A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

The given question is incomplete. The complete question is as follows.

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A) If the net force acting on the particle is N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Enter your answers numerically separated by commas.

Explanation:

The given data is as follows.

           Q =

           E = 1300 N/C in the +x direction

           B = 1.02 T in the +z direction

and,     in the +x direction

Also,      

                         = qE - qvB

Now, we will calculate the value of v as follows.

             v =

                 =

                v = 458.507 m/s

Using the value for velocity, we need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for .

Using the right hand rule where:

your right thumb goes toward the , then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.

Thus, we can conclude that = 0, -(458.507), 0.

answer: i don't know the answer good luck

explanation:

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west