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
Distance, since distance represents how far something has travelled, which would be in our case 2.5m.
Answer:
IGNEOUS ROCKS
Explanation: Igneous rocks are those rocks that solidify from magma.
Igneous rock is divided into two ,they are:
1. Intrusive
Igneous rocks crystallized belowearth"s crust. Its cooling material is called lava.
2 Extrusive igneous rock is also known as known as volcanic rocks
Answer:
Specific heat capacity is an intensive property and does not depend on sample size.
Explanation:
Answer:
62.06 g/mol
Explanation:
We are given that a solution containing 10 g of an unknown liquid and 90 g
Given mass of solute =
=10 g
Given mass of solvent=
=90 g
Freezing point of solution =-3.33
C
Freezing point of solvent =
C
Change in freezing point =Depression in freezing point
=Freezing point of solvent - freezing point of solution=0+3.33=
Hence, molar mass of unknown liquid is 62.06g/mol.
