Answer:
A) 27209506.5 N
B) 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
Explanation:
Force du to depth of water is
F = pghA
P = density of salt water = 1025 kg/m3
g = acceleration due to gravity 9.81 m/s2
h = depth of water 11000 m
A = area pressure acts
Area = ¶r^2 = 3.142 x 0.280^2 = 0.246 m^2
Therefore
F = 1025 x 9.81 x 11000 x 0.246
= 27209506.5 N
Weight of a jetliner with mass 2.44 × 10^5 kg is,
2.44×10^5 x 9.81 = 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
Answer:
pascal
Explanation:
its obtained after either division or multiplication
Answer:
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Answer:
14.2 m/s
Explanation:
Given data:
Speed of the stream, v₁ = 7.1 m/s
let the cross section area at initial point be A₁
now area at the second point, A₂ = (1/2)A₁ = 0.5A₁
now, from the continuity equation, we have
A₁v₁ = A₂v₂
where, v₂ is the velocity at the narrowed portion
thus, on substituting the values, we get
A₁ × 7.1 = 0.5A₁ × v₂
or
v₂ = 14.2 m/s