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3 years ago

Why are Watson’s experiments considered controversial?

2 answers:

fomenos3 years ago

5 0

John Watson's experiments are considered controversial because D. They would be considered unethical today.

The Little Albert experiment exposed children, usually infants, first to things like rabbits, puppies, and hairballs. The children did not show fear of the items. However, once they were exposed to sudden loud sounds, rats, and stressful controlled situations. The goal of this experiment was to "condition phobia into a mentally stable child". However, these experiments are also failed in modern scientific testing. However, that doesn't make the Little Albert experiment as controversial as the unethical aspect does.

Hope I could help!

iris [78.8K]3 years ago

4 0

D, it is considered unethical today

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The diagram below shows the structural formula and 3D model of the same substance. Which substance is represented by the diagram

Agata [3.3K]

I feel like its C but mmmm idk

4 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago

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igomit [66]

Answer:

See below

Explanation:

Find the NET forces on the objects

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B 0

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D 15==>

So biggest accel = C because it has the most force acting on it

next is A because it has the next biggest force

next is D then B ...B has no net force acting on it

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2 years ago

As we move ACROSS the periodic table in a row, the electronegativity increases decreases

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It increases across a period but it decreases down a group.

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