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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V= 1/3 π r²h
this is the formula for a cone hope this helps :)
Answer:
7.5 m/s
Explanation:
We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:
where v = final velocity
u = initial velocity
a = acceleration
s = distance traveled
From the question:
u = 28 m/s
a = -4 
s = 91 m
Therefore:
The velocity of the boat when it reaches the buoy is 7.5 m/s.
Answer:
Weight of the car, normal force, drag force
Explanation:
The forces acting on the car are:
- The normal force which acts perpendicularly to the downhill plane
- The weight of the car which acts vertically downwards
- The drag force due to air resistance which acts in opposition to the motion of the car
Friction is ignored, so the force due to friction is assumed negligible
