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shepuryov [24]
3 years ago
5

What is the Milgram experiment?

Physics
1 answer:
lora16 [44]3 years ago
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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Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2
Nookie1986 [14]
<h2>Answers:</h2>

<h2>(a) </h2>

According to Newton's Law of Gravitation, the Gravity Force is:

F=\frac{GMm}{{r}^{2}}     (1)

This expression can also be written as:

F=GMm{r}^{-2}    (2)

If we derive this force F respect to the distance r between the two masses:

\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr     (3)

Taking into account GMm are constants:

\frac{dF}{dr}dFdr=-2GMm{r}^{-3}     (4)

Or

\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}     (5)

<h2> (b) dF/dr represents the rate of change of the force with respect to the distance between the bodies.  </h2><h2 />

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

<h2>(c) The minus sign indicates that the bodies are being forced in the negative direction.  </h2>

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

<h2>(d) </h2>

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit N/km:

\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}     (5)

We have a force that decreases with a rate 1 \frac{dF_{1}}{dr}dFdr=4N/km when r=20000km:

4N/km=-2\frac{GMm}{{(20000km)}^{3}}     (6)

Isolating -2GMm:

-2GMm=(4N/km)({(20000km)}^{3})     (7)

In addition, we have another force that decreases with a rate 2 \frac{dF_{2}}{dr}dFdr=X when r=10000km:

XN/km=-2\frac{GMm}{{(10000km)}^{3}}     (8)

Isolating -2GMm:

-2GMm=X({(10000km)}^{3})     (9)

Making (7)=(9):

(4N/km)({(20000km)}^{3})=X({(10000km)}^{3}       (10)

Then isolating X:

X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}  

Solving and taking into account the units, we finally have:

X=-32N/km>>>>This is how fast this force changes when r=10000 km

7 0
4 years ago
Read 2 more answers
Over the years most technology has gotten?
WITCHER [35]
More expensive & probably less complicated
3 0
4 years ago
A tank is full of oil weighing 40 lb/ft^3. The tank is an inverted right circular cone (with the base at the top) with a height
krok68 [10]

Answer:

26945.6 ft⋅lbf

Explanation:

Volume of Right Circular Cone = pi*(radius^2)*(height/3)

Pi*(4)*(5/3) = 20.94 ft^3

Density = Mass / Volume

Mass = Density*Volume

Mass = (40)*(20.94)

Mass = 837.6 lb

Work = Force*Height

Force = Mass*Acceleration

Acceleration will be gravitational acceleration

Work = (837.6)*(32.17)*(1)

Work = 26945.6 ft⋅lbf

8 0
3 years ago
As a woman carries her suitcase up a flight of stairs, she does work against
Rashid [163]

Answer:

C. 1176 J

Explanation:

12 x 9.8 x 10

Hope it helpss

4 0
3 years ago
A transmission line consisting of two concentric circular cylinders of metal, with radii a, and b, with b &gt; a, is filled with
AURORKA [14]

Answer:

P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.

Explanation:

So, we will be making use of the data or parameters given in the question above;

=>" radii a, and b, with b > a, is filled with a uniform dielectric material with permittivity ε and permeability μ0."

=> " A TEM mode is propagated along the line and the peak value of magnetic field when rho = a is B0."

So, we will be making use of the two equations below;

Ë = ( λ/ 2πEP) × P'. --------------------------(1)..

B' = √ μE × ( λ/ 2πEP) × P' --------------(2).

Where equation (1) and (2) represent Gauss' law and magnetic field equation respectively.

Jo = 1/√ μE × λ/2 × π × a.

When we solve for charge per unit length, we have;

λ = 2 × π × Jo × a × √ μE.

The energy flux,s = E' × J'= √μE× |Jo(Z) |^2 × a^2/b^2 { cos^2 kz - wt + Avg Jo} Z'.

Hence, the time. Average power flux = 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z'.

Therefore, P = ∫z' . <s> da

P = ∫ ∫ 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z' pd pd p Θ.

(Take limit on the first at second integration as : 2π,0 and b,a).

P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.

5 0
3 years ago
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