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3 years ago

What includes the hippocampus, amygdala and hypothalamus?

Physics

2 answers:

grigory [225]3 years ago

8 0

Answer:

limbic system

Explanation:

The "limbic system" is situated<em> under the cerebrum. </em>It is responsible for the person's <u>emotions</u> and<u> memories.</u> Its primary structures include: <em>hippocampus, amygdala, thalamus and hypothalamus.</em>

The "hippocampus" is responsible for obtaining<em> new memories.</em> It is also associated with mental functions like<em> "learning."</em>

The "amygdala" is responsible for <u><em>emotional memories.</em></u>

The "thalamus" and "hypothalamus" are responsible for <u>reactions towards the emotions.</u>

marysya [2.9K]3 years ago

6 0

Answer:

the limbic system has its input and processing side (the limbic cortex, amygdala and hippocampus) and an output side (the septal nuclei and hypothalamus).

Explanation:

hope it helps

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A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley below. Find the final velocity and displ

Anestetic [448]

Answer:

a. -39.2 m/s; -78.4 m

b. -31.2 m/s; -46.4 m

c. -47.2 m/s; -110.4 m

Explanation:

We are given/can infer these variables:

t = 4.0 s
a = -9.8 m/s²
v_0 = 0 m/s

We want to find the displacement and the final velocity of the rock.

Δx = ?
v = ?

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = 0 + (-9.8)(4.0)
v = -9.8 * 4.0
v = -39.2 m/s

The final velocity of the rock is -39.2 m/s.

Now we can use this equation to find the displacement of the rock:

Δx = v_0 t + 1/2at²

Plug in the known variables into this equation.

Δx = 0 * 4.0 + 1/2(-9.8)(4.0)²
Δx = 1/2(-9.8)(4.0)²
Δx = -4.9 * 16
Δx = -78.4 m

The displacement of the rock is -78.4 m.

We are given/can infer these variables:

v_0 = 8.0 m/s
a = -9.8 m/s²
t = 4.0 s

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = 8.0 + (-9.8)(4.0)
v = 8.0 + -39.2
v = -31.2 m/s

The final velocity of the rock is -31.2 m/s.

We can use this equation to find the displacement:

Δx = v_0 t + 1/2at²

Plug in known variables:

Δx = 8.0(4.0) + 1/2(-9.8)(4.0)²
Δx = 32 - 4.9(16)
Δx = -46.4 m

The displacement of the rock is -46.4 m.

We are given/can infer these variables:

v_0 = -8.0 m/s
a = -9.8 m/s²
t = 4.0 s

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = -8.0 + (-9.8)(4.0)
v = -8.0 - 39.2
v = -47.2 m/s

The final velocity of the rock is -47.2 m/s.

We can use this equation to find the displacement:

Δx = v_0 t + 1/2at²

Plug in known variables:

Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²
Δx = -32 - 4.9(16)
Δx = -110.4 m

The displacement of the rock is -110.4 m.

8 0

3 years ago

Oil having a specific gravity of 0.9 is pumped as illustrated with a water jet pump. The water volume flowrate is 1 m3 /s. The w

love history [14]

Answer:

The rate at which the pump moves oil is 1 m³/s

Explanation:

Assumptions:

there is steady-state flow

oil and water are incompressible

first fluid is water, second fluid is oil and third fluid is the mixture of oil and water.

$\rho_1Q_1 + \rho_2Q_2 = \rho_3Q_3 -------equation (i)$

where;

ρ is the fluid density

Q is the volumetric flow rate

$Q_1 + Q_2 = Q_3--------equation (ii)$

Substitute in Q₃ in equation i

$\rho_1Q_1 + \rho_2Q_2 = \rho_3(Q_1 +Q_2)$

divide through by ρ₁

$\frac{\rho_1Q_1}{\rho_1}+ \frac{\rho_2Q_2}{\rho_1} =\frac{ \rho_3(Q_1 +Q_2)}{\rho_1}\\\\Note; \frac{\rho_2}{\rho_1} = \gamma_2 \ and \ \frac{\rho_3}{\rho_1} = \gamma_3\\\\Q_1 + \gamma_2Q_2 = \gamma_3(Q_1+Q_2)$

Make Q₂ the subject of the formula

$Q_2 = \frac{Q_1(1- \gamma_3)}{\gamma_3-\gamma_2} = \frac{1 (\frac{m^3}{s}) (1-0.95)}{0.95-0.9} = 1 \ \frac{m^3}{s}$

Therefore, the rate at which the pump moves oil is 1 m³/s

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3 years ago

How far does a car travel in 10 s if its initial velocity is 3 m/s and its acceleration

victus00 [196]

To learn more about Acceleration please visit -
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4 0

2 years ago

A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the loc

sattari [20]

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

$\overrightarrow{v}_{s,w}=6.5 \widehat{j}$

velocity of water with respect to earth = 1.5 m/s at 40° north of east

$\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)$

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

$\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}$

$\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j} \right )$

$\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}$

The magnitude of the velocity of ship relative to earth is $\sqrt{1.15^{2}+5.54^{2}}$ = 5.66 m/s

5 0

3 years ago

The motion analysis derived for use within a translating reference frame is not capable of including motion of the point occurri

Natali [406]

Answer:

False

Explanation:

The motion analysis that is derived to be used within a translating reference frame is capable of including the motion of a point that occurs within the translation or we say in the translating frame because of the fact that for inertial frame, any motion can be implemented in translating frame provided that the frame is strictly inertial frame

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4 years ago