Answer:
2.4 m/s
Explanation:
Given:
Velocity of the object moving north = 2.1 m/s
Velocity of the river moving eastward = 1.2 m/s
The resultant velocity is the vector sum of the velocities of object and river.
Since the directions of velocity of object and river are perpendicular to each other, the magnitude of the resultant velocity is obtained using Pythagoras Theorem.
The velocities are the legs of the right angled triangle and the resultant velocity is the hypotenuse.
The magnitude of the resultant velocity (R) is given as:

Therefore, the resultant velocity has a magnitude of 2.4 m/s.
B a bar magnet has a north and a south pole.
Answer:
This is due to impulse
Explanation:
Impulse equal to mΔv and FΔt
You can set these equal as mΔv = FΔt
When a boxer punches a tissue, it is like punching a cushion or a pillow. The time that the hit takes is much grater than if they were to hit something solid. In addition, the change in velocity of the boxer's arm would be much greater when they hit a punching bag. In this equation, the greater the time, the less force that is needed.
Answer:
a.After
second Mr Comer's speed

b.Distance travelled by Mr.Comer in
seconds

Explanation:
a. Lets recall our first equation of motion 
Now we know that
,
and

Plugging the values we have.




Then Mr.Comer's speed after
sec

b.
Lets find the distance and recall our third equation of motion.

So
distance covered.
Dividing both sides with 2a we have.

Plugging the values.


So Mr.Comer will travel a distance of
.