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3 years ago

Complete the statement about memory improvement.

1 answer:

8 0

Elaborate rehearsal involves organizing and breaking down information into easier groups to expand capacity. Rehearsal is the verbal repetition of information. These techniques are especially important for the improvement of long-term memory.

Six strategies of elaborate rehearsal are:

1. Rephrase information into your own words.

2. Create your own study questions and answer them.

3. Use images to help you.

4. Group terms that are of the same topic together.

5. Use a mnemonic strategy.

6. Space our your learning and don't try to cram in on sitting.

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If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average

Nina [5.8K]

Answer:

Average accelation = -4V

Explanation:

$a=\frac{V-V0}{t}$

V=0 m/s (because the frog stopped)

V0 = V (average velocity)

t= 0,25 s

So;

$a=\frac{V-V0}{t}=\frac{0-V}{0.25}=-4V$

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3 years ago

Two electric charges A and B were placed facing each other at a distance of separation "r". The common electrostatic force betwe

lutik1710 [3]

Answer:

From the formula of force:

$F = \frac{kAB}{ {r}^{2} } \\$

since AB and k are constants:

$F \: \alpha \: \frac{1}{ {r}^{2} } \\ \\ F = \frac{x}{ {r}^{2} }$

x is a constant of proportionality

• when force is 4N, separation distance is 1

$4 = \frac{x}{1} \\ x = 4$

therefore, equation becomes

$F = \frac{4}{ {r}^{2} } \\$

when r is doubled, r becomes 2. find F:

$F = \frac{4}{ {2}^{2} } \\ \\ F = \frac{4}{4} \\ \\ { \underline{force \: is \: 1N}}$

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3 years ago

Select all that apply. The force that opposes the start of motion is referred to as _____.

77julia77 [94]

The correct answers are <span>starting friction and </span>static friction

Friction slows down all forces, but starting friction slows down or stops completely the start of motion.

8 0

3 years ago

Read 2 more answers

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

Basile [38]

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3 years ago

5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal whil

Lelu [443]

Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.

By Newton's second law, the net vertical force is

• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0

where a is the acceleration of the wagon.

Solve for n (the magnitude of the normal force) :

n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N

Then

f = 0.500 (58.0 N) = 29.0 N

Meanwhile, the horizontal component of the applied force has magnitude

(80.0 N) cos(30.0°) ≈ 69.3 N

Now calculate the work done by either force.

• friction: -(29.0 N) (10.0 m) = -290. J

• pull: (69.3 N) (10.0 m) = 693 J

7 0

3 years ago