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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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A 60 W lightbulb is on for 24 hours. A stereo with a power of 150 W is on for 2

Helen [10]

Answer:

B) The lightbulb uses 4,104,000 J more than the stereo.

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3 years ago

A light ray travels in the +x direction and strikes a slanted surface with an angle of 62° between its normal and the ty axis. T

Elena-2011 [213]

Answer:

- 0.6

Explanation:

Given that angle between normal y axis is 62° so angle between normal

and x axis will be 90- 62 = 28 °. Since incident ray is along x axis , 28 ° will be the angle between incident ray and normal ie it will be angle of incidence

Angle of incidence = 28 °

angle of reflection = 28°

Angle between incident ray and reflected ray = 28 + 28 = 56 °

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- 0.6

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3 years ago

Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a

nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

$Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi$

$Q = \frac{\pi^2 r^4}{2}}$

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0

3 years ago

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zaharov [31]

Divide (25,000) by (the number of miles you can walk in 1 hour).
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3 years ago

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Anit [1.1K]

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3 years ago