1. $85047
2. $113,080
3. $ 75,820
4. $41595
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
At critical temperature, the resistivity of the superconductor
B. It suddenly drops to zero
Explanation:
Materials can be classified into three different types depending on their resistance:
- Conductors: these materials have generally low resistance and allow electricity to pass through easily. The resistance of a conductor increases linearly with the temperature
- Insulators: these materials do not allow electricity to pass through - so they have very high resistance
- Semi-conductors: these are materials that are insulators are room temperature, however they becomes conductors when heated. Therefore, the resistance of a semiconductor decreases when the temperature increases
- Superconductors: these are special materials that are normally conductors; however, at very low temperatures (we are talking about temperature very near to 0 K), their resistance becomes suddenly zero.
Therefore, the correct answer is:
B. It suddenly drops to zero
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