Einstein's special theory of relativity explains that the electric and magnetic fields are both can formulate together in mathematically.
It is given Einstein's special theory of relativity.
It is find the Einstein's special theory of relativity explains the perpendicular behavior of moving charges without recourse to invoking the concept of a magnetic field.
<h2>What is Einstein's special theory of
relativity?</h2>
As we know that one charge creates a field and its that field that actually exerts a force on the other charge. Here we it gives the relationship of two fields like electric field and magnetic field and gives the formula for electromagnetic objects.
Special relativity fixes the problem by the points that the magnetic force in one frame of reference easily be an electric force in other and also some of the combination of them in a frame.
Thus, Einstein's special theory of relativity explains that the electric and magnetic fields are both can formulae together in mathematically.
Learn more about magnetic field here:
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The current decreases as the overall resistance increases. In addition, if one bulb is removed from the “chain” the other bulbs go out. ... If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the “ladder”.
(a) 
The change in potential energy of the electron is given by:
where
is the magnitude of the electron's charge
is the magnitude of the electric field
d = 520 m is the distance through which the electron has moved
Substituting into the equation, we find
(b) 78 kV
The potential difference the electron has moved through is given by
where
is the magnitude of the electric field
d = 520 m is the distance through which the electron has moved
Substituting into the equation, we find
Density is equal to mass divided by volume; that said, you would divide 38.6 by 2 to get your answer
Answer:
24.3 degrees
Explanation:
A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:
Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.
Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.
So 
