Answer:
0.25938 W
Step-by-step explanation:
c = Speed of light =
![3* 10^8\ m/s](https://img.qammunity.org/2020/formulas/physics/high-school/ks86dmym7sxdf8u3g31nulpxe5c16w96hm.png)
= Frequency = 3 GHz
d = Diameter of lossless antenna = 1 m
r = Radius =
![(d)/(2)=(1)/(2)=0.5\ m](https://img.qammunity.org/2020/formulas/physics/college/eukom3qfjavr7rzw0s31zzgkmjwctfbt5w.png)
= Area of transmitter
= Area of receiver
R = Distance between the antennae = 40 km
= Power of receiver =
![10* 10^(-9)\ W](https://img.qammunity.org/2020/formulas/physics/college/g6zgnc3o2xdrf39le6ezrrykteq1v4ry5d.png)
= Power of Transmitter
Wavelength
![\lambda=(c)/(\\u)\\\Rightarrow \lambda=(3* 10^8)/(3* 10^9)\\\Rightarrow \lambda=0.1\ m](https://img.qammunity.org/2020/formulas/physics/college/kl71bhvplcd8qudg8x62jmv9hlfrasqk3i.png)
From Friis transmission formula we have
![(P_t)/(P_r)=(\lambda^2R^2)/(A_tA_r)\\\Rightarrow P_t=P_r(\lambda^2R^2)/(A_tA_r)\\\Rightarrow P_t=10* 10^(-9)(0.1^2* (40* 10^3)^2)/(\pi 0.5^2* \pi 0.5^2)\\\Rightarrow P_t=0.25938\ W](https://img.qammunity.org/2020/formulas/physics/college/m5yegijhc8i16qoren40mjbu0eclkz7kn7.png)
The power that should be transmitted is 0.25938 W