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:
Explanation:
The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:
Where:
Final pressure is:
Answer:
Omqnp
Explanation:
I have done this before and got it correct
A. Solid
Solid structures are actually made of repeating patterns.
Answer:
the girl must sit 2 cm from the pivot at the opposite end of the seesaw.
Explanation:
Given;
length of the seesaw, L = 4.0 m
weight of the boy, W₁ = 400 N
position of the boy from the pivot, d₁ = 1.5 m
weight of her sister, W₂ = 300 N
First, make a sketch of this information given;
0---0.5m---------------------Δ--------------------------4m
↓<--------1.5m-------> <---------x--------->↓
400 N 300N
Apply the principle of moment about the pivot, to determine the value of x;
Sum of anticlockwise moment = sum of clockwise moment
400(1.5) = 300(x)
600 = 300x
x = 600/300
x = 2 cm
Thus, the girl must sit 2 cm from the pivot at the opposite end of the seesaw.
