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scZoUnD [109]
2 years ago
15

Why would researchers not be allowed to recreate the Little Albert experiment today?

Physics
1 answer:
wariber [46]2 years ago
3 0

Answer:

Explanation:

En la historia de la ciencia se han dado auténticas barbaridades. Pruebas con animales que hoy no perdonaría nadie, o investigaciones de conducta con personas como la de la cárcel de Stanford, que se han saldado como una especie de pasado incómodo sobre los límites de la experimentación. Sin embargo, pocos se pueden acercar por su carácter perturbador al denominado experimento de Little Albert o Pequeño Albert: El salvaje intento por probar con un bebé que las fobias pueden ser condicionadas y aprendidas. Y lo que es peor, conseguirlo.

Esta idea surgió de la mente de John Broadus Watson, reconocido padre de la rama conductista de la psicología, que desde 1913 había comenzado a probar en animales sus tesis. Estas bebían directamente del los estudios de Iván Pavlov, fisiólogo ruso que ganó el Nobel en 1904 por sus estudios sobre el sistema digestivo, pero que también sentó precedentes sobre la psicología.

link por si te interesa:

https://hipertextual.com/2017/10/pequeno-albert

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__________ is the school of psychology that believes perception is more than the sum of its parts—it involves a whole pattern. A
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A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

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