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lisov135 [29]
3 years ago
11

What is the advantage of nuclear plants?

Physics
2 answers:
DerKrebs [107]3 years ago
8 0

They emit low leavls of carbon dioxide (CO2) and the emissions of green house gases and therefore the contribution of nuclear power plants to global warming is therefore pretty little

Marat540 [252]3 years ago
3 0

...............................................................

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What is globalization? A) an organization of people competing against each other in physical activities B) learning about belief
SOVA2 [1]
Globalization would be c

3 0
3 years ago
Read 2 more answers
At a certain altitude above the Earth's surface, the electric field has a magnitude of 135 V/m. How much energy is stored in 1.0
Paul [167]

Answer:

Explanation:

Given that,

Electric field E=135V/m

Energy stored in 1m³of air=?

The energy stored in an electric field is given as

u = ½ εo E²

Where

U is the energy stored

εo is permissivity and it value is 8.85×10^-12C²/N..m²

And E is the electric field

Then,

U=½×8.85×10^-12×135²

U=8.06×10^-8J/m³

Then, the energy stored in 1m³ of air is 8.06×10^-8 J/m³

4 0
3 years ago
The speed of a car is 40m/s after 10 sec suddenly there was a crowd and the driver reduces its speed to 20m/s. What is the decel
aliya0001 [1]

Explanation:

the number 0.4 in P4 from east

4 0
3 years ago
Which was an idea use to distribute the theory of plate tectonics
xxMikexx [17]

Answer:

continental drift

Explanation:

Wegener proposed that the continents plowed through crust of ocean basins, which would explain why the outlines of many coastlines look like they fit together like a puzzle

8 0
4 years ago
A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radi
svetlana [45]

Answer:

E=0 at r < R;

E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}} at 2R > r > R;

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}} at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon} (integral over a closed surface)

where,

E = Electric field

Q_{enclosed} = charged enclosed within the closed surface

\epsilon = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

Q_{enclosed} = 0 and hence E = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

Q_{enclosed} = Q,

therefore,

E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}      

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to 4\pi r^{2})

or, E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}

at r >= 2R

Q_{enclosed} = 2Q

Hence, by similar calculations, we get,

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}

4 0
4 years ago
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