Answer:
19063.6051 g
Explanation:
Pressure = Atmospheric pressure + Gauge Pressure
Atmospheric pressure = 97 kPa
Gauge pressure = 500 kPa
Total pressure = 500 + 97 kPa = 597 kPa
Also, P (kPa) = 1/101.325 P(atm)
Pressure = 5.89193 atm
Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)
Temperature = 28 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28.2 + 273.15) K = 301.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K
⇒n = 595.76 moles
Molar mass of oxygen gas = 31.9988 g/mol
Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g
Answer:
β = = 0.7071 ≈ 1 ( damping condition )
closed-form expression for the response is attached below
Explanation:
Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s
computing a solution :
M = 1,
c = 2,
k = 2,
Wn = =
next we determine the damping condition using the damping formula
β = = 0.7071 ≈ 1
from the condition above it can be said that the damping condition indicates underdamping
attached below is the closed form expression for the response
Answer:
The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
Explanation:
Length of curve is given as
is given as
The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as
A is given as
With initial grade, the elevation of PVC is
The station is given as
Low point is given as
The station of low point is given as
The elevation is given as
So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.