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3 years ago

Explain how cognitive psychologists combine traditional conditioning models with cognitive processes.

2 answers:

6 0

One area in which cognitive psychologists have influenced traditional conditioning practices is modern education. Education previously employed conditioning models to teach students with reward (good scores) and punishment (bad scores). Modern educators use metacognition (a concept from cognitive psychology) to aid students in evaluating their personal knowledge. That in turn helps students to strategize for improvement in school subjects where they are weak.

user100 [1]3 years ago

4 0

Behaviorists generally claimed that conditioning occurred without thinking or reasoning ans was simply a result of consequences or reinforcement. Cognitive psychologists demonstrated that thinking and reasoning (cognition) influences the conditioning processes and that many behaviors that are conditioned depend on the type of cognitive reasoning that occurs during conditioning. Therefore, as one is being conditioned to respond to environmental stimuli or is responding to a consequence, they are also pondering and thinking about the process occuring. Cognition is often the reason individuals are not all conditioned in the same manner.

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4 years ago

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hichkok12 [17]

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3 years ago

Read 2 more answers

A group of students are provided with three objects all of the same mass and radius. The objects include a solid cylinder, a thi

SOVA2 [1]

Answer:

Sphere, cylinder hoop

Explanation:

To analyze Which student is right it is best to propose the solution of the problem. Let's look for the speed of the center of mass. Let's use the concept of mechanical energy

In the highest part of the ramp

Em₀ = U = mg y

In the lowest part

Here the energy has part of translation and part of rotation

$E_{mf}$ = $K_{T}$ + $K_{R}$

$E_{mf}$ = ½ m $v_{cm}$ ² + ½ I w²

Where I is the moment of inertia of the body and w the angular velocity that relates to the velocity of the center of mass

$v_{cm}$ = w r

w = $v_{cm}$ / r

Let's replace

$E_{mf}$ = ½ I ( $v_{cm}$ / r)²

Energy is conserved

mg y = ½ m $v_{cm}$ ² + ½ I $v_{cm}$ ² / r2

½ (m + I / r²) $v_{cm}$ ² = m g y

½ (1 + I / m r²) $v_{cm}$ ² = g y

$v_{cm}$ = √ [2gy / (1 + I / mr²)]

This is the velocity of the center of mass of the bodies, as they all have the same radius with comparing this point is sufficient. Now let's use the speed definition

v = d / t

t = d / v

t = d / (√ [2gy / (1 + I / mr²)])

t = (d / √ 2gy) √(1 + I / m r²)

Therefore we see that time is proportional to the square root. All quantities are constant and the one that varies is the moment of inertia.

The moments of inertia of

Sphere is Is = 2/5 M r²

Cylinder Ic = ½ M r²

Hoop Ih = M r²

Let's replace each one and calculate the time

Sphere

ts = (d / √2gy) √ (1 + 2/5 Mr² / mr²)

ts = (d / √ 2gy) √ (1 +2/5) = (d / √ 2gy) √(1.4)

ts = (d / √ 2gy) 1.1

Cylinder

tc = (d / √2gy) √ (1 + 1/2 Mr² / Mr²)

tc = (d / √2gy) √ (1 + ½) = (d / √ 2gy) √ 1.5

tc = (d / √ 2gy) 1.2

Hoop

th = (d / √2gy) √ (1 + mr² / mr²)

th = (d / √2gy) √(1 + 1) = (d / √ 2gy) √ 2

th = (d / √ 2gy) 1.41

We have the results for the time the body that arrives the fastest is the sphere and the one that is the most hoop. Therefore the correct answer is

ts < tc < th

Sphere, cylinder hoop

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3 years ago