The density of the material would be 4.1 g/cm³.
Density is calculated by dividing the mass by the volume.
D=m÷v
D=45 g÷11 cm³
D=4.1 g/cm³
Answer:
A) ≥ 325Kpa
B) ( 265 < Pe < 325 ) Kpa
C) (94 < Pe < 265 )Kpa
D) Pe < 94 Kpa
Explanation:
Given data :
A large Tank : Pressures are at 400kPa and 450 K
Throat area = 4cm^2 , exit area = 5cm^2
<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>
The range of flow of back pressures that will make the flow entirely subsonic
will be ≥ 325Kpa
attached below is the detailed solution
<u>B) Have a shock wave</u>
The range of back pressures for there to be shock wave inside the nozzle
= ( 265 < Pe < 325 ) Kpa
attached below is a detailed solution
C) Have oblique shocks outside the exit
= (94 < Pe < 265 )Kpa
D) Have supersonic expansion waves outside the exit
= Pe < 94 Kpa
Answer:
The change in momentum = -20000 kg m/s.
Explanation:
Mass m = 1000 kg
speed v₁ = 20 m/s
speed v₂ = 0 m/s
We know that,
The change in momentum
ΔP = m (Δv)
ΔP = m (v₂ - v₁)
= 1000 (0 - 20)
= 1000 (-20)
= -20000 kg m/s
Thus, the change in momentum = -20000 kg m/s.
Note: negative sign indicates that the velocity is reducing when it hits the barrier.