E S *
The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).
The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .<span>00001581 light years
D'e,*'(Distance between earth and Proxima) = </span><span>4.243 light years
Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are </span>9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.
To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters<span>
D'e,*'(Distance between earth and Proxima) = </span><span>4.014 * 10^18 centimeters
</span>
Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (<span>1.496 * 10^13 ).
Why? because that how proportions work.
As a result, you get a mere 268335.7 centimeters.
To put that into perspective, that's only about 1.7 miles
A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.</span>
Answer:
A) A negative charge of value Q is induced on sphere B
B) there is an attraction between sphere
C) The charge of sphere A is distributed between the two spheres,
Explanation:
This is an electrostatic problem, in general charges of the same sign attract and repel each other.
with this principle let's analyze the different situations
A) The sphere A that is insulating has a charge on its surface and zero charge is its interior
The conducting sphere B has zero charge, but the sphere A creates an attraction in the electrons, therefore a negative charge of the same value as the charge of the sphere A is induced in the part closest and in the part farther away than one that a positive charge.
A negative charge of value Q is induced on sphere B
B) In this case there is an attraction between sphere A with positive charge and sphere B with negative induced charge
C) When the two spheres come into contact, the charge of sphere A is distributed between the two spheres, therefore each one has a positive charge of value half of the initial charge, as now we have net positive charges in the two spheres charges of the same sign repel each other so the spheres separate
Answer:
U= 
J
u= 0.708 
Explanation:
diameter= 2 cm
distance= 0.5 mm
Area A= 
Volume v= 
Constant vacuum permittivity
a).
F
b).
