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SIZIF [17.4K]
2 years ago
7

In Milgram's experiment, compliance, or doing what the experimenter asked,

Physics
2 answers:
-BARSIC- [3]2 years ago
8 0

Answer:

Option C

Explanation:

In his experiment, Milgram wanted to test the extent of obedience a person can go as per the instructions of the authority.  

In this he concluded the experiment saying that several factor when changed in the second round of experiment led to the dropping of obedience level and these factors are as follows –  

a) Conducting the experiment at some ordinary place instead of Yale university campus

b) When the teachers were asked to force the hand of learner to touch the shock plate and they were placed in the same room

c) Two participants in the same room – If one participants showed disobedience then the other participant’s disobedience level will fall automatically

d) When the authority giving instructions is not nearby or is absent

Hence, option C is correct

Orlov [11]2 years ago
5 0

Answer:

In Milgram's experiment, compliance, or doing what the experimenter asked,

the teacher and the learner were in the same room. -C.

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A student walks to the right 25-m along the 800 hall in 15-s. They turn around and walk 15-m to the left in 8.0-s. Calculate the
Ludmilka [50]

Answer:

Explanation:

Total distance covered  = 25 + 15 = 40 m

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3 years ago
The density of aluminum is 2.70 g/ml. A piece of aluminum foil has a mass of 44 g. What is the volume of this piece of aluminum
anyanavicka [17]

Answer:

C) 16.3 ml

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Density is equal to the ratio between the mass of an object and its volume:

d=\frac{m}{V}

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- mass of the aluminium foil: m=44 g

So we can re-arrange the equation above and use these data to find the volume of the piece of aluminium foil:

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7 0
3 years ago
What is the direction of transfer of energy in the waves produced?
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A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.13 s. W
vitfil [10]

Answer:

Part a)

T = 0.52 s

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As we know that the particle move from its maximum displacement to its mean position in t = 0.13 s

so total time period of the particle is given as

T = 4\times 0.13 = 0.52 s

now we have

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T = time to complete one oscillation

so here it will move to and fro for one complete oscillation

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Part b)

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