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2 years ago

What is the Milgram experiment?

1 answer:

6 0

Stanley Milgram, a psychologist at Yale University, conducted anexperiment<span> focusing on the conflict between obedience to authority and personal conscience. He examined justifications for acts of genocide offered by those accused at the World War II Nuremberg War Criminal trials.</span>

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An upward force act on a proton as it moves with a speed of 2.0 x 10^5 meters/seconds through a magnetic field of 8.5 x 10^2

Gekata [30.6K]

Force on a moving charge is given by formula

$\vec F = q(\vec v \times \vec B)$

here we know that this force will be maximum when velocity is perpendicular to magnetic field

$\vec F = qvB$

here we know that

$v = 2.0 \times 10^5 m/s$

$q = 1.6 \times 10^{-19} C$

$B = 8.5 \times 10^2 T$

now we have

$F = (1.6 \times 10^{-19})(2 \times 10^5)(8.5 \times 10^2)$

$F = 2.72 \times 10^{-11} N$

7 0

2 years ago

What type of wave travels through matter only??

Olenka [21]

A mechanical wave can only travel through matter.

7 0

2 years ago

Read 2 more answers

A scientist that applies the laws of science to the needs of communities is called _____.

s2008m [1.1K]

Answer:

The experimental scientist

5 0

1 year ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a

Vesna [10]

Answer:

Tangential speed=5.4 m/s

Radial acceleration= $88.6m/s^2$

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution= $2\pi rad$

2.59 rev= $2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s$

By using $\pi=3.14$

Angular velocity= $\omega=16.27rad/s$

Distance from axis=r=0.329 m

Tangential speed= $r\omega=16.27\times 0.329=5.4m/s$

Radial acceleration= $\frac{v^2}{r}$

Radial acceleration= $\frac{(5.4)^2}{0.329}=88.6m/s^2$

6 0

2 years ago

Why is the following situation impossible? A skater glides along a circular path. She defines a certain point on the circle as h

Arturiano [62]

Answer:

A skater glides along a circular path. She defines a certain point on the circle as her origin. Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin.

So here in circular motion of the skater we can see that the total path length of the skater is along the arc of the circle while we can say that displacement is defined as the shortest distance between initial and final position of the object.

So it is not possible in any circle that arc-length is less than the chord joining the two points on the circle

As we know that arc length is given as

$L = R\theta$