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3 years ago

How is reflection of light used in research

1 answer:

7 0

In order to read the publications of his peers, or read his own notes of the work
that he did on the previous day, or find his coffee mug on his desk in the lab, the
research scientist must arrange to have each of them illuminated with visible
wavelengths of light, and then he must catch the light reflected from each of them
with his eyes.

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Which statement relates most directly to the second law?

alukav5142 [94]

Statement three i do believe

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4 years ago

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A slinky is stretched by two forces. When the forces are removed, the slinky returns to its original length. The slinky has been

e-lub [12.9K]

Answer: elastically deformed or non-permanently deformed

Explanation:

According to classical mechanics, there are two types of deformations:

-Plastic deformation (also called irreversible or permanent deformation), in which the material does not return to its original form after removing the applied force, therefore it is said that the material was permanently deformed.

This is because the material undergoes irreversible thermodynamic changes while it is subjected to the applied forces.

-Elastic deformation (also called reversible or non-permanent deformation), in which the material returns to its original shape after removing the applied force that caused the deformation.

In this case the material also undergoes thermodynamic changes, but these are reversible, causing an increase in its internal energy by transforming it into elastic potential energy.

Therefore, the situation described in the question is related to elastic deformation.

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3 years ago

A 10kg roller coaster does a loop-de-loop and it’s your job to make sure it is safe. The people will fall out if the cart’s spee

Bingel [31]

Answer:

Explanation:

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3 years ago

Do Number 8 Please respond fast WILL MARK BRAINLIEST (THIS INVOLVES INCLINED PLANES)

jasenka [17]

Answer:

We need to apply a force of magnitude 32.13 N

Explanation:

The forces acting on this crate are:

1) the crate's weight (which we can decompose in two components: one parallel to the incline, and the other one perpendicular to it) The parallel component of the weight is the product of the mass, times g (9.8 m/s^2) times sin(35). This component is pointing down the incline, and let's call it the "x" component. Its magnitude is given by:

$w_x=m\,\,g\,\,sin(35^o)$

The magnitude of the perpendicular component of the weight (let's call it the y-component) is given by the expression:

$w_y=m\,\,g\,\,cos(35^o)$

2) The force of friction between the crate and the surface of the incline (acting parallel to the incline and opposite the direction of the parallel component of the weight ( $w_x$ )

3) The extra force we need to apply perpendicular to the incline and towards it (lets call it "E") so the crate doesn't slide down.

4) The normal force that the incline applies on the crate as reaction. This force is pointing away from the incline.

For the block not to move (slide down the incline), we need that the parallel component of the weight equals the force of friction with the surface of the incline. Let's call this force of friction $f_s$ , and recall that it is defined as the product of the normal force times the coefficient of friction (given as 0.3 in value). This in equation form becomes:

$f_s=m\,g\,sin(35^o)\\\mu\,\,n=m\,g\,sin(35^o)\\n=\frac{m\,g\,sin(35^o)}{\mu} \\n=\frac{3\,*\,9.8\,\,sin(35^o)}{\0.3}\\n=56.21 \,\,N$

We solved for the needed magnitude of the normal force in order to keep the crate from sliding.

Now we study the forces vertical to the incline, which should also be balanced since the crate is not moving in this direction. We add all the forces acting towards the incline (the perpendicular component of the crate's weight, and the extra force "E"), and make them equal to the only force coming outwards (the normal force):

$n=E+m\,\,g\,\,cos(35^o)$

and we can solve for the magnitude of the extra force we need to apply by replacing all other known values:

$n=E+m\,\,g\,\,cos(35^o)\\56.21\,N=E+3\,*\,9.8\,\,cos(35^o)\,N\\E= [56.21-3\,*\,9.8\,\,cos(35^o)]\,N\\E=32.13\,N$

So this is the magnitude of the force we need to apply in order to keep the crate from sliding down the incline.

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3 years ago

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dezoksy [38]

Cost of vehicle, cost per launch, capacity of each vehicle, re-usability, size of vehicle, etc

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3 years ago