If a radio wave has a period of 1 μs what is the wave's period in seconds

The magnetic field is calculated to be -6.137 × T. Explanation: Given the radio wave wavelength of λ = 0.3 m and an intensity of I = 45 W/m² at times t = 0 and t = 1.5 ns, we determine Bz at the origin. We use the intensity formula relating to the electric field, which incorporates the known intensity of 45, the speed of light c = 3 × m/s², and ∈o as 8.85 × C²/N.m², leading us to E = 184.15. Consequently, applying the equations, we find B = -6.137 × T at the z-axis.

ke prop to v^2

ke1/v1^2=ke2/v2^2

400/50x50=joules/100x100

400x2x2

1600j

Result:

,

Explanation:

The electromagnetic attraction between the electron and the proton in the nucleus is equivalent to the centripetal force:

where

k represents the Coulomb constant

e denotes the charge of the electron

e denotes the charge of the proton in the nucleus

r signifies the distance from the electron to the nucleus

v indicates the velocity of the electron

is the mass of the electron

Rearranging for v, we determine

Inside a hydrogen atom, the distance separating the electron from the nucleus is roughly

while the mass of the electron is

and the charge is

By plugging in the values into the formula, we achieve

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

where

represents the charge's magnitude

and signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at