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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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How many minutes of daylight do we gain after winter solstice

serg [7]

Answer:

Depends on which hemisphere you are belong to and how much distance you are away from Ecuador.

Explanation:

Minutes of daylight is equal on everywhere only on the equinox days (21 March and 23 September). On other days it depends on the place that you are belong to. On winter solstice, places on Ecuador have 12 hours daylight. North side of Ecuador have less, south side of Ecuador have more hour of daylight.

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3 years ago

At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has radius of 2 in? The depth of the soda dec

evablogger [386]

Answer:

The soda is being sucket out at a rate of 3.14 cubic inches/second.

Explanation:

R= 2in

S= π*R²= 12.56 inch²

rate= 0.25 in/sec

rate of soda sucked out= rate* S

rate of soda sucked out= 3.14 inch³/sec

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4 years ago

Newton's second law problem. The spring constant can be determined from the balance of forces in equilibrium. Suppose the spring

mario62 [17]

Answer:

K =6.697 Kg/s²

Explanation:

Given:

delta m =41 g = 0.041 kg

delta x = 6cm = 0.06m

g = 9.8 m/s²

according to the given formula

K = delta m g /delta x

K = (0.041 kg × 9.8 m/s²) / 0.06m

K =6.697 Kg/s²

6 0

3 years ago

The acceleration due to gravity on the surface of Jupiter is about 2.5 times the acceleration due to gravity on Earth’s surface.

Ann [662]

Answer: The correct answer is option C.

Explanation:

Weight = Mass × Acceleration

Let the mass of the space probe be m

Acceleration due to gravity on the earth = g

Weight of the space probe on earth = W

$W=m\times g$

Acceleration due to gravity on the Jupiter = g' = 2.5g

Weight of the space probe on earth = W'

$W'=mg'=m\times 2.5g$

$\frac{W'}{W}=\frac{m\times 2.5g}{m\times g}$

$W'=2.5\times W$

The weight of the space probe on the Jupiter will be 2.5 times the weight of the space probe on earth.

Hence, the correct answer is option C.

6 0

3 years ago

1) Calculate the ideal efficiency of a heat engine that takes in energy at 800 K and expels heat to a

masha68 [24]

Answer:

Could you explain that more better?

Explanation:

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3 years ago