Answer:
Explanation:
he answer is 540 because you need to do the area of the wall times six. 15 times 6 times 6 equals 540
The rest energy of a particle is

where

is the rest mass of the particle and c is the speed of light.
The total energy of a relativistic particle is

where v is the speed of the particle.
We want the total energy of the particle to be twice its rest energy, so that

which means:


From which we find the ratio between the speed of the particle v and the speed of light c:

So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.
Answer:
1234285.7 m or 1234.3 km
Explanation:
Let the distance be
, the time taken by P waves be
and the time taken by the S waves be
.


For the P waves,


For the S waves,


Equating the
,

Divide both sides of the equation by 500 to reduce the terms.

Since S waves arrive 2 minutes (= 120 seconds) after P waves,


Substitute this in the equation of the distance.




Substitute this in the equation for
involving
.



The correct answer is:
<span>C: in the protons and neutrons of an atom
In fact, the nuclear energy refers to the binding energy of the nucleons (protons and neutrons) of an atom. The protons and the neutrons are held together by the strong nuclear interaction, one of the four fundamental forces of nature, and the energy associated to this interaction is called nuclear energy.
</span>
Answer:
a) 
b) 
c) 
Explanation:
Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:


Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.
When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.
The new surface charge density can be calculated as follows:

c) The electric field outside the cylinder can be found by Gauss' Law:

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.
