A small pad subjected to a shearing force is deformed at the top of the pad 0.12 in. The heigfit of the pad is 1.15 in. What is
1 answer:
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Answer:
Speed of aircraft ; (V_1) = 83.9 m/s
Explanation:
The height at which aircraft is flying = 3000 m
The differential pressure = 3200 N/m²
From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3
Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.
Thus, let's apply the Bernoulli equation :
P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2
Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.
We'll obtain ;
P1/ρg + (V_1)²/2g = P2/ρg
Let's make V_1 the subject;
(V_1)² = 2(P1 - P2)/ρ
(V_1) = √(2(P1 - P2)/ρ)
P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question
Thus,
(V_1) = √(2 x 3200)/0.909)
(V_1) = 83.9 m/s
Answer:
BOD5 = 200 mg/L
Explanation:
given data
diluted = 1% = 0.01
time = 5 day
oxygen consumption = 2.00 mg · L−1
solution
we get here BOD5 that is BOD after 5 day
and here total volume is 100% = 1
so dilution factor is = 100
so BOD5 is
BOD5 = oxygen consumption × dilution factor
BOD5 = 2 × 100
BOD5 = 200 mg/L