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damaskus [11]
3 years ago
14

The psychologist known for latent learning and cognitive maps is _________. A. Robert Rescorla B. Edward Tolman C. William James

D. B. F. Skinner
Physics
2 answers:
Vitek1552 [10]3 years ago
7 0

Answer:

The correct answer is B. The psychologist known for latent learning and cognitive maps is Edward Tolman.

Explanation:

Edward Tolman was an American psychologist. He became known for his work in the field of learning theory and is considered a pioneer of cognitivism.

He was mainly interested in the problems of learning in the context of behaviorism. Behavior, Tolman argues, cannot be reduced to the stimulus-response pattern. The organism is not only "reactive"; it acts according to a specific aim. In his theory, Tolman takes into account behaviorism, functionalism and the psychology of form. His idea is that one cannot do without the notion of "design", of "pursued goal", neither in humans nor in animals.  Tolman also developed the concept of the “cognitive map”.

Lubov Fominskaja [6]3 years ago
4 0

Answer:

B

Explanation:

B. Edward Tolman

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Answer:

(a) 6650246.305 N/C

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Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

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Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

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(a)

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Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

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E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

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Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

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E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

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