Answer:
The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
Explanation:
Given the data in the question;
Using the Clapeyron equation
where 
is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu
T is the temperature ( 15 + 460 )R
m is the mass of water ( 0.5 Ibm )
is specific volume ( 1.5 ft³ )
we substitute
/ 
272.98 Ibf-ft²/R
Now,
where P₁ is the initial pressure ( 50 psia )
P₂ is the final pressure ( 60 psia )
T₁ is the initial temperature ( 15 + 460 )R
T₂ is the final temperature = ?
we substitute;
480.275 R
Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
Answer:
friction help to slow motion in other word it oppose motion, but in a frictionless environment object would move with difficult stopping point.
Answer:
a = 2.275 10⁻⁴ m
Explanation:
This is a diffraction problem that is described by the equation
a sin θ = m λ
The first dark minimum occurs for m = 1
a = λ / sin θ
The angle can be found by trigonometry,
tan θ = y / x
θ = tan⁻¹ y / x
Let's reduce the magnitudes to the SI system
y = 8.24 mm = 8.24 10⁻³ m
λ = 625 nm = 625 10⁻⁹ m
θ = tan⁻¹ 8.24 10⁻³ / 3.00
θ = 0.002747 rad
We calculate
a = 625 10⁻⁹ / sin 0.002747
a = 2.275 10⁻⁴ m
Answer:
Explanation:
Given:
P = 6.35 atm
= 1.01 × 10^5 × 6.35
= 6.434 × 10^5 N/m^2
As = 975 cm^2
D = 3.8 g/cm^2
M = 320 kg
Since the propellant volume is equal to the cross sectional area, As times the fuel length, the volumetric propellant consumption rate is the cross section area times the linear burn rate, bs , and the instantaneous mass flow rate of combustion, ms gases generated is equal to the volumetric rate times the fuel density, D
ms = D × As × bs
ms ÷ bs = M/L
M/L = 3.8 × 975
= 3705 g/cm
= 3.705 × 10^6 kg/m^3
Pressure = mass × g/area
= mass/length × time^2
t = sqrt(3.705 × 10^6/6.43 × 10^5)
= 2.4 s
