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3 years ago

Why society might initially reject a new scientific theory?

1 answer:

6 0

I was about to say: because people generally get comfortable with
what they think they know, and don't like the discomfort of being told
that they have to change something they're comfortable with.

But then I thought about it a little bit more, and I have a different answer.

"Society" might initially reject a new scientific theory, because 'society'
is totally unequipped to render judgement of any kind regarding any
development in Science.

First of all, 'Society' is a thing that's made of a bunch of people, so it's
inherently unequipped to deal with scientific news. Anything that 'Society'
decides has a lot of the mob psychology in it, and a public opinion poll or
a popularity contest are terrible ways to evaluate a scientific discovery.

Second, let's face it. The main ingredient that comprises 'Society' ... people ...
are generally uneducated, unknowledgeable, unqualified, and clueless in the
substance, the history, and the methods of scientific inquiry and reporting.

There may be very good reasons that some particular a new scientific theory
should be rejected, or at least seriously questioned. But believe me, 'Society'
doesn't have them.

That's pretty much why.

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A pumkin is dropped, after 5 seconds, its velocity is 47 m/s. What is its acceleration?

EastWind [94]

Since the pumpkin is dropped, that means it is in freefall and is accelerating at the acceleration due to gravity. The acceleration due to gravity is 9.81m/s down.

7 0

3 years ago

Where did we use rotational and irrotational flow

WARRIOR [948]

Answer:

The term rotational and irrotational flow is associated withe the flow of particles in fluid.

The common example of irrrotational flow can be seen on the carriages of the Ferris wheel (giant wheel).

Explanation:

If the fluid is rotating along its axis with the streamline flow of its particles,then this type of flow is rotational flow.
Similarly if fluid particles do not rotate along its axis while flowing in a stream line flow then it is considered as the irrotational flow.
In majority, if the flow of fluid is viscid then it is rotational.
Fluid in a rotating cylinder is an example of rotating flow.

3 0

3 years ago

An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to

Amiraneli [1.4K]

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

$\theta_1$ is the angle of incidence (in the first medium)

$\theta_2$ is the angle of refraction (in the second medium)

Let's now consider a situation in which

$n_1 > n_2$

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

Where $\frac{n_1}{n_2}$ is a number greater than 1. This means that above a certain value of the angle of incidence $\theta_1$ , the term on the right can become greater than 1. So this would mean

$sin \theta_2 > 1$

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

$\theta_c =sin^{-1}(\frac{n_2}{n_1})$

8 0

3 years ago

You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m

Ann [662]

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

$v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y$

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

$|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}$ ≈ $6.32 m/s$

And the angle with respect to the east should be with:

$arctan(\frac{2}{6} )=18.43 \°$

8 0

3 years ago

Define output work and input work

Olenka [21]

Answer: Input work is the work done on a machine as the input force acts through the input distance.Other ways it would be the work done on a body or system, that is, forces that are applied to the body or system. This is in contrast to output work which is a force that is applied by the body or system to something else.Output work is the work done by a machine as the output force acts through the output distance. The machine does to the object to increase the output distance.

Explanation:

6 0

3 years ago