Answer:
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
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,
must be in the north direction.
Let
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
and head of the vector
must lie on the north-south line.
Now, for this condition, from the triangle OAB




Hence, the kayaker must paddle in the direction of
in the north of east direction.
Answer:
No the given statement is not necessarily true.
Explanation:
We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

Similarly the momentum is given by 
For 2 particles with masses
and moving with velocities
respectively the respective kinetic energies is given by


Similarly For 2 particles with masses
and moving with velocities
respectively the respective momenta are given by


Now since it is given that the two kinetic energies are equal thus we have

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.
<span>v is perpendicular to both E and B and has a magnitude E/B</span>
A human hair is approximately 50 μm = 5 × 10⁻⁵ m in diameter