Assume that we have a BS with a 6-dB antenna gain and an MS with antenna gain of 2 dB, at heights 10 m and 1.5 m, respectively,
operating in an environment where the ground plane can be treated as perfectly conducting. The lengths of the two antennas are 0.5 m and 15 cm, respectively. The BS transmits with a maximum power of 40 W and the mobile with a power of 0.1 W. The center frequency of the links (duplex) are both at 900 MHz, even if in practice they are separated by a small duplex distance (frequency difference).(a) Assuming that holds, calculate how much received power is available at the output of the receive antenna (BS antenna and MS antenna, respectively), as a function of distance d?(b) Plot the received powers for all valid distances d holds and the far field condition of the antennas is fulfilled.
1 answer:
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Answer:
D) 1.04 Btu/s from the liquid to the surroundings.
Explanation:
Given that:
flow rate (m) = 2 lb/s
liquid specific enthalpy at the inlet ( Btu/lb)
liquid specific enthalpy at the exit ( Btu/lb)
initial elevation ()
final elevation ()
acceleration due to gravity (g) = 32.174 ft/s²
= 3 Btu/s
The energy balance equation is given as:
Since kinetic energy effects are negligible, the equation becomes:
Substituting values:
The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.
Answer:
eccentrcity of orbit is 0.22
Explanation:
GIVEN DATA:
Initial velocity of satellite = 8333.3 m/s
distance from the sun is 600 km
radius of earth is 6378 km
as satellite is acting parallel to the earth therefore
and radial component of given velocity is zero
we have
h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s
we know that
so
solvingt for
therefore eccentrcity of orbit is 0.22