Answer:
D (density) = Mass / Volume
V (ice) = Mass / Density = 50000 g / .92 kg / L
V = 50 kg / .92 kg / L
1 Liter of water weighs 1 kilogram = 50 L weighs 50000 g
V = 54.3 L the mass does not change upon the change of phase (freezing)
Explanation:
Below is an attachment containing the solution.
Answer:
B: leaving metal outside in the rain until rust forms an it's surface
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