A traveling electromagnetic wave in a vacuum has an electric field amplitude of 90.5 V/m. 90.5 V/m. Calculate the intensity S S of this wave. Then, determine the amount of energy U U that flows through area of 0.0291 m2 0.0291 m2 over an interval of 19.3 s, 19.3 s, assuming that the area is perpendicular to the direction of wave propagation

The intensity of this wave is 10.87 W/m²

The energy flowing through the given area is 6.10J

Explanation:

The expression for the intensity of the electromagnetic wave is,

Here, is the permittivity of the free space,

 is the electric field amplitude

c is the speed of the light.

Substitute   for c, N.m² for ,  and 81.1 V/m for

Therefore, the intensity of this wave is 10.87 W/m²

The expression for the energy is,

Here, I is the intensity of the electromagnetic wave, A is the area, and t is the time.

Substitute 10.87W/m² for I, 0.0291 m² for A, and 19.3 s for t.

E = (10.87)(0.0291)(19.3)

E = 6.10J

Hence, The energy flowing through the given area is 6.10J

multiply 2kg by latent heat of vap

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft