A small boat sailed straight north out of a harbor in strong east wind (blowing from west to east). After sailing for 120 minute
1 answer:
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?.
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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Answer:
A. Its translational kinetic energy is larger than its rotational kinetic energy.
Explanation:
Given that
Radius = R
Mass = M
We know that mass moment of inertia for the solid sphere
Lets take angular speed =ω
Linear speed =V
Condition for pure rolling , V= ω R
Rotation energy ,RE
RE= 0.2 MV²
The transnational kinetic energy TE
TE= 0.5 MV²
From above we can say that transnational energy is more than rotational energy.
Therefore the answer is A.