<span>René Descartes suggests this.</span>
Remember, half of the energy in an EM wave is in the E field, the rest is in the B field.
Thus, multiply E field energy by 2.
To calculate the energy of the wave you must then use the following equation: W = A*t*c*2*(1/2*E^2*Eo). Where, A = Area, t = time, c = speed of light (which is a constant), E = Electric field, E0 = vacuum permittivity (8.85*10^-12 Nm^2/C^2). Substituting W =(0.320)*(26)*(3*10^8)*(2)*((1/2)*(1.95*10^-2)^2*(8.854*10^-12)) = 8.40*10^-6 J
It expands and pushes the crack further aprt
Answer: 0.817A
Explanation:
Assuming , that one coulomb per second of negative charge alone flow through a conductor and no positive charges flow. I.e Q=It
It means a current of one A flow in the opposite direction.
This is similar to one coulomb per second of positive charge flowing through and there is no negative charge,
In addition, the one coulomb per second of positive charge flows. This is flowing in the current direction of the previous one. Then, the total current is 2 A. Since 2 coulomb of positive charges flow through one due to real positive charge and another due to the negative charge flowing in opposite direction.The charges cannot cancel each other, because even before the current flow the conductor was neutral.
According to this, the current in the given problem is
[2.7 + 2.4] x 10 ^ 18 * 1.602 x 10^ [-19] C/s
= 0.817 A
Can you please post the picture of the graph and i will gladly answers it.
