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:
c. 60 feet is the correct answer
Explanation:
what is the contour interval of this map? a.20 b.-20 c. 60 feet 11
Answer:
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As some people have it wrong, waves in the water (ocean) are not waves of moving water, rather the wave is moving through the water. A wave is a disturbance of a medium not the meduim moving.
We will make the comparison between each of the sizes against the known wavelengths.
In the case of the <em>hydrogen atom</em>, we know that this is equivalent to m on average, which corresponds to the wavelength corresponding to X-rays.
In the case of the <em>Virus</em> we know that it is oscillating in a size of 30nm to 200 nm, so the size of the virus is equivalent to the range of the wavelength of an ultraviolet ray.
In the case of <em>height</em>, it fluctuates in a person around to m, which falls to the wavelength of a radio wave.
Answer:
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