Ahemispherical surface (half of a spherical surface) of radius r is located in a uniform electric field of magnitude e that is parallel to the axis of the hemisphere. what is the magnitude of the electric flux through the hemisphere surface?

π*R²*E

Explanation:

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis,  and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒Flux = E*π*R²

50 inches should be the resultant vector

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

don't remove my answer im trying my best to answer them

explanation: