91影视

radius of dish(r): 44/2 cm= 0.22m

$s_0=10km=10000m$

v = 50km/h = 50000m/h = 13.88m/s

$P_{source}=2.5kW=2500W$

Time = 1hr

Energy received by dish is power received over the course of 1hr time period.

$E={鈭�}_0^{1hr}{P}_{dish}dt$ ____ eq (1)

Now, Intensity received by dish is $I=\frac{P}{A}$,

Where, P = power

A = Area

So, P = IA

$E={鈭�}_0^{1hr}\frac{P_{source}}{4{\mathrm{蟺谤}}^2}Adt$____ eq(3)

Now we know that the truck is moving away from the station at a constant speed. So truck's distance from station at any time t is

$r=s_0+vt$

Substituting this in equation 3 we get,

$$\begin{gathered} E={鈭�}_0^{1hr}\frac{P_{source}A}{4\mathrm{蟺}{({\mathrm{s}}_0+\mathrm{vt})}^2}dt \\ =\frac{P_{source}A}{4\mathrm{蟺}}{鈭�}_0^{1hr}\frac{1}{{({\mathrm{s}}_0+\mathrm{vt})}^2}dt \\ =\frac{P_{source}A}{4\mathrm{蟺惫}}{\left[\right(-1\left){({\mathrm{s}}_0+\mathrm{vt})}^{-1}\right]}_0^{1hr}=\frac{-P_{source}A}{4\mathrm{蟺惫}}[\frac{1}{s_0+v\left(1hr\right)}-\frac{1}{s_0}] \end{gathered}$$

Here A = ${\mathrm{蟺谤}}^2$

Substituting this value in above equation we get:

$E=\frac{-P_{Source}{r}^2}{4\mathrm{v}}[\frac{1}{S_0+v\left(1hr\right)}-\frac{1}{S_0}]$

$$\begin{gathered} E=\frac{-\left(2500W\right){(0.22m)}^2}{4(13.88m/s)}[\frac{1}{10000+50000}-\frac{1}{10000}] \\ E=-2.179脳(-8.333脳{10}^{-5)} \\ E=0.000181J \end{gathered}$$