Answer:
hi
Explanation:
The answer is Responding Variables Hope this helps :D
Answer:
1.98 m/s
Explanation:
To solve this, we would be using the law of conservation of energy, i.e total initial energy is equal to total final energy.
E(i) = E(f)
mgh = ½Iw² + ½mv²
Recall, v = wr, thus, w = v/r
Also, I = ½mr²
I = 0.5 * 5 * 2²
I = 10 kgm²
Remember,
mgh = ½Iw² + ½mv²
Substituting w for v/r, we have
mgh = ½I(v/r)² + ½mv²
Now, putting the values in the equation, we have
5 * 9.8 * 0.3 = ½ * 10 * (v/2)² + ½ * 5 * v²
14.7 = 1.25 v² + 2.5 v²
14.7 = 3.75 v²
v² = 14.7/3.75
v² = 3.92
v = √3.92
v = 1.98 m/s
Thus, the speed is 1.98 m/s
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