ})}=35.6 N

8 0
3 years ago
If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m. s
Sladkaya [172]
 the orbital period is 5170 s
6 0
3 years ago
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