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2 years ago

2. According to research, what is the most reliable form of identifying potentially effective reinforcers?

1 answer:

6 0

The most reliable form of identifying potentially effective reinforcers is known as Preference Assessment. It is an adaptive procedure vastly used in the field of reinforcers.

Explanation:

Preference Assessment identifies items that are likely to be effective as reinforcers by identifying a particular learner's preference for them.

Reinforcement is an important part of being an effective behavior technician and the preference assessments are crucial.
Plenty of research has shown that when you give the learner an opportunity to show you what they want that is a much better indication of what will work as a reinforcer rather than asking family.
Preference assessments identify attention, objects or activities that have a strong potential for serving as reinforcers for target clients. There are ethical reasons to use them.
Beyond the scope of the guide is the exhaustive review of the Preference assessment methodology and literature. However, interested readers go through Virus-Ortega et al(2014) for decision- making process and selecting an appropriate method of preference assessment for individuals with disabilities.

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Answer:

Inhibit the flow of electrons

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Why can an object still be seen when it is at absolute zero?

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2 years ago

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

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Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

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