Question

14.) What fraction of the power radiated by the sun is intercepted by the planet Neptune? The radius of Mercury is 2.48 107 m, and its mean distance from the sun is 4.50 1012 m. Assume that the sun radiates uniformly in all directions

4.) In a certain UHF radio wave, the shortest distance between positions at which the electric and magnetic fields are zero is 0.342 m. Determine the frequency of this UHF radio wave.

Answer 1

14)
let P is the power radiated by the sun.

Intensity of the sun at the location of Neptune, I = Power/Area

= P/(4*pi*d^2)

= P/(4*pi*(4.5*10^12)^2)

= 3.93*10^-27*P

Power intercepted by the Netun, P' = I*Aera of the nuptune

= I*pi*r^2

= 3.93*10^-27*pi*(2.48*10^7)^2*P

= 7.95*10^-12*P

P'/P = 7.95*10^-12   <<<<<<<<--------------Answer

4) from the given data

wavelength, lamda = 0.342 m

use, c = lamda*f

==> f = c/lamda

= 3*10^8/0.342

= 8.77*10^8 Hz <<<<<<<<--------------Answer