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3 years ago

What evidence supports the information consolidation theory?

Physics

1 answer:

sladkih [1.3K]3 years ago

3 0

Explanation:

Memory loss in retrograde amnesia has long been held to be larger for recent periods than for remote periods, a pattern usually referred to as the Ribot gradient. One explanation for this gradient is consolidation of long-term memories. Several computational models of such a process have shown how consolidation can explain characteristics of amnesia, but they have not elucidated how consolidation must be envisaged. Here findings are reviewed that shed light on how consolidation may be implemented in the brain. Moreover, consolidation is contrasted with alternative theories of the Ribot gradient. Consolidation theory, multiple trace theory, and semantization can all handle some findings well but not others.

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An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

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3 years ago

An electric motor rotating a workshop grinding wheel at 1.00 × 10² rev/min is switched off. Assume the wheel has a constant nega

ololo11 [35]

An electric engine turning a workshop sanding rotation at 1.00 × 10² rev/min is switched off. Take the wheel includes a regular negative angular acceleration of volume 2.00 rad/s². 5.25 moments long it takes the grinding rotation to control.

<h3>What is negative angular acceleration?</h3>

A particle that has a negative angular velocity rotates counterclockwise.
Negative angular acceleration () is a "push" that is hence counterclockwise.
The body will speed up or slow down depending on whether and have the same sign (and eventually go in reverse).
For instance, when an object rotating counterclockwise slows down, acceleration would be negative.
If a rotating body's angular speed is seen to grow in a clockwise direction and decrease in a counterclockwise direction, it is given a negative sign.
It is known that a change in the linear acceleration correlates to a change in the linear velocity.

Let t be the time taken to stop.

ω = 0 rad/s

Use the first equation of motion for rotational motion

ω = ωo + α t

0 = 10.5 - 2 x t

t = 5.25 second

To learn more about angular acceleration, refer to:

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7 0

2 years ago

What kind of wave exists when a musician plays a note on a trombone?

kykrilka [37]

The trombone is a wind musical instrument and as all musical instruments can produce a standing (or stationary) wave.

This kind of waves is the result of the composition of two waves that produces a pattern that looks like it is not moving but just vibrating. Some points of the wave look like they are not even vibrating, they just stand still, and they are called nodes. Other points of the wave vibrate from the maximum positive value to the maximum negative value and are called antinodes.

5 0

4 years ago

Which best explains why infrared waves are ineffective for treating cancer? They have frequencies that are too high to kill canc

Roman55 [17]

Infrared waves are ineffective for treating cancer because they do not transfer enough energy to destroy cancer cells.

<h3>Importance of high frequency waves</h3>

One of the uses of a high frequency wave is in the treatment of cancer. These high frequency waves transfer enough energy to destroy the cancer cells.

E = hf

where;

f is frequency of the wave

Infrareds have low frequency, and hence low energy.

Thus, infrared waves are ineffective for treating cancer because they do not transfer enough energy to destroy cancer cells.

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5 0

2 years ago

Read 2 more answers

How does a balanced chemical equation demonstrate the Law of Conservation of Mass?

lbvjy [14]

Answer:

It shows the same number of atoms of each element on both sides of the equation

A balanced equation demonstrates the conservation of mass by having the same number of each type of atom on both sides of the arrow.

Explanation:

Matter cannot be created or destroyed in chemical reactions. This is the law of conservation of mass. In every chemical reaction, the same mass of matter must end up in the products as started in the reactants. Balanced chemical equations show that mass is conserved in chemical reactions.

6 0

3 years ago