The cube measures 3.0-ft on all sides and has a density of 3.1 slugs/ft3. How much does it weigh?
1 answer:
Answer:
W = 2695.14 lb
Explanation:
given,
side of cube = 3 ft
density of the cube = 3.1 slugs/ft³
we know,
mass = density x volume
volume = 3³ = 27 ft³
mass = 3.1 x 27
m = 83.7 slugs.
weight calculation
converting mass from slug to pound
weight of 1 slug is equal to 32.2 lb
now,
weight of the cube is equal to
W = 83.7 slugs x 32.2 lb/slug
W = 2695.14 lb
hence, weight is equal to W = 2695.14 lb
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Explanation:
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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.
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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)
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(V_1) = √(2 x 3200)/0.909)
(V_1) = 83.9 m/s
