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borishaifa [10]
3 years ago
7

If scientists could create a sheet of glass with a refractive index so high that it took light months or years to travel through

the glass, what kind of things do you think they could do with it?
Physics
1 answer:
omeli [17]3 years ago
4 0

Wow ! They could set up sheets of "slow glass" beside beautiful forests with rivers and squirrels and deer and grassy fields, and load a year of this scene into the glass, and then sell it to people who live next to dirty brick walls or ugly empty lots, and those people could install the slow glass in their windows and have beautiful scenery, until it all worked its way out of the glass.

This is a great idea ! If you possibly can, find and read the sci-fi short story "Light of Other Days" written by Bob Shaw, published in Analog Science Fiction in 1966. It's all about this exact type of glass. I read this story in 1966 and I never forgot it ! (Not yet anyway.)

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A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the
Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

Substitute numerical values:

E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

#SPJ4

6 0
2 years ago
The kinetic theory states that the particles in matter are always in?
OlgaM077 [116]
Motion is the correct word that fits in.
Hope this helps.
7 0
3 years ago
A heliocentric system is _____-centered.
aniked [119]
A heliocentric system is a sun-centered 
3 0
3 years ago
Read 2 more answers
Please someone answer this ASAP❗️❗️‼️‼️
kogti [31]
Starting making jokes and rapping
5 0
3 years ago
A van de Graaff generator accelerates electrons so that they have energies equivalent to that attained by falling through a pote
Marat540 [252]

Answer:

Hello your question is incomplete hence I will give you a general answer on how A van de Graaff generator works

answer :

If the electrons falls through a PD of 150mV the electron will gain energy of   150MeV

Explanation:

when a Van de Graff generator is used to accelerate an electron through a PD ( potential difference ) of any value the particle ( electron )  the electron will gain energy  ( eV )  which is is equivalent in value of the PD it accelerated through

hence if the electrons falls through a PD of 150mV the electron will gain energy of   150MeV

4 0
3 years ago
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