Answer:
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Answer:
C = 292 Mbps
Explanation:
Given:
- Signal Transmitted Power P = 250mW
- The noise in channel N = 10 uW
- The signal bandwidth W = 20 MHz
Find:
what is the maximum capacity of the channel?
Solution:
-The capacity of the channel is given by Shannon's Formula:
C = W*log_2 ( 1 + P/N)
- Plug the values in:
C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)
C = (20*10^6)*log_2 (25001)
C = (20*10^6)*14.6096
C = 292 Mbps
Answer:
A selective medium, a differential medium, and a complex medium.
Explanation:
A selective media is a microbiological media which only support the growth of a particular specie or types of species of microorganisms,this media acts in such a way to inhibit or hinder the growth of other microorganisms.
Differential media are media that acts to Identifying particular strains of microorganisms of similar species.
Complex media are media used for the growth of microorganisms this which contains complex or a wide range of nutrients with chemical composition which may be difficult to determine.
Question
Determine the average water exit velocity
Answer:
53.05 m/s
Explanation:
Given information
Volume flow rate, 
Diameter d= 8cm= 0.08 m
Assumptions
- The flow is jet flow hence momentum-flux correction factor is unity
- Gravitational force is not considered
- The flow is steady, frictionless and incompressible
- Water is discharged to the atmosphere hence pressure is ignored
We know that Q=AV and making v the subject then
where V is the exit velocity and A is area
Area, 
where d is the diameter
By substitution
To convert v to m/s from m/s, we simply divide it by 60 hence
