Aproton moving at 5.00 106 m/s through a magnetic field of magnitude 1.78 t experiences a magnetic force of magnitude 7.40 10-13 n. what is the angle between the proton's velocity and the field? (enter both possible answers from smallest to largest. enter only positive values between 0 and 360.)

The magnetic part using the Lorentz force is: F = q v x B, 
where v and B are vectors and v x B is the vector cross product. 

Magnitude of the force: F = q v B sin(α) 

So, sin(α) = F/( e v B), with e the proton charge. 

This will give you a value for sin(α), and two potentials for its opposite.

You will now look for: 

sin(α) = 7.40 10^-13/( 1.60 10^-19 * 5 10^6 * 1.78) 
= 0.520

So either sin(α) = 0.502 or sin(α) = -0.502 
The 1st α = 30.1 degrees or α = 150 degrees. 
The 2nd α = 210 degrees or α = 330 degrees. 
So we can say that 30.1 degrees and 330 degrees would be minimum and biggest on [0,360]

think about the oscillation of the toy shown in the video. when the toy is at the highest point of its motion, its velocity is zero and its acceleration is negative.

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down