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navik [9.2K]
3 years ago
8

Ann has a geography test next Friday. She needs to know all the US states and their capitals. Provide three effective techniques

she could use to
learn this information and why they would be effective. Be sure to point out at which stage she would be enhancing, encoding, storing, original
retrieving memory.
Physics
1 answer:
Mumz [18]3 years ago
8 0

Answer: THIS IS THE EXACT ANSWER

Three effective techniques Ann could use to learn all the US states and their state capitals are visualization and association, creating a mnemonic device, and rhyming. She could learn some of the states and their capitals by visualizing her own experiences in or about the state. As an example of association, if Ann visited Idaho last winter to see her twin brothers, she might think “I went to Idaho to visit the boys” (sounds like Boise). By employing association, she connects her own life experience to the information she wants to remember to make recalling the information easier. To remember the capital of Washington, Ann might create a mnemonic device like “George Washington went to the Olympics” (Washington’s capital is Olympia). Mnemonic devices organize information to encode it in a way that helps with recall. To effectively use rhyming, she might remember Michigan is a northern state and use “Reindeer prancing in Lansing.” Rhyming provides a cue that helps with recall. All three of these techniques work in the encoding and retrieval stages.

Explanation:

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Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

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The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

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          d = \frac{1}{4} \  2\pi  r

           d =\frac{\pi }{2r}

we substitute

           v =  \frac{\pi  r}{2t}

           r = \frac{2 \ t  \ v}{\pi }

           

let's calculate

           r =\frac{2 \ 2.2  \ 10^{-3} \ 88}{\pi } 2 2.2 10-3 88 /πpi

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let's substitute the values

           q = \frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}7.2 10-8 88 / 0.6 123.25

            q = 8.57 10⁻⁸ C

Let's reduce to mC

           q = 8.57 10⁻⁸ C (10³ mC / 1C)

           q = 8.57 10⁻⁵ mC

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