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
Static electricity<span> is caused by the build up of </span>electrical<span> charges on the surface of objects, while </span>current electricity<span> is a phenomenon from the flow of electrons along a conductor. 2. When objects are rubbed, a loss and/or gain of electrons occurs, which results in the phenomenon of </span>static electricity<span>.</span>
Answer:
567.321nm
Explanation:
See attached handwritten document for more details
Answer:
Thus the time taken is calculated as 387.69 years
Solution:
As per the question:
Half life of 
= 28.5 yrs
Now,
To calculate the time, t in which the 99.99% of the release in the reactor:
By using the formula:
where
N = No. of nuclei left after time t
= No. of nuclei initially started with
(Since, 100% - 99.99% = 0.01%)
Thus
Taking log on both the sides:
t = 387.69 yrs
