Answer:
a) - 72.5°c
b) pressure = 3625.13 Pa
c) density = 0.063 kg/m^3
d) it is a subsonic aircraft
Explanation:
a) Determine Temperature
Temperature at 19.5 km ( 19500 m )
T = -131 + ( 0.003 * altitude in meters )
= -131 + ( 0.003 * 19500 ) = - 72.5°c
b) Determine pressure and density at 19.5 km altitude
Given :
Po (atmospheric pressure at sea level ) = 101kpa
R ( gas constant of air ) = 0.287 KJ/Kgk
T = -72.5°c ≈ 200.5 k
pressure = 3625.13 Pa
hence density = 0.063 kg/m^3
attached below is the remaining part of the solution
C) determine if the aircraft is subsonic or super sonic
Velocity ( v ) = 
= 
= 283.8 m/s
hence it is a subsonic aircraft
Answer:
y = 67.6 feet, y = 114.4/ (22 - 3t)
Explanation:
For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram
Large triangle Projector up to the screen
tan θ = y / L
For the small triangle. Projector up to the person
tan θ = y₀ / (L-d)
The angle is the same, so we equate the two equations
y₀ / (L -d) = y / L
y = y₀ L / (L-d)
The distance from the screen (d), we look for it with kinematics
v = d / t
d = v t
we replace
y = y₀ L / (L - v t)
y = 5.2 22 / (22 - 3 t)
y = 114.4 (22 - 3t)⁻¹
This is the equation of the shadow height change as a function of time
For the suggested distance the shadow has a height of
y = 114.4 / (22-13)
y = 67.6 feet
Answer:
2m head start or else you done for
Explanation:
you cant even out run a bear they run at 35mph the fastest human is 25
A voltmeter is the instrument used to measure a potential difference between two points in an electric circuit
Answer:
Rebounce angle is 345°
Rebounce speed is 989.95m/s
Explanation:
Calculate the x component of the velocity of the bullet before impact by using the following relation:
Vbx= Vb Cos thetha
Here, is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°
Substituting
Vbx = Cos15 ×1400 = 1352.30m/s
Calculate the y component using the relation:
Vby = Vo Sin theta
Vby = sin 15° × 1400
Vby = 362.35m/s
The rebounce angle = 360 - incidence angle
Rebounce angle =( 360 - 15)° = 345°
The rebound speed V' = Vby - Vbx
V' = (1352.30 - 362.35)m/s
V' = 989.95 m/s
