Kinetic energy = (1/2) (mass) (speed²).
A Physicist in the canoe, or on a raft floating downriver next to the canoe, will say that the canoe's kinetic energy is zero.
A Physicist on the riverbank, watching the canoe drift by at 1 m/s, will say that its kinetic energy is 9 Joules.
They're both correct.
it can be said that the speed of the east wind is
v=0.3608m/s
From the question we are told
A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).
After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,
- what is the speed of the east wind?.
<h3> the speed of the east wind</h3>
Generally the equation for the distance is mathematically given as
BA=3000sin60
BA=2598.07m
Therefore
the speed of the east wind

v=0.3608
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The applicable equation:
P = F/A
P = pressure
F = Force or weight
A = surface area
Pressure on each cylinder = (W/n)/A
Where n = number of cylinders. Additionally, pressure in the reservoir is equivalent to the pressure in each cylinder.
Net pressure = 75 - 14.7 = 60.3 psi
Therefore,
60.3 = (W/n)/A = (450/n)/(πD^2/4) = (450/n)/(π*1.5^2/4) = (450/n)/(1.7671)
60.3*1.7671 = 450/n
106.03 = 450/n
n = 450/106.3 = 4.244 ≈ 5
The number of cylinders is 5.