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:
2.25in³
Explanation:
For a 12 awg conductor the minimum volume allowance as stated by the NEC is 2.25in³
See attached Table 314.16(B) from NEC 2011
Answer:
The velocity of the ship relative to the earth V = 9.05 
Explanation:
The local ocean current is = 1.52 m/s
Direction
= 40°
Velocity component in X - direction
= 1.52
°
= 1.164 
Velocity component in Y - direction
= 8 + 1.52
°
= 8.97 
The velocity of the ship relative to the earth

Put the values of
and
we get,
⇒ 
⇒ V = 9.05 
This is the velocity of the ship relative to the earth.