Answer:
V = 0.30787 m³/s
m = 2.6963 kg/s
v2 = 0.3705 m³/s
v2 = 6.017 m/s
Explanation:
given data
diameter = 28 cm
steadily =200 kPa
temperature = 20°C
velocity = 5 m/s
solution
we know mass flow rate is
m = ρ A v
floe rate V = Av
m = ρ V
flow rate = V = 
V = Av = 
V = 
V = 0.30787 m³/s
and
mass flow rate of the refrigerant is
m = ρ A v
m = ρ V
m = 
= 
m = 2.6963 kg/s
and
velocity and volume flow rate at exit
velocity = mass × v
v2 = 2.6963 × 0.13741 = 0.3705 m³/s
and
v2 = A2×v2
v2 = 
v2 = 
v2 = 6.017 m/s
Answer:
Explanation:
The detailed steps is as shown in the attachment.
Answer: Let us use the pickled file - DeckOfCardsList.dat.
Explanation: So that our possible outcome becomes
7♥, A♦, Q♠, 4♣, 8♠, 8♥, K♠, 2♦, 10♦, 9♦, K♥, Q♦, Q♣
HPC (High Point Count) = 16
Answer:
12
Explanation:
5 • 3 = 15
3 • 3 = 9
4 • 3 = 12
The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.
<h3>What is altitude?</h3>
Altitude or height exists as distance measurement, usually in the vertical or "up" approach, between a reference datum and a point or object. The exact meaning and reference datum change according to the context.
The MOCA exists in the lower published altitude in effect between fixes on VOR airways, off-airway routes, or route segments that satisfy obstacle support conditions for the whole route segment. This altitude also ensures acceptable navigational signal coverage only within 22 NM of a VOR.
The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.
Therefore, the correct answer is 22 NM of a VOR.
To learn more about altitudes refer to:
brainly.com/question/1159693
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