To solve the problem it is necessary to take into account the concepts related to the Magnetic Force, which is given by,

Where
F= Magnetic force
q= charge of proton
v= velocity
B Magnetic field
Angle between the velocity and the magnetic field.
Re-arrange the equation to find the angle we have,

Replacing our values we have,

Then,

The angle between 0 to 180 degrees would be,

Therefore the two angles required are 13.63° and 166.36°
Answer:
Their change in momentum is the same in magnitude and opposite in direction
Explanation:
The momentum of an object is defined as:

where
m is the mass of the object
v is the velocity of the object
Therefore, the change in momentum of an object is

where
is the change in velocity.
During a collision, the force experienced by an object is equal to the rate of change of momentum:

where
is the duration of the collision.
According to Newton's third law of motion, the force exerted by vehicle 1 on vehicle 2 during the collision is equal (and opposite) to the force exerted by vehicle 2 on vehicle 1, so

Which means

And since the duration of the collision is the same for the two vehicles, this becomes

Answer:
W = 2.74 J
Explanation:
The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.
This is the electrostatic equivalent of the work-energy theorem.

where the potential energy is defined as follows

Let's first calculate the distance 'r' for both positions.

Now, we can calculate the potential energies for both positions.

Finally, the total work done on the moving particle can be calculated.

We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)