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3 years ago

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

Physics

2 answers:

-BARSIC- [3]3 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]3 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 water tank is filled with up to 3.5 m height.Calculate the pressure given by the tank at its bottom

anyanavicka [17]

Answer:

96.04Pa

Explanation:

height=3.5m

gravity=9.8%

density=9.8/3.5

=2.8

Preassure=h×g×d

=3.5×9.8×2.8

=96.04Pa

6 0

3 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

and

$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

where:

x = 1322 ft is the final position of the shell when it strikes the hill

$u=156 ft/s$ is the initial velocity of the shell

$\theta=49.0^{\circ}$ is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

$v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s$

Instead, the vertical component of the velocity is given by

$v_y=usin \theta -gt$

And substituting at t = 12.9 s (time at which the shell strikes the hill),

$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0

3 years ago

1) A force of 20 Newton acts on a bar having a cross sectional area of 0.8m^2 and length 10cm.calculate the stress developed in

Elanso [62]

Answer:25N/M^2

Explanation:

Force=20N Area=0.8M^2

Stress=force/area

Stress=20/0.8

Stress=25N/M^2

4 0

3 years ago

Why is this a scientific question?

defon

The 3rd one. The question can be tested by a systematic procedure

7 0

3 years ago

Read 2 more answers

Given that the frequency of an EM wave is 4THz, what is its wavelength?

Vilka [71]

Answer:

The wavelength of the EM wave is 7.5 * 10⁻⁴ m

Explanation:

The velocity of a wave is related to its wavelength by the following formula;

velocity = wavelength * frequency

For an electromagnetic (EM) wave, its velocity is equal to the velocity of light, c = 3.0 * 10⁸ m/s

Given that the frequency and veloity of the given EM wave in the question is known, its wavelength is calculated as follows:

wavelength = velocity/frequency

where velocity of the EM wave = 3.0 * 10⁸ m/s;

frequency = 4THz = 4 * 10¹² Hz

wavelength = 3.0 * 10⁸m/s / 4 * 10¹² Hz

wavelength = 7.5 * 10⁻⁴ m

Therefore, the wavelength of the EM wave is 7.5 * 10⁻⁴ m

8 0

3 years ago