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3 years ago

What is similar about analogical and symbolic representations?

Physics

2 answers:

Otrada [13]3 years ago

7 0

<h2>Answer:</h2>

<u>Both of them are treated by brain in augmented form</u>

<h2>Explanation:</h2>

Symbolic representation is the representation in mind where the image of the object is made in brain such as if we say the word flower then the image of flower comes to our mind so the same thing can be done in practical form for the people who are deaf. Gestures are also the examples of symbolic representation. On the other hand analogical representation is usually a comparison between one thing and another, typically for the purpose of explanation or clarification. The image is not shown but the pre recorded image of the object is recalled in the memory and the brain realizes it. For example the word flower relates to the word beauty.

ICE Princess25 [194]3 years ago

3 0

Symbolic representations are mental pictures that have no direct relationship to the actual object you are thinking about. Instead, these mental pictures are connected by themes that are meaningful to you. Anytime you think of words and descriptions for one main concept, you're thinking symbolically. Analogical representations are mental pictures that have a direct relationship to the actual object you are thinking about. Do cows have their ears on the top or side of their heads? Rather than thinking symbolically about a cow and his ears (tiny, smelly, leather), analogical representations of the cow's ears involve thinking of an actual cow in your head.

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nalin [4]

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puteri [66]

Answer:

Explanation:

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GREYUIT [131]

Answer:

The found acceleration in terms of h and t is:

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Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

Constant acceleration, starts from rest.

Distance = $y = \frac{1}{2}a(t_1)^2$

Velocity = $v_1=at_1$

<h3>Stage 2</h3>

Constant velocity where

Velocity = $v_o=v_1=at_1$

Distance =

<h3> $y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\$ </h3><h3 /><h3>Stage 3</h3>

Constant deceleration where

Velocity = $v_0=v_1=at_1$

Distance =

$y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2$

<h3>Total Height</h3>

Total height = y₁ + y₂ + y₃

Total height = $\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2$

<h3 /><h3>Acceleration</h3>

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

$a=\frac{h}{5(t_1)^2}$

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zavuch27 [327]

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Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.

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