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meriva
3 years ago
13

______ involves organizing and breaking down information into easier groups to expand capacity. Rehearsal is the verbal repetiti

on of information. These techniques are especially important for the improvement of_____ memory.
Physics
1 answer:
adell [148]3 years ago
6 0

Hi!


The answers would be <u>chunking</u> & <u>short-term</u>


1. <u>Chunking </u>involves organizing and breaking down information into easier groups to expand capacity.

<h3>Explanation:</h3>

Chunking is a mental process that is observed to increase short-term memory by taking the information and categorizing it into small groups. For instance, a longer number taken as a single unit is harder to recall then when it is divided into smaller units. 235469350 is harder to instantly recall as compared to when it is chunked into 3 groups: 235 469 350.

This allows more information to be stored in, thereby increasing the capacity of the mind to store information.


2. Rehearsal is the verbal repetition of information. These techniques are especially important for the improvement of <u>short-term</u> memory.

<h3>Explanation: </h3>

Short-term memory is lost after a couple of seconds or minutes, for instance even if you chunk the information, you might not recall it after 30 seconds. Rehearsing or repetition of information, either loudly or mentally, extends the time a particular information is retained.

So you depending on the number of times you repeat the number 235 469 350, the more your short term memory improves .


Hope this helps!


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On what factor does speed of sound depend​
velikii [3]

Answer:

For sound waves to travel, there is a requirement of medium and density of the medium is considered to be one of the factors on which the speed of sound depends. When the medium is dense, the molecules in the medium are closely packed which means that the sound travels faster.

Explanation:

4 0
3 years ago
A mass on a spring with k=88.7 N/m oscillates 15 times in 9.24s. what is the objects mass? unit=kg?
sweet [91]

The mass on the spring is 0.86 kg

Explanation:

The period of a mass-spring system is given by the equation

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant

In this problem, we have:

k = 88.7 N/m is the spring constant

The system makes 15 oscillations in 9.24 s: therefore, the period of the system is

T=\frac{9.24}{15}=0.62 s

Now we can re-arrange the first equation  to solve for the mass:

m=k(\frac{T}{2\pi})^2=(88.7)(\frac{0.62}{2\pi})^2=0.86 kg

Learn more about period:

brainly.com/question/5438962

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3 0
3 years ago
Two thin 80.0-cm rods are oriented at right angles to each other. Each rod has one end at the origin of the coordinates, and one
kogti [31]

Answer:

The net force on the electron is given as:

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

Explanation:

Given:

charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C

charge on rod along y-axis = Q₂ = 15 x 10⁻⁶ C

distance of electron from rod 1 = r₁ = 0.4 m

distance of electron from rod 1 = r₂ = 0.4 m

charge on electron = q = -1.6 x 10⁻¹⁹ C

ε° = 8.85 x 10⁻¹² C²/Nm²

Electric force on charge due to rod 1:

F₁ = qE = 1/4πε°(qQ₁/r₁²)

F₁ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x -15 x 10⁻⁶)/0.4²

F₁ = 1.35 x 10⁻¹³ N

Negative negative repels each other so the rod will Force the electron in positive y-direction.

F₁ = 1.35 x 10⁻¹³ N j

Electric force on charge due to rod 2:

F₂ = qE = 1/4πε°(qQ₂/r₂²)

F₂ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x 15 x 10⁻⁶)/0.4²

F₂ = - 1.35 x 10⁻¹³ N

Opposite charges attract each other so the rod will force the electron in negative x-direction.

F₂ =  - 1.35 x 10⁻¹³ N i

Net Force:

F = F₁ + F₂

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

4 0
3 years ago
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d1i1m1o1n [39]
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3 years ago
Read 2 more answers
A glass capillary tube with a diameter of 8.5 mm and length 8 cm is filled with a salt solution with a resistivity of 2.5 ?m. Wh
MA_775_DIABLO [31]

Answer:

The resistance is 3.5\times10^{-4}\ \Omega

Explanation:

Given that,

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Resistivity = 2.5 m

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The resistance is equal to the product of the resistivity and length divided by the area of cross section .

In mathematical form,

R = \dfrac{\rho\times l}{A}

Where, \rho=resistivity

l = length

A = area of cross section

Put the value into the formula

R = \dfrac{2.5\times8\times10^{-2}}{3.14\times(\dfrac{8.5}{2}\times10^{-3})^2}

R=3526.32\ \Omega

R=3.5\times10^{-4}\ \Omega

Hence, The resistance is 3.5\times10^{-4}\ \Omega

6 0
3 years ago
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