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kozerog [31]
3 years ago
9

A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the ea

st at 2.0 m/s. The kayaker can paddle with a speed of 3.0 m/s. In which direction should he paddle in order to travel straight across the harbor? Express your answer in degrees measured north of east.

Physics
1 answer:
Savatey [412]3 years ago
3 0

Answer:

41.81^{\circ}

Explanation:

The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

Let Vector \overrightarrow{OA} is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, \vec {R} must be in the north direction.

Let \overrightarrow{AB} is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector \overrightarrow {OA} and head of the vector \overrightarrow{AB} must lie on the north-south line.

Now, for this condition, from the triangle OAB

|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|

\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3

\Rightarrow \theta=\sin^{-1}\frac23

\Rightarrow \theta=41.81^{\circ}

Hence, the kayaker must paddle in the direction of 41.81^{\circ}  in the north of east direction.

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Two cylinders A&amp;B at the same temperature contains the same quantity of the same kind of gas. Cylinder A has three times the
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3.

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