A believe that’s called a reference point.
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have:
<span>theta = arcsin(3/20) = approx. 8.63° </span>
<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>
<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>
<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
<span>= approx. 0.025 hr </span>
<span>= approx. 91 seconds</span>
If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.
The resultant of the two forces F1 and F2 is

with direction at

north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be

south-east.
a una velocidad de
22 m/s, quien lo golpea y devuelve en la misma
dirección con una velocidad de 14 m/s. Si el
tiempo de contacto del balón con la jugadora es
de 0,03 s, ¿con qué fuerza golpeó la jugadora el
balón?
21 Una bala de 0,8 g, está en la recámara de un rifl e
cuando se g
Answer:

Explanation:
The magnitude of the net force exerted on q is known, we have the values and positions for
and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by
on q. Then we can know the magnitude of the force exerted by
about q, finally this will allow us to know the magnitude of 
exerts a force on q in +y direction, and
exerts a force on q in -y direction.

The net force on q is:

Rewriting for
:
