Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
Answer:
A. Time Division multiple Access (TDMA)
Explanation:
Time-division multiple access (TDMA) is a channel access method for shared-medium networks
It allows several users to share the same frequency channel by dividing the signal into different time slots. The users transmit in rapid succession, one after the other, each using its own time slot.
Answer:
The length of the tube is 3.92 m.
Explanation:
Given that,
Electric potential = 100 MV
Length = 4 m
Energy = 100 MeV
We need to calculate the value of 
Using formula of relativistic energy

Put the value into the formula


Here, 



We need to calculate the length
Using formula of length

Put the value into the formula


Hence, The length of the tube is 3.92 m.
Answer:
2.41 L
Explanation:
We can solve the problem by using the ideal gas equation, which can be rewritten as:

where we have:
(initial pressure is stp pressure)
is the initial volume
is the initial temperature (stp temperature)
is the final pressure
is the final volume
is the final temperature
By substituting the numbers inside the formula and solving for V2, we find the final volume:

which corresponds to 2.41 L.